Chapter 1

1. Introduction

1.1 Equation (a) of the problem statement is used to solve for h as

(a) )( ∞−

=TTA

Qh&

The Principle of Dimensional Homogeneity is used to determine the dimensions of the heat transfer coefficient. Using the F-L-T system dimensions of the quantities in Equation (a) are

[ ] (b) T

LF⎥⎦⎤

⎢⎣⎡ ⋅

=Q&

[ ] [ ] (b) L2=A [ ] [ ] (c) Θ=− ∞TT

Thus from Equations (a)-(d) the dimensions of the heat transfer coefficient are

[ ]

(d) LT

F

LTLF

2

⎥⎦⎤

⎢⎣⎡

⋅Θ⋅=

⎥⎦⎤

⎢⎣⎡

⋅Θ⋅⋅

=h

Possible units for the heat transfer coefficient using the SI system are Ksm

N ⋅⋅

while

possible units using the English system are Rsft

lb⋅⋅

.

1.2 The Reynolds number is defined as

(a) Reµ

ρVD=

The dimensions of the quantities on the left-hand side of Equation (a) are obtained using Table 1.2 as

[ ]

[ ]

[ ] [ ]

[ ] (e) TL

M(d) LD

(c) TLV

(b) L3

⎥⎦⎤

⎢⎣⎡

⋅=

=

⎥⎦⎤

⎢⎣⎡=

⎥⎦⎤

⎢⎣⎡=

µ

ρ M

Substituting Equations (b)-(e) in Equation (a) leads to

1

Chapter 1

[ ]

[ ] (f) 1 TLMTLM

TLM

LTL

LM

Re

3

3

3

=

⎥⎦

⎤⎢⎣

⎡⋅⋅⋅⋅

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⋅

⋅⋅=

Equation (f) shows that the Reynolds number is dimensionless.

1.3 The capacitance of a capacitor is defined by

(a)

dtdviC =

The dimension of is that of electric current, which is a basic dimension. The dimensions of electric potential are obtained from Table 1.2 as

i

[ ] (b) TiLF⎥⎦⎤

⎢⎣⎡⋅⋅

=v

Thus the dimensions of the time rate of change of electric potential are

(c) TiLF

2 ⎥⎦⎤

⎢⎣⎡⋅⋅

=⎥⎦⎤

⎢⎣⎡

dtdv

Use of Equation (c) in Equation (a) leads to

[ ]

(d) LFTi

TiLF

i

22

2

⎥⎦

⎤⎢⎣

⎡⋅⋅

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⋅⋅

=C

1.4 (a) The natural frequency of a mass-spring system is

(a) mk

n =ω

where m is mass with dimension [M] and k is stiffness with dimensions in the M-L-T

system of ⎥⎦⎤

⎢⎣⎡

2TM . Thus the dimensions of natural frequency are

2

Chapter 1

[ ]

(b) T1

MTM 2

1

2

⎥⎦⎤

⎢⎣⎡=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

=nω

(b) The natural frequency of the system is 100 Hz, which for calculations must be converted to r/s,

(c) sr 7.125

cyclesr2

scycles 20

scycles 20

=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛=

=

π

ωn

Equation (a) is rearranged as (d) 2

nmk ω= Substitution of known values into Equation (d) leads to

( )

(e) mN 1058.1

sr 7.125kg 1.0

3

2

x

k

=

⎟⎠⎞

⎜⎝⎛=

1.5 (a) The mass of the carbon nanotube is calculated as ( )

( ) (kg 3.78x10

m 1080m 1034.0mkg 1300

23-

9293

2

−−⎟⎠⎞

⎜⎝⎛=

==

xx

LrALm

π

πρρ

)

(b) Conversion between TPa and psi leads to

28

22

212

212

inlb 1060.1

in 12ft 1

ft 3.28m 1

Nlb 225.0

mN 1.1x10

mN 1.1x10TPa 1.1

x

E

=

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛=

==

(c) Calculation of the natural frequency leads to

3

Chapter 1

( )

( ) ( )

sr 1.73x10

m 10801034.0mkg 1300

m 1034.04m

N 101.137.22

37.22

10

49293

492

12

4

=

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

=

=

−−

−

xx

xx

ALEI

π

π

ρω

Converting to Hz gives

Hz 1075.2 r 2

cycle 1sr 1073.1

9

10

x

x

=

⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛=

πω

1.6 The power of the motor is calculated as

(a) kW 5.37 hr 24

hrkW 900

=

⋅=P

The power is converted to English units using the conversions of Table 1.1

(b) slbft 1077.2

sm

ft 3.28mN

lb 0.225N105.37

smN 37.5x10

W105.37

4

3

3

3

⋅=

⎟⎠⎞

⎜⎝⎛⋅⎟

⎠⎞

⎜⎝⎛

=

⋅=

=

x

x

xP

Conversion to horsepower leads to

(c) hp 3.50

slbft 550

hp 1slbft 1077.2 4

=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⋅⋅

= xP

1.7 The conversion of density from English units to SI units is

4

Chapter 1

(a) mkg 1099.9

m 1ft 28.3

slugs 0.00685kg 1

ftslugs 94.1

ftslugs 94.1

33

3

3

3

x=

⎟⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

=ρ

1.8 The constant acceleration of the train is

(a) sm 6 2−=a

The velocity is obtained using Equation (a) as (b) 6)( Cttv +−=

The constant of integration is evaluated by requiring

(c) sm 50

s 3600hr 1

kmm 1000

hrkm 180

hrkm 180)0(

=

⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛=

==tv

Using Equation (c) in Equation (b) leads to

(d) sm 506)( +−= ttv

The train stops when its velocity is zero,

(e) s 33.85060

=+−=

tt

The distance traveled is obtained by integrating Equation (d) and assuming x(0)=0, leading to

(f) 503)( 2 tttx +−= The distance traveled before the train stops is

(g) m 3.208 )33.8(50)33.8(3)33.8( 2

=+−=x

1.9 The differential equation for the angular velocity of a shaft is

(a) TcdtdJ t =+ ωω

Each term in Equation (a) has the same dimensions, those of torque or [ . The

dimensions of angular velocity are

]LF ⋅

⎥⎦⎤

⎢⎣⎡T1 . Thus the dimensions of are tc

5

Chapter 1

[ ]

[ ] (b) TLF T1LF

⋅⋅=⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡⋅

=tc

1.10 The equation for the torque applied to the armature is (a) faa iiKT =

Equation (a) is rearranged as

(b) fa

a iiTK =

The dimensions of torque are [ ]LF ⋅ thus the dimensions of the constant are

[ ] (c) i

LF2 ⎥⎦⎤

⎢⎣⎡ ⋅

=aK

The equation for the back emf is (d) ωfviKv =

Equation (d) is rearranged as

(e) ωf

v ivK =

The dimensions of voltage are ⎥⎦⎤

⎢⎣⎡⋅⋅TiLF and the dimensions of angular velocity are ⎥⎦

⎤⎢⎣⎡T1 .

The dimensions of the constant are vK

[ ]

(f) i

LF

T1i

TiLF

2 ⎥⎦⎤

⎢⎣⎡ ⋅

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡⋅⋅

=vK

It is clear from Equations (c) and (f) that the dimensions of [ ] [ va KK and ]are the same.

These dimensions are the same as those of inductance (Table 1.2).

1.11 (a) The dimensions of Q&are determined from Equation (a) ( ) (a) 44

bTTAQ −= εσ&

[ ][ ] (b) T

LFTL

LF 4242 ⎥⎦

⎤⎢⎣⎡ ⋅

=Θ⎥⎦⎤

⎢⎣⎡

Θ⋅⋅⋅ L

(b) The differential equations governing the temperature in the body is

(c) 0)( 44 =−+ bTTdtdTc σερ

The perturbation in temperature in the radiating body is defined by

6

Chapter 1

(d) 1bbsb TTT += This leads to a perturbation in the temperature of the receiving body defined as

(e) 1TTT s += Substitution of equations (d) and (e) in Equation (c) leads to

( ) ( ) ( )[ ] (f) 041

411 =+−+++ bbsss TTTTTT

dtdc σερ

Simplifying Equation (f) gives

(g) 0114

144

141 =⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+−⎟⎟

⎠

⎞⎜⎜⎝

⎛++

bs

bbs

ss T

TT

TTT

dtdTc σερ

Expanding the nonlinear terms, keeping only through the linear terms and noting that bss TT =

(h) 44

044

13

131

14141

bbss

bs

bbs

ss

TTTTdtdT

c

TT

TTTT

dtdTc

σεσερ

σερ

=+

=⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛+

1.12 The differential equation is linearized by using the small angle assumption which implies θθ ≈sin and 1cos ≈θ . Using these approximations in the differential equation leads to the linearized approximation as

(a) 041

31 222 =++ θθθ kLcLmL &&&

1.13 The differential equation is linearized by using the small angle assumption which implies θθ ≈sin and 1cos ≈θ . Using these approximations in the differential equation leads to the linearized approximation as

(a) 23

1 2 xLyLmgmL &&&&&& =⎟⎠⎞

⎜⎝⎛ ++ θθ

1.14 The nonlinear differential equations governing the concentration of the reactant and temperature are

( )

(b)

(a)

)/(

)/(

dtdTVcCeVQTqcTqc

qCCVeqdt

dCV

pARTE

pip

AiARTEA

ραλρρ

α

=+−−

=++

−

−

&

The reactor is operating at a steady-state when a perturbation in flow rate occurs according to

(c) )(tqqq ps += The flow rate perturbation induces perturbations in concentration and temperature according to

7

Chapter 1

(e) )(

(d) )(

tTTT

tCCC

ps

ApAsA

+=

+=

The steady-state conditions are defined by setting time derivatives to zero in Equation (a) leading to

( )(g) 0

(f) )/(

)/(

=+−−

=+−

−

s

s

s

ARTE

spsip

AissARTE

s

CeVQTcqTqc

CqCVeq

αλρρ

α&

Substitution of Equations (d) and (e) into Equations (a) and (b) leads to

[ ]( )( ) ( )

( ) ( ) ( ) [ ]( ) )i(

)h(

)(/

)(/

dtdT

VcCCeVQTTcqqTcqq

CqqCCVeqqdt

dCV

ppApA

TTREpsppsipps

AipsApAsTTRE

psAp

s

ps

ps

ραλρρ

α

=++−++−+

+=++++

+−

+−

&

It is noted from Equation (f) of Example (1.6) that a linearization of the exponential terms in Equations (h) and (i) is

(j) 2)(

pRTE

s

RTE

TTRE

TeRT

Eee ssps−−

+−

+=

Use of Equation (j) in Equations (h) and (i) and rearrangement leads to

( ) ( )

( ) ( ) ( ) ( )

(l)

(k)

2

2

dtdT

Vc

CCTeRT

EeVQTTcqqTcqq

CqqCCTeRT

EeVqqdt

dCV

pp

ApApRTE

s

RTE

psppsipps

AipsApAspRTE

s

RTE

psAp

s

ss

ss

ρ

αλρρ

α

=

+⎥⎥⎦

⎤

⎢⎢⎣

⎡++−++−+

+=+⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛++++

−−

−−

&

Equations (g) and (h) are used to simplify Equations (k) and (l) to

( )

( ) ( )

(n)

(m)

2

2

dtdT

Vc

CCTeRT

EVCeVTqTqTqcTcq

Cq

CCTeRT

EVCVeCqCqCqdt

dCV

pp

ApApRTE

sAp

RTE

pppssppipp

Aip

ApAspRTE

sAp

RTE

AppAspApsAp

s

ss

ss

ρ

αλαλρρ

αα

=

+⎥⎥⎦

⎤

⎢⎢⎣

⎡++++−

=

+⎟⎟

⎠

⎞

⎜⎜

⎝

⎛+++++

−−

−−

Neglecting products of perturbations Equations (m) and (n) are rearranged as

8

Chapter 1

( ) (p) 0

(o)

2

2

=⎥⎥⎦

⎤

⎢⎢⎣

⎡++++−

−=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛++++

−−

−−

AspRTE

sAp

RTE

pssppippp

p

AspAip

AspRTE

sAp

RTE

AppApsAp

CTeRT

EVCeVTqTqcTcqdt

dTVc

CqCq

CTeRT

EVCVeCqCqdt

dCV

ss

ss

αλαλρρρ

αα

1.15 The specific heat is related to temperature by

(a) 6.23

5.121 TATAAcp ++=

The transient temperature is the steady-state temperature plus a perturbation, (b) ps TTT +=

Substituting Equation (b) into Equation (a) leads to ( ) ( )

(c) 11

6.2

6.23

5.15.1

21

6.25.121

⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟⎟

⎠

⎞⎜⎜⎝

⎛++=

++++=

s

ps

s

ps

pspsP

TT

TATT

TAA

TTTTAAc

Using the binominal expansion to linearize Equation (c) leads to

(d) 6.215.11 6.23

5.121 ⎟⎟

⎠

⎞⎜⎜⎝

⎛++⎟⎟

⎠

⎞⎜⎜⎝

⎛++=

s

ps

s

psp T

TTA

TT

TAAc

The differential equation for the time-dependent temperature is

(e) 11∞=+ T

RT

RdtdTc p

Substituting Equations (b) and (d) into Equation (e) along with ps TTT ∞∞∞ += leads to

( ) ( ) ( )(f) 116.215.11 6.23

5.121 pspsps

s

ps

s

ps TT

RTT

RTT

dtd

TT

TATT

TAA ∞∞ +=+++⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟⎟

⎠

⎞⎜⎜⎝

⎛++

Noting that the steady-state is defined by 0=dt

dTs and ss TT ∞= reduces Equation (f) to

(g) 116.215.11 6.23

5.121 pp

p

s

ps

s

ps T

RT

RdtdT

TT

TATT

TAA ∞=+⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟⎟

⎠

⎞⎜⎜⎝

⎛++

Terms such as dt

dTT p

p are nonlinear. Equation (g) is linearized by noting that 1<<s

p

TT

( ) (h) 116.23

5.121 pp

pss T

RT

RdtdT

TATAA ∞=+++

1.16 The force acting on the piston at any instant is (a) pAF =

where A is the area of the piston head. The pressure is related to the density by (b) γρCp =

9

Chapter 1

The mass of air in the cylinder is constant and is calculated when the piston is in equilibrium as

(c) 0 Ahm ρ= where 0ρ is the density of the air in equilibrium. Using Equation (b) in Equation (c) leads to

(d)

1

0 AhCp

mγ⎟⎠⎞

⎜⎝⎛=

where is the pressure in the cylinder when the piston is in equilibrium. At any instant the mass is calculated as

0p

(e) )(

)(1

xhACp

xhAm

−⎟⎠⎞

⎜⎝⎛=

−=

γ

ρ

Since the mass is constant, Equations (d) and (e) are equated leading to

(f) 0

γ

⎟⎠⎞

⎜⎝⎛

−=

xhhpp

Substitution of Equation (f) into Equation (a) leads to

(g) 0

γ

⎟⎠⎞

⎜⎝⎛

−=

xhhApF

(b) Equation (g) is rearranged as

(h) 10

γ−

⎟⎠⎞

⎜⎝⎛ −=

hxApF

Since 1<hx a binomial expansion can be used on the right-hand side of Equation (h).

Using the binomial expansion keeping only through the linear term leads to

(e) 00 x

hAp

ApFγ

+=

The linear stiffness is obtained from Equation (e) as

(f) 0

hAp

kγ

=

10

Chapter 1

1.17 The appropriate superposition of the voltage in Figure P1.17 is illustrated below

The mathematical representation of the voltage source is

[ ] ( )[ ])4()2(624)2()(12)( −−−−+−−= tututtututv

1.18 The superposition of the force of Figure P1.18 is illustrated below. The

mathematical representation of the force is (a) )]30()15([300)]15()([300)( −−−−−−= tututututF

11

Chapter 1

1.19 The superposition of the cam displacement over one period is shown below

(a) The mathematical representation of the displacement over one period is

[ ](a) )]3.0()25.0()[04.0012.0(

)25.0()05.0([002.0)05.0()(04.0)(−−−−+

−−−+−−=tutut

tututututx

(b) The period of the cycle is 0.5 s. Thus the displacement over the second period is obtained by replacing t by t+0.5 in Equation (a). The displacement over the kth period is obtained by replacing t by t+(k-1)(0.5) in Equation (a). The total displacement is obtained by summing over all periods

[ ]

( )[ ] (b) ]2.05.0()25.5.0(04.0008.002.0

)]25.5.()45.5.0([002.)45.5.()5.5.(04.0)(1

+−−+−−−+

+−−+−++−−+−= ∑=

ktuktutk

ktuktuktuktutxK

k

where K is the smallest integer greater than t/(0.05). 1.20 Integration of Newton’s second law with respect to time leads to the principle of impulse and momentum

(a) )( 12

2

1

vvmFdtIt

t

−== ∫

where the total impulse applied between is . The 12 impulse is

applied instantaneously to the 4-kg particle when it is at rest. Application of the principle of impulse and momentum leads to

21 tand t ∫2

1

t

t

Fdt sN ⋅

12

Chapter 1

( )

(b) sm 3

kg 4sN 12kg 4sN 12

2

2

=

⋅=

=⋅

v

v

1.21 The equation for the voltage drop across an inductor is

(a) dtdiLv =

Integration of Equation (a) with respect to time leads to

( ) (b) 120

iiLvdtt

−=∫

The initial current is zero. Solving Equation (b) for leads to 2i

(c) A 50 H 0.4

sV 20

02

=

⋅=

=∫

L

vdti

t

1.22 The mathematical representation of the force is (a) )8.3(50)5.2(150)(100)( −+−+= ttttF δδδ

1.23 The MATLAB file Problem1_23 which determines the steady-state response of a series LRC circuit is listed below % Problem1_23.m % Steady-state response of seties LRC circuit clear disp('Steady-state response of series LRC circuit') % Input parameters disp('Input resistance in ohms') R=input('>> ') disp('Input capacitance in farads') C=input('>>') disp('Input inductance in henrys') L=input('>> ') disp('Input source frequency in r/s') om=input('>> ') disp('Input source amplitude in V') V0=input('>> ') % Calculates parameters disp('Natural freqeuncy in r/s =')

13

Chapter 1

omn=1/(L*C)^0.5 disp('Dimensionless damping ratio =') zeta=R/2*(C/L)^0.5 disp('Phase angle in rad=') C1=om^2-omn^2; C2=2*zeta*om*omn; phi=atan2(C1,C2) disp('Steady-state amplitude in A =') C3=V0*om/L; C4=1/(C1^2+C2^2)^0.5; I=C3*C4 tf=10*pi/om; dt=tf/200; for k=1:201 t(k)=(k-1)*dt; i(k)=I*sin(om*t(k)+phi); end plot(t,i) xlabel('t (s)') ylabel('i (A)') title('Steady-state response of series LRC circuit') str1=['R=',num2str(R),' \Omega']; str2=['C=',num2str(C),' F']; str3=['L=',num2str(L),' H']; str4=['\omega=',num2str(om),' r/s']; str5=['V_0=',num2str(V0),' V']; text(0.9*tf,I,str1) text(0.9*tf,0.8*I,str2) text(0.9*tf,0.6*I,str3) text(0.9*tf,0.4*I,str4) text(0.9*tf,0.2*I,str5) The MATLAB workspace from a sample execution of Problem1_23.m is >> Problem1_23 Steady-state response of series LRC circuit Input resistance in ohms >> 100 R = 100 Input capacitance in farads >>0.2e-6

14

Chapter 1

C = 2.0000e-007 Input inductance in henrys >> 0.5 L = 0.5000 Input source frequency in r/s >> 2000 om = 2000 Input source amplitude in V >> 120 V0 = 120 Natural freqeuncy in r/s = omn = 3.1623e+003 Dimensionless damping ratio = zeta = 0.0316 Phase angle in rad= phi = -1.5042 Steady-state amplitude in A = I =

15

Chapter 1

0.0798 >> The resulting steady-state plot is

1.24 The MATALB file Prolbem1_24.m is listed below % Problem1_24.m %(a) Input two five by five matrices disp('Please input matrix A by row') for i=1:5 for j=1:5 str=['Enter A(',num2str(i),num2str(j),')']; disp(str) A(i,j)=input('>> '); end end disp('Please input matrix B by row') for i=1:5 for j=1:5 str=['Enter B(',num2str(i),num2str(j),')']; disp(str) B(i,j)=input('>> '); end end A

16

Chapter 1

B % (b) =A+B C=A+B % (c) D=A*B D=A*B % (d) det(A) detA=det(A) % eigenvalues and eigenvectors of A [x,Y]=eigs(A); disp('Eigenvalues of A') Y disp('Matrix of eigenvalues of A') x A sample output from execution of the file is shown below >> clear >> Problem1_24 Please input matrix A by row 'Enter A(11)' >> 1 'Enter A(12)' >> 0 'Enter A(13)' >> 12 'Enter A(14)' >> -1 'Enter A(15)' >> 21 'Enter A(21)' >> 14 'Enter A(22)' >> -3 'Enter A(23)' >> 2 'Enter A(24)' >> 0

17

Chapter 1

'Enter A(25)' >> -22 'Enter A(31)' >> 11 'Enter A(32)' >> 12 'Enter A(33)' >> 10 'Enter A(34)' >> -4 'Enter A(35)' >> 12 'Enter A(41)' >> 10 'Enter A(42)' >> 11 'Enter A(43)' >> 18 'Enter A(44)' >> 12 'Enter A(45)' >> 21 'Enter A(51)' >> 10 'Enter A(52)' >> 11 'Enter A(53)' >> 31 'Enter A(54)' >> 21 'Enter A(55)'

18

Chapter 1

>> 11 Please input matrix B by row 'Enter B(11)' >> 21 'Enter B(12)' >> -21 'Enter B(13)' >> 21 'Enter B(14)' >> 10 'Enter B(15)' >> 9 'Enter B(21)' >> 8 'Enter B(22)' >> 2 'Enter B(23)' >> 2 'Enter B(24)' >> 4 'Enter B(25)' >> -5 'Enter B(31)' >> 16 'Enter B(32)' >> 12 'Enter B(33)' >> 11 'Enter B(34)' >> 18 'Enter B(35)'

19

Chapter 1

>> 11 'Enter B(41)' >> 21 'Enter B(42)' >> 32 'Enter B(43)' >> 14 'Enter B(44)' >> 19 'Enter B(45)' >> 12 'Enter B(51)' >> 12 'Enter B(52)' >> 9 'Enter B(53)' >> -5 'Enter B(54)' >> 13 'Enter B(55)' >> 21 A = 1 0 12 -1 21 14 -3 2 0 -22 11 12 10 -4 12 10 11 18 12 21 10 11 31 21 11

20

Chapter 1

B = 21 -21 21 10 9 8 2 2 4 -5 16 12 11 18 11 21 32 14 19 12 12 9 -5 13 21 C = 22 -21 33 9 30 22 -1 4 4 -27 27 24 21 14 23 31 43 32 31 33 22 20 26 34 32 D = 444 280 34 480 570 38 -474 420 -122 -299 547 -107 249 418 353 1090 601 493 969 818 1367 955 812 1244 859 detA = -1171825 Eigenvalues of A Y = 43.3949 0 0 0 0 0 -18.9247 0 0 0 0 0 -1.8896 -11.6121i 0 0 0 0 0 -1.8896 +11.6121i 0 0 0 0 0 10.3091 Matrix of eigenvalues of A x = -0.3664 -0.1632 -0.3379 - 0.4681i -0.3379 + 0.4681i 0.2658

21

Chapter 1

0.1869 0.7510 0.7187 0.7187 -0.4106 -0.2133 -0.4463 0.0528 + 0.2510i 0.0528 - 0.2510i -0.4042 -0.6060 -0.2256 -0.0845 - 0.0753i -0.0845 + 0.0753i 0.6726 -0.6466 0.3991 -0.2465 + 0.1043i -0.2465 - 0.1043i 0.3808 1.25 A MATLAB file to calculate and plot ),( ζrΛ is given below % Plots the function LAMBDA(r,zeta) as a function of r for several values of % zeta % Specify four values of zeta zeta1=0.1; zeta2=0.4; zeta3=0.8; zeta4=1.5; % Define values of r for calculations for i=1:400 r(i)=(i-1)*.01; % Calculate function LAMBDA1(i)=r(i)^2/((1-r(i)^2)^2+(2*zeta1*r(i))^2)^0.5; LAMBDA2(i)=r(i)^2/((1-r(i)^2)^2+(2*zeta2*r(i))^2)^0.5; LAMBDA3(i)=r(i)^2/((1-r(i)^2)^2+(2*zeta3*r(i))^2)^0.5; LAMBDA4(i)=r(i)^2/((1-r(i)^2)^2+(2*zeta4*r(i))^2)^0.5; end plot(r,LAMBDA1,'-',r,LAMBDA2,'.',r,LAMBDA3,'-.',r,LAMBDA4,'--') xlabel('r') ylabel('\Lambda') str1=['\zeta=',num2str(zeta1)]; str2=['\zeta=',num2str(zeta2)]; str3=['\zeta=',num2str(zeta3)]; str4=['\zeta=',num2str(zeta4)]; legend(str1,str2,str3,str4) title('\Lambda vs. r') The resulting output from execution of the .m file is the following plot

22

Chapter 1

1.26 The MATALB .m file Problem1_26 which determines and plots the step response of an underdamped mechanical system is shown below.

% Problem1_26.m % Step response of an underdamped mechanical system % Input natural frequency and damping ratio clear disp('Step response of underdamped mechanical system') disp('Please input natural frequency in r/s') om=input('>> ') disp('Please input the dimensionless damping ratio') zeta=input('>> ') % Damped natural frequency omd=om*(1-zeta^2)^0.5; C1=zeta*om/omd; C2=1/om^2; C3=zeta*om; tf=10*pi/omd; dt=tf/500; for i=1:501 t(i)=(i-1)*dt; x(i)=C2*(1-exp(-C3*t(i))*(C1*sin(omd*t(i))-cos(omd*t(i)))); end plot(t,x) xlabel('t (s)') ylabel('x (m)')

23

Chapter 1

str1=['Step response of underdamped mechancial system with \omega_n=',num2str(om),'and \zeta=',num2str(zeta)] title(str1) str2=['x(t)=',num2str(C2),'[1-e^-^',num2str(C3),'^t(',num2str(C1),'sin(',num2str(omd),'t)-cos(',num2str(omd),'t))]'] text(tf/4,C2/2,str2)

Output from execution of Problem1_26 follows

>> Problem1_26 Step response of underdamped mechanical system Please input natural frequency in r/s >> 100

om =

100

Please input the dimensionless damping ratio >> 0.1

zeta =

0.1000

24

Chapter 1

1.27 The perturbation in liquid level is ( ) )(1)( )/( aeqRth RAt −−=

(a) Since the argument of a transcendental function must be dimensionless the dimensions of the product of resistance and area must be time. Thus the dimensions of

resistance must be ⎥⎦⎤

⎢⎣⎡

2LT

(b) Note that the steady-state value of the liquid-level perturbation is . The MATLAB file Problem1_27.m which calculates and plots h(t) from t=0 until h is within 1 percent of its steady-state value is given below

qR

disp('Please enter resistance in s/m^2 ') R=input('>> ') % Final value of h hf=0.99*q*R; dt=0.01*R*A; h1=0; h(1)=0; t(1)=0; i=1; while h1<hf i=i+1; t(i)=t(i-1)+dt; h(i)=q*R*(1-exp(-t(i)/(R*A))); h1=h(i); end plot(t,h) xlabel('t (s)') ylabel('h (m)') title('Perturbation flow rate vs time') str1=['A=',num2str(A),' m^3/s'] str2=['R=',num2str(R),' s/m^2'] str3=['q=',num2str(q),' m^3/s'] text(0.5*t(i),0.5*h(i),str1); text(0.5*t(i),0.4*h(i),str2); text(0.5*t(i),0.3*h(i),str3);

Sample output from execution of Problem1_27.m is given below

>> Please enter tank area in m^2 >> 100

A =

100

Please enter flow rate in m^3/s

25

Chapter 1

>> 0.2

q =

0.2000

Please enter resistance in s/m^2 >> 15

R =

15

str1 =

A=100 m^3/s

str2 =

R=15 s/m^2

str3 =

q=0.2 m^3/s

>>

26

Chapter 2

2. Mechanical Systems

2.1 The mass moment of inertia of the annular cylinder is obtained by subtracting the moment of inertia of the hollow part from the moment of inertia of a solid cylinder.

(a) hs III −= The mass of a solid cylinder of radius r and length L is

( ) (b) 2 Lrm πρ= The mass of the solid cylinder is

( ) ( )

(c) kg 1058.4

m 2.1m 4.0mkg 7600

3

23

x

ms

=

⎟⎠⎞

⎜⎝⎛= π

The mass missing from the hollow part of the cylinder is ( ) ( )

(d) kg 1058.2

m 2.1m 3.0mkg 7600

3

23

x

msh

=

⎟⎠⎞

⎜⎝⎛= π

The total mass of the cylinder is

(e) kg 1001.2 3x

mmm hs

=

−=

(a) The mass moment of inertia of a solid cylinder about a centroidal axis parallel to the x axis is obtained using Table 2.1 as

(f) )3(

121 22 LrmI xx +=

The moment of inertia of the solid cylinder about the cetnroidal axis is

( )[ ](g) mkg 5.733

m) 2.1(m) 4.0(3kg 1058.4121

2

223

⋅=

+= xI xxs

The moment of inertia of the hollow part of the cylinder abut the centroidal x axis is

( )[ ](h) mkg 5.367

m) 2.1(m) 3.0(3kg 1058.2121

2

223

⋅=

+= xI xxh

The moment of inertia of the cylinder about the centroidal x axis is

(i) mkg 0.366 2⋅=

−= xxhxxsxx III

The parallel axis theorem is used to determine the moment of inertia about the x axis as shown in Figure P2.1 is

27

Chapter 2

( )( )(j) mkg 1009.1

m 6.0kg 1001.2mkg 0.366

2

23

232

2

⋅=

+⋅=

⎟⎠⎞

⎜⎝⎛+=

xx

LmII xxxx

(b) The mass moment of inertia of a cylinder about a longitudinal axis through its center is

(k) 21 2mrI yy =

The mass moment of inertia of the solid cylinder about the y axis is

( )(l) mkg 7.366

m) 4.0(kg 1058.421

2

23

⋅=

= xI yys

The moment of inertia of the hollow part of the cylinder about the y axis is

( )(l) mkg 0.116

m) 3.0(kg 1058.421

2

23

⋅=

= xI yyh

The moment of inertia of the annular cylinder about the y axis is

(m) mkg 7.250 2⋅=

−= yyhyysyy III

2.2 The total mass moment of inertia about an axis through A is the sum of the moments of inertia of the plate, the bar, and the cylinder,

(a) AcAbApA IIII ++= (a) The moment of inertia of the thin plate about an axis perpendicular to the page through its centroid is determined using Table 2.1 as

( ) ( )[ ](b) mkg 1083.6

m) 4.0(m 2.0kg 1.4121

)(121

22

22

22

⋅=

+=

+=

−x

hwmI zp

The parallel-axis theorem is used to determine its moment of inertia about an axis through A as

( ) ( )( )(c) mkg 10.8

m 4.1kg 1.4mkg 1083.6 2

222

2

⋅=

+⋅=

+=−x

mdII zpzAp

For the slender bar

28

Chapter 2

( )( )

(d) mkg 1007.8

m 1.1kg 8.0121

121

22

2

2

⋅=

=

=

−x

mLI zb

( ) ( )(e) mkg 531.0

m) 75.0(kg 8.0mkg 1067.6 2

222

2

⋅=

+⋅=

+=−x

mdII zbzAb

For the cylinder

( )

( )[ ](f) mkg 1080.2

m) 3.0(m) 1.0(3kg 8.2121

3121

22

22

22

⋅=

+=

+=

−x

LrmI zc

( ) ( )( )(g) mkg 091.0

m 15.0kg 8.2mkg 1080.2 2

222

2

⋅=

+⋅=

+=−x

mdII zczAc

The moment of inertia of the total assembly about an axis perpendicular to the page at A is obtained using Equations (a), (c), (e), and (g) as

( ) ( )(h) mkg 73.8

mkg 091.0)mkg 531.0(mkg 10.82

222

⋅=

⋅+⋅+⋅=zAI

(a) A horizontal axis through A is a centroidal axis for each component. Thus application of the parallel axis theorem is not required. Using Table 2.1 for the plate

( )( )

(i) mkg 0547.0

m 4.0kg 1.4121

121

2

2

2

⋅=

=

= mhI xp

The moment of inertia of the slender bar about a horizontal axis is approximately zero. For the cylinder

( )( )

(j) mkg 0.014

m 1.0kg 8.221

21

2

2

2

⋅=

=

= mrIxc Thus the moment of inertia of the assembly about a horizontal axis through A is

29

Chapter 2

( ) ( ) ( )(k) mkg 0687.0

mkg 104.1mkg 0mkg 104.52

22222

⋅=

⋅+⋅+⋅= −− xxI x

2.3 The total moment of inertia of the assembly of Figure P2.3 about an axis through O is the sum of the moment of the inertia of the bar about O and the moments of inertia of the particles about O.

(a) 21 OpOpObO IIII ++= The moment of inertia of the bar about its own centroidal axis is

( )

(b) mkg 26.7

m) 4(kg 20121

121

2

2

2

⋅=

=

= mLIb

The parallel axis theorem is used to calculate the moment of inertia of the bar about O as

( )(c) mkg 7.46

m) 1(kg 20mkg 7.26 2

22

2

⋅=

+⋅=

+= mdII bOb

The moment of inertia of a particle about its own centroidal axis is zero. The parallel axis theorem is used to calculate the moment of inertia of a particle about another axis. For the 5-kg particle

(c) mkg 5

m) kg)(1 5(2

21

⋅=

=OpI

The moment of inertia of the 10-kg particle about O is

(d) mkg 90

m) kg)(3 10(2

22

⋅=

=OpI

Combining Equations (a),(b),(c), and (d) leads to

(e) mkg 7.141

mkg 90mkg 5mkg 7.462

222

⋅=

⋅+⋅+⋅=OI

2.4 The velocity of the 2-kg particle is given as

[ ] (a) sm 2)3cos(4)3sin(2 kjiv ++= tt

(a) Noting that the particle is at the origin at t=0, the position vector of the particle at an arbitrary time t is obtained as

30

Chapter 2

[ ]

[ ] (b) m 2)3sin(34)3cos(1

32

2)3sin(34)3cos(

32

2)3cos(4)3sin(2

0

0

0

⎭⎬⎫

⎩⎨⎧ ++−=

⎥⎦⎤

⎢⎣⎡ ++−=

++=

=

=

∫

∫

kji

kji

kji

vr

ttt

ttt

dttt

dt

t

t

t

t

(b) The acceleration of the particle at an arbitrary time is

[ ] (c) sm )3sin(12)3cos(6 2ji

va

tt

dtd

−=

=

(a) The velocity of the particle at t=1 is obtained from Equation (a) as

[ ]

[ ] (d) sm 296.3282.0

sm 2)3cos(42sin(3)s) 1(

kji

kjiv

+−=

++==t

(b) The magnitude of the velocity at t=1 s is

( ) ( )[ ](e)

sm 45.4

)2(96.3282.0s) 1( 21

222

=

+−+==tv

(c) The kinetic energy of the particle at t=1 s is

( )

(f) mN 8.19 sm 45.4kg 2

21

s) 1(21

2

2

⋅=

⎟⎠⎞

⎜⎝⎛=

== tmT v

(d) The linear momentum of the particle at t=1 s is

( )

( ) (g) sN 492.7564.0 sm 296.30.282kg) 2(

s)1(

⋅+−=

⎥⎦⎤

⎢⎣⎡ +−=

==

kji

kji

vL tm

2.5 (a) The velocity of the mass center of the bar at the instant shown is obtained using the relative velocity equation

(a) / OGOG rkvv ×+= ω

31

Chapter 2

where O is located at the pin support. Since O is a fixed axis of rotation 0av o ==o . Thus

( )

(b) sm 15

m 5.1sr 10

j

ik0vG

⎟⎠⎞

⎜⎝⎛=

×⎟⎠⎞

⎜⎝⎛+=

(b) The acceleration of the mass center is obtained using the relative acceleration equation

( ) (c) sm 5.4 150

sm 15

sr 10m) 5.1(

sr 3

)(

2

2

//

ji

jkik0

rkkrkaa OGOGOG

−−=

⎟⎠⎞

⎜⎝⎛×⎟

⎠⎞

⎜⎝⎛+×⎟

⎠⎞

⎜⎝⎛−+=

××+×+= ωωα

(c) The kinetic energy of the bar at this instant is

( ) ( )( )

(d) mN 1300 sr 10m 5kg 6

121

21

sm 15kg 6

21

21

21

22

2

22

⋅=

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=

+= ωImT Gv

(d)The angular momentum of the bar about the centroidal axis is

( )

(e) m sN 125 sr 10m) 5(kg 6

121 2

⋅⋅=

⎟⎠⎞

⎜⎝⎛=

= ωIHG

2.6 Since the center of the fan is a fixed axis of rotation the kinetic energy of each blade is

(a) 21 2ωOb IT =

The parallel axis theorem is used to determine the moment of inertia of each blade about O

(b) mkg 592.0 m) kg)(0.45 2.1(mkg 4.0

2

22

2

⋅=

+⋅=

+= mdIIO

Using the result of Equation (b) in Equation (a) leads to

( )(c) mN 9.202

s 60min 1

revrad 2

minrev 250mkg 592.0

21

22

⋅=

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛⋅= πbT

The fan consists of four blades, thus the total kinetic energy of the fan is

32

Chapter 2

(d) mN 811.5 4

⋅== bTT

2.7 (a) The total kinetic energy of the engine is the sum of the kinetic energies of the components

(a) pistonrodcrsnk TTTT ++= The crank rotates about a fixed axis at O, thus its kinetic energy is

(b) 21 2

crankOcrank IT ω=

where and 2mkg 4.0 ⋅=OI

(c) sr 2.26

sec 60min 1

revrad 2

minrev 250

=

⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛=

πωcrank

Substituting these values into Equation (b) leads to

( )(d) mN 3.171

sr 2.26mkg 5.0

21 2

2

⋅=

⎟⎠⎞

⎜⎝⎛⋅=crankT

The connecting rod does not rotate about a fixed axis. Its kinetic energy is calculated by

(e) 21

21 22 ωIvmTrod +=

The connecting rod is a slender rod of mass 0.4 kg and length 30 cm. Its centroidal mass moment of inertia is obtained using Table 2.1 as

( )( )

(f) mkg 100.3

m 3.0kg 4.0121

121

23

2

2

⋅=

=

=

−x

mLI rod

Rigid-body kinematics is used to determine the angular velocity and velocity of the mass center of the connecting rod. The angle made by the connecting rod with the vertical is determined using the law of sines as illustrated below

(g) 4.315209.0

)10sin(3)sin(m 1.0

)10sin(m 3.0

)sin(

°==

°=

°=

b

b

b

Let A be point at which the crank is pinned to the connecting rod. Application of the relative velocity equation leads to

33

Chapter 2

( )[ ]

( ) (h) sm 36.123.2

)4.31cos()4.31sin(m 1.0sr 2.26

/

ji

jik0

rkvv OAOA

+=

°+°−×⎟⎠⎞

⎜⎝⎛−+=

×+= crankω

The piston, P, is constrained to move in the y direction. Application of the relative velocity equation between A and P on the connecting rod gives

( ) ( )[ ]( ) (i) )0521.036.1(0.295-2.23

)10cos()10sin(m 3.0sm 36.123.2

rod

/

ji

jikji

rkvj APA

rod

rod

rodPv

ωω

ω

ω

++=

°+°×++=

×+=

Equating the x components of the vectors on both sides of Equation (i) leads to

(j) sr 56.7

295.023.20

=

−=

rod

rod

ω

ω

Equating the y components of the vectors on both sides of Equation (i) leads to

(k) sm 76.1

0521.036.1

=

+= rodPv ω

The velocity of the mass center of the connecting rod is calculated using the relative velocity equation on the connecting rod between A and the mass center

( ) ( )[ ]

( ) (l) sm 56.118.1

)10cos()10sin(m 15.0sr 56.7

sm 36.123.2

/

ji

jikji

rkvv AG

+=

°+°×⎟⎠⎞

⎜⎝⎛++=

×+= rodArod ω

The magnitude of the velocity of the mass center of the connecting rod is

(m) sm 92.1

sm 56.1

sm 18.1

22

=

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=rodv

The use of Equations (f), (j), and (m) in Equation (e) gives

( ) ( )(n) mN 823.0

sr 56.7mkg 100.3

21

sm 92.1kg 4.0

21 2

232

⋅=

⎟⎠⎞

⎜⎝⎛⋅+⎟

⎠⎞

⎜⎝⎛= −xTrod

The kinetic energy of the piston is

34

Chapter 2

( )

(o) mkg 465.0 sm 76.1kg 3.0

21

21

2

2

2

⋅=

⎟⎠⎞

⎜⎝⎛=

= Ppiston mvT

The total kinetic energy of the engine is calculated by substituting Equations (d), (n) and (o) in Equation (a) leading to

(p) mN 172.6 mN 0.465mN 0.823mN 3.171

⋅=⋅+⋅+⋅=T

(b) The acceleration of the piston is obtained using the relative acceleration equations. The crank rotates with a constant angular velocity, thus its angular acceleration is zero. Application of the relative acceleration equation between O and A on the crank leads to

( )[ ]

( ) (q) sm 5.587.35

)4.31cos()4.31sin(m 1.0sr 2.26

2

2/

2/

j

ji00

rrkaa

−=

°+°−⎟⎠⎞

⎜⎝⎛−+=

−×+= OAcrankOAcrankOA ωα

Application of the relative acceleration equation on the connecting rod between A and P leads to

( ) ( )[ ]

( )[ ]

( ) ( ) (r) 0521.04.75295.07.32

)10cos()10sin(m 3.0sr 56.7

)10cos()10sin(m 3.0sm 5.587.35

2

2

/2

/

ji

ji

jikji

rrkaj AP

rodrod

rod

rodAProdAPa

αα

α

ωα

+−+−=

°+°⎟⎠⎞

⎜⎝⎛−

°+°×+−=

−×+=

Equating x components of the vectors on both sides of Equation (r) leads to

(s) sr 7.110

0295.07.32

2=

=−

rod

rod

α

α

Equating the y components of the vectors on both sides of Equation (r) leads to ( )

(t) sm 6.69

7.110)0521.0(4.75

2−=

+−=Pa

2.8 The total kinetic energy of the system is the sum of the kinetic energy of the three system components

(a) 321 TTTT ++= where is the kinetic energy of the 5-kg block, is the kinetic energy of the disk and

is the kinetic energy of the 10-kg block. 1T 2T

3T

35

Chapter 2

(a) At an arbitrary instant the kinetic energy of the 5-kg block is

(b) 2.5

kg) 5(21

21

2

2

211

x

x

xmT

&

&

&

=

=

=

The kinetic energy of the disk is

(c) 21 2

2 θ&IT =

where θ& is the angular velocity of the disk. The velocity of the particle on the disk where the cable which connects the 5-kg block is attached must be the same as the velocity of the cable and thus the 5-kg block. Thus

( )

(d) 101.0

m 1.0

xxx

&&&

&&

==

=

θ

θ

Substitution of Equation (d) into Equation (c) leads to

( )( )

(e) 10

102.021

2

22

x

xT

&

&

=

=

Letting y be the upward displacement of the 10-kg block, its kinetic energy is

(f) 21 2

33 ymT &=

Using the same kinematic arguments as above

(g) 1.5 )(0.15)(10

m) 15.0(

xx

y

&&

&&

=== θ

Substitution of Equation (g) into Equation (f) leads to

( )( )

(h) 25.11

5.1kg 1021

2

23

x

xT

&

&

=

=

Thus the total kinetic energy of the system is

(i) 23.75 25.11105.2

2

222

xxxxT

&

&&&

=

++=

(b) Substitution of Equation (d) Into Equation (i) leads to ( )

(j) 0.2375

1.075.232

2

θ

θ&

&

=

=T

2.9 The kinetic energy of a body in pure rotation is

(a) 21 2ωOIT =

36

Chapter 2

The angular velocity of shaft AB of 250 rpm is converted to r/s as

(b) 2.26 s 60

min 1rev

r 2minrev 250

=

⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛=

πωAB

The angular velocities of meshing gears are related by

(c) 2

112

2211

nn

nn

ωω

ωω

=

=

where represents the number of teeth of a gear. The angular velocities of the shafts connected by gears to shaft AB are calculated using Equation (c). The angular velocity of the shaft with the 25 tooth gear is

n

(d) sr 47.10

2510

sr 2.261

=

⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛=ω

The angular velocity of the remaining shaft is

(e) sr 5.78

1030

sr 2.263

=

⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛=ω

The total kinetic energy of the system is

( )

( )

( )(f) J 1067.2

sr 5.78mkg 6mkg 1

21

sr 2.26mkg 5.3mkg 8mkg 4.1

21

sr 47.10mkg 1.2mkg 10

21

4

222

2222

222

x

T

=

⎟⎠⎞

⎜⎝⎛⋅+⋅+

⎟⎠⎞

⎜⎝⎛⋅+⋅+⋅+

⎟⎠⎞

⎜⎝⎛⋅+⋅=

2.10 The parallel combination of springs is replaced by a spring of stiffness

mN 103

mN 101

mN 102 555

1 xxxk =⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=

The two series combinations are replaced by springs of stiffnesses

mN 1071.1

mN 104

1

mN 103

11 5

55

2 x

xx

k =+

=

37

Chapter 2

mN 1008.3

mN 108

1

mN 105

11 5

55

3 x

xx

k =+

=

The remaining springs are in parallel leading to an equivalent stiffness of

mN 1079.4

mN 1008.3

mN 1.71x10

5

55

32

x

x

kkkeq

=

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=

+=

2.11 Define θ as the clockwise angular displacement of the bar, measured from its equilibrium position. Let be the torsional stiffness of a torsional spring attached to a particle on the bar such that the potential energy of the torsional spring is equal to the potential energy of the system of springs at any instant assuming small

eqtk ,

θ . The potential energy of the equivalent torsional spring at any instant is

(a) 21 2

, θeqtkV =

The total potential energy of the existing springs is

( )( ) ( )( ) ( )

( ) (b) 1098.421

18002123000

2134000

21

24

222

θ

θθθ

x

V

=

++=

Equating the potential energies of Equations (a) and (b) leads to

(c) rmN 1098.4 4

,⋅

= xk eqt

2.12 Let θ be the clockwise angular displacement of bar AB and let φ be the counterclockwise angular displacement of bar CD. It is desired to replace the springs by a single spring of an equivalent stiffness located at A as illustrated.

38

Chapter 2

The potential energy of this spring, for an arbitrary θ is

( ) (a) 1.021 2θeqkV =

The potential energy of the springs as illustrated in Figure P2.12 is

( ) ( ) (b) 0.1mN 8000

213.0

mN 5000

21 22 φθ ⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=V

The two bars are connected by a rigid massless link. The displacements of the particles at each end of the link are equal,

(c) 98

9.08.0

θφ

φθ

=

=

Substitution of Equation (c) into Equation (b) gives

( ) (d) 1013.121

988000)3.0(5000

21

24

22

2

θ

θ

x

V

=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛+=

Equating the left-hand sides of Equations (a) and (d) leads to

(e) mN 1013.1

01.1013.1

6

4

xk

kx

eq

eq

=

=

2.13 Two springs in series are illustrated below.

Let represent the change in length of the spring of stiffness and let represent the change in length of the spring of stiffness . The total potential energy of the two springs is

1x 1k 2x

2k

(a) 21

21 2

22211 xkxkV +=

The displacement of the block, x, is the sum of the changes in length of the springs (b) 21 xxx +=

39

Chapter 2

The force in each spring is the same, (c) 2211 xkxk =

Equation (c) is rearranged as

(d) 12

12 x

kkx =

Substitution of Equation (d) into Equation (b) leads to

(e) 1 12

1 xkkx ⎟⎟⎠

⎞⎜⎜⎝

⎛+=

Equation (e) is rearranged resulting in

(f) 21

21 x

kkkx+

=

Equation (f) is used in Equation (f) giving

(g) 21

12 x

kkkx+

=

Substitution of Equations (f) and (g) in Equation (a) leads to

( )

(h) 21

)()(

21

21

21

21

2

21

21

22

21

2121

22

21

212

221

2

21

12

2

21

21

xkk

kk

xkk

kkkk

xkk

kkkk

xkk

kkxkk

kkV

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

⎥⎦

⎤⎢⎣

⎡+

+=

⎥⎦

⎤⎢⎣

⎡

++

=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

+⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

The equivalent stiffness is obtained from Equation (h) as

(i) 21

21

kkkkkeq +

=

2.14 The spring of stiffness is in series with the parallel combination of the spring of stiffness and the viscous damper. The force in each series component is the same as F,

as

illustrated below

1k

2k

40

Chapter 2

Let represent the change in length of the spring of stiffness and let represent the change in length of the parallel combination. Thus

1x 1k 2x

(a) 11xkF = and

(b) 222 xcxkF &+= The change in length of series components is the sum of the changes in lengths of the individual components

(c) 21 xxx += Equating the force in Equations (a) and (b) leads to

(d) 22211 xcxkxk &+= Solving Equation (c) for and substituting into Equation (d) leads to 2x

(e) )()()(

21121

11211

xcxkxcxkkxxcxxkxk&&&&

+=++−+−=

Noting from Equation (a) that 1

1 kFx = , Equation (e) is rewritten as

( )

(f) 1 211

2

211

21

xcxkFkcF

kk

xcxkkFc

kFkk

&&

&&

+=+⎟⎟⎠

⎞⎜⎜⎝

⎛+

+=++

2.15 A free-body diagram of the disk is illustrated below.

Application of the rigid-body force equation to the disk gives

41

Chapter 2

(a) amPFamF=+−

=∑

Summing moments about the mass center of the disk leads to

(b) 21 2α

α

mrFr

IM G

=

=∑

If the disk rolls without slip then (c) αra =

Substituting Equations (b) and (c) into Equation (a) leads to

(d) 32

23

21

mrP

Pmr

mrPmr

=

=

=+−

α

α

αα

Substitution of Equation (d) into Equation (b) gives

(e) 32

32

21

P

mrPmrF

=

⎟⎠⎞

⎜⎝⎛=

(a) For the assumption of no slip to be valid, mgF µ< . Thus the criterion for no slip is

(f) 23

32

mgP

mgP

µ

µ

<

<

Substitution of given values into Equation (f) leads to the largest value of P such that the disk rolls without slip as

(g) N 41.4 sm 9.81kg) 8.1)(25.0(

23

2max

=

⎟⎠⎞

⎜⎝⎛=P

(b) The angular acceleration of the disk is calculated using Equation (d) as

(h) sr 18.8

m) kg)(0.2 3(1.8

N) 2(4.41

32

2=

=

=mrPα

42

Chapter 2

2.16 Let θ be the counterclockwise angular rotation of the disk measured from the system’s equilibrium position. The work done by a viscous damper of damping coefficient attached to the block of mass is eqc Am

( ) ( )( )∫∫

−=

−=

(a) 2 θθ

θθ

drc

rdrcW

eq

eq

&

&

It is shown in Example 2.4 that the two viscous dampers in series can be replaced by a single viscous damper of damping coefficient

(b) 21

214 cc

ccc+

=

The work done by this equivalent viscous damper as the disk rotates through an angle θ is

(c) 49

23

23

24

4

∫

∫

⎟⎠⎞

⎜⎝⎛−=

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛−=

θθ

θθ

drc

rdrcW

&

&

The work done by the viscous damper of damping coefficient as the disk rotates through an angle

3cθ is

(d) 49

23

23

23

3

∫

∫

⎟⎠⎞

⎜⎝⎛−=

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛−=

θθ

θθ

drc

rdrcW

&

&

The total work done by the attached viscous dampers is obtained by adding Equations (c) and (d)

∫ ⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+

+−= (e) 49 2

21

213 θθdr

cccc

cW &

The equivalent viscous damping coefficient is obtained by requiring that the work done by the equivalent viscous damper, Equation (a) is equal to the work done by that attached viscous dampers, Equation (e), for all θθ & and . Thus

(f) 49

21

213 ⎟⎟

⎠

⎞⎜⎜⎝

⎛+

+=cc

cccceq

2.17 A free-body diagram of the block is illustrated below.

43

Chapter 2

(a) In order to initiate motion the applied force P must be greater than the friction force developed using the static coefficient of friction,

(a) N 4.132 sm9.81kg) (0.30)(45 2

=

⎟⎠⎞

⎜⎝⎛=

> mgP sµ

When motion occurs the friction force is (b) mgF kµ=

Application of Newton’s second law to the free-body diagram leads to (c) maFP ==

The acceleration of the particle is obtained from Equation (c) as

( )

(d) sm 294.0

sm 9.81kg) (0.27)(45N 4.132

kg 451

1

2

2

=

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−=

−= mgPm

a kµ

2.18 The solution follows that of Example 2.6, but with

(a) 32

1 =nn

(a) The power delivered to the system is the same as calculated in Example 2.6 (b) kW 4.209=P

(b) Use of Equation (c) of Example 2.6 with rpm 1000=ω gives rpm 3000=CDω . Then from Equation (g) of Example 2.6

( )

(c) mN 7.666

mN 2000rpm 3000rpm 1000

⋅=

⋅⎟⎟⎠

⎞⎜⎜⎝

⎛=

= ABCD

ABCD TT

ωω

2.19 The force of Figure P2.19 has the following superposition

44

Chapter 2

The mathematical representation of the force is

(a) )2()2040()1()4040()(20 )2()2040()1()2040()1(20)(20)(

−−−−−+=−−−−−+−−=

tuttutttututtutttuttutF

Assuming that is the only force applied to the particle, application of Newton’s second law to the particle leads to

)(tF

Noting that dtdva = the velocity of the particle is obtained by integrating Equation (b)

[ ] (b) )2()2040()1()4040()(2000∫∫ −−−−−+=tt

dttuttutttudtdtdv

Noting that

(c) )()()()(0

∫∫ −=−t

a

t

dttgatudtatutg

the integrals in Equation (b) are evaluated leading to [ ] [ ]

(d) )2()401040()1()202040()(105.2)(1040)2(2040)1()(10)0()(

2222

21

22

−−−−−−−++=

−−−−−+=− ==

tutttutttuttvtttutttututvtv t

ttt

(a) Evaluation of Equation (a) for t=0.5 s gives

(e) sm 5.0

)5.0(105.2)5.0( 2

=

+=v

(b) Evaluation of Equation (d) at t=2.5 s gives [ ]

(f) sm 22.5

40)5.2(10)5.2(4020)5.2(20)5.2(40)5.2(105.2)5.2( 222

=

−−−−−++=v

2.20 The superposition of the displacement over one period is illustrated below

45

Chapter 2

(a) The mathematical representation of the displacement over one period is

(a) )4.0()02.0008.0( )3.0()02.0006.0()1.0()02.0002.0()(02.0

)4.0()2.0008.0()3.0()02.0008.0( )3.0(002.0)1.0(002.0)1.0(02.0)(02.0)(

−−−−−+−−+=

−−−−−+−−−+−−=

tuttuttutttu

tuttuttututtuttutx

(b) The velocity over one period is obtained as

(b) 0.4)(t0.02t)(0.008 )4.0(02.0)3.0()02.0006.0()3.0(02.0 )1.0()02.0002.0()1.0(02.0)(02.002.0

−−−−+−−+−−−−+−−+=

=

δδ

δδtutttutttutt

dtdxv

Note that the terms multiplying the unit impulse functions are zero at the times where the impulses are applied and thus these terms are zero. Thus the velocity reduces to

(c) )4.0(02.0)3.0(02.0 )1.0(02.002.0)( −+−−−−= tutututv

2.21 A free-body diagram of the system at an arbitrary instant is shown below. It is assumed the block slides on a frictionless surface.

46

Chapter 2

Application of Newton’s second law to the free-body diagram leads to

(a) 2502000104101 55 xxxxxx

maF

&&&=−−−

=∑

Equation (a) is rearranged leading to the second-order differential equation (b) 01052000250 5 =++ xxxx &&&&

2.22 Let x be the distance traveled by the vehicle along the incline after the driver spots the stalled car. Thus from the definition of x and the information given appropriate initial conditions are

(a2) sm 60)0(

(a1) 0)0(

=

=

x

x

& A free-body diagram of the vehicle, drawn at an arbitrary instant after the braking force is applied, is illustrated below.

Application of Newton’s second law to the free-body diagram of the vehicle in the direction of motion leads to

(b) sin xmmgFmaF

B

xx

&&=+−

=∑θ

Rearranging Equation (b) and substituting given values leads to

47

Chapter 2

(c) 365.1105

)8sin(sm 81.9

2000

sin

4

2

+−=

°⎟⎠⎞

⎜⎝⎛+−=

+−=

−B

B

B

Fx

F

gmFx θ&&

Since the braking force is constant, Equation (c) shows that the acceleration is a constant. Integration of Equation (c) leads to the time-dependent velocity as

( ) (d) 105365.1 14 CtFxx B +−= −&

The constant of integration is evaluated through application of Equation (a2) leading to

sm 601 =C . Integration of Equation (d) then gives

( ) (e) 60105365.121

224 CttFxx B ++−= −

Application of Equation (a1) leads to 02 =C . Thus

( )( ) (g) 60 105365.1)(

(f) 60105365.121)(

4

24

+−=

+−=

−

−

tFxtx

ttFxtx

B

B

&

The vehicle stops when its velocity is equal to zero. Thus the minimum braking force required is the value of such that BF 0=x& when . Define

. The time required for the vehicle to stop is obtained from Equation (b) as

m 300=x

BFxa 4105365.1 −−=

(h) 60a

ts −=

Requiring , using Equation (h) in Equation (f) gives 300)( =stx

(i) 360021

60606021300

2

a

aaa

−=

⎟⎠⎞

⎜⎝⎛−+⎟

⎠⎞

⎜⎝⎛−=

Equation (i) is solved giving 2sm 6−=a . Thus the minimum braking force is determined

from

(j) N 1047.1

105365.164

4

xF

Fx

B

B

=

−=− −

2.23 The free-body diagram of the bolt is illustrated below

48

Chapter 2

As it turns the bolt rotates about a fixed axis through its center. Application of the moment equation to the free-body diagram of the bolt, assuming positive moments counterclockwise, leads to

( )(a)

sr 4.571

mkg 035.0mN 480mN 500

2

2

=

⋅=⋅−⋅

=∑

α

α

αOO IM

The angular acceleration is constant. The angular velocity of the bolt at an arbitrary time, noting that the angular velocity is zero at t=0, is

(b) 4.57100

t

dtdtdtd tt

=

= ∫∫ω

αω

The angular displacement of the bolt is

(c) 7.285 200

t

dtdtdtd tt

=

= ∫∫θ

ωθ

It takes 50 revolutions to remove the bolt, thus

( )

(d) s 05.1285.7tr 2.314

7.285rev

r 2rev 50

2

2

==

=⎟⎠⎞

⎜⎝⎛

t

tπ

2.24 The projectile motion equations obtained in Example 2.11 are

( )

(b) )sin(21

(a) cos

02

0

tvgty

tvx

θ

θ

+−=

= (a) The projectile lands when m 100−=y . Substituting given values into Equation (b) leads to

( )

(c) 010040090.4

)30sin(80081.921100

2

2

=−−

°+−=−

tt

tt

The quadratic formula is applied to solve Equation (c) for the time

49

Chapter 2

(d) 9.81 ,249.0 )90.4(2

)100)(90.4(4)400(400 2

−=

−−−±=t

The negative root is discarded implying that the particle impacts the ground at t=81.9 s. (a) The horizontal range of the projectile is the value of x at the time the particle impacts the ground. From Equation (a)

(e) m 1067.5

s) 9.81)(30cos(sm 800)9.81(

4x

x

=

°⎟⎠⎞

⎜⎝⎛=

(b) The projectile achieves its maximum altitude when the y component of velocity is zero,

(f) s 77.40sm 81.9

)30sin(sm 800

sinsin0

2

0

0

=

°⎟⎠⎞

⎜⎝⎛

=

=

+−=

t

gv

t

vgtθ

θ

The maximum altitude is the value of y at 40.77 s. From Equation (b)

( )

(g) m 1015.8

s) 77.40)(30sin(sm 800s 77.40

sm 81.9

21)77.40(

3

22

x

y

=

°⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛−=

2.25 A free-body diagram of the projectile at an arbitrary instant shows the gravity force and the drag force.

(a) Summing forces in the x direction leads to

( )

(a) sm 10

kg 0.2N 2

2−=

=−

=∑

x

x

maF xx

&&

&&

Assuming the projectile is fired at the origin of the coordinate system the initial conditions for determining x(t) are

50

Chapter 2

(c) sm 6.459

)40cos(sm 600)0(

(b) 0)0(

=

°⎟⎠⎞

⎜⎝⎛=

=

x

x

&

Integrating Equation (a) with respect to time leads to (d) 10)( 1Cttx +−=&

Application of Equation (c) to Equation (d) leads to 6.4591 =C and (e) 6.45910 +−= tx&

Integration of Equation (e) with respect to time and application of Equation (b) leads to (f) 6.4595)( 2 tttx +−=

Summing forces in the y direction leads to

(g) 0==−

=∑

yymmg

maF yy

&&&&

The initial conditions which are applied to determine y(t) are

(i) sm 7.385

)40sin(sm 600)0(

(h) 0)0(

=

°⎟⎠⎞

⎜⎝⎛=

=

y

y

&

Integration of Equation (g) with application of Equation (i) leads to (j) 7.38581.9)( +−= tty&

Integration of Equation (j) with application of Equation (h) leads to (b) The maximum altitude occurs when which from Equation (j) occurs at 0=y& s 3.39=t . The maximum altitude is obtained from Equation (k) as

(l) m 1059.7 )3.31(7.385)3.31(9.4)3.39(

3

2

xy

=

+−=

The time at which the projectile lands is when y=0, which is obtained using Equation (k) as . The range of the projectile is obtained from Equation (f) as s 6.78=t

m 1024.5 6.459)6.78(5)6.78(

3

2

xx

=

+−=

(k) 7.3859.4)( 2 ttty +−=

2.26 A free-body diagram of the disk, drawn at an arbitrary instant, is shown below

51

Chapter 2

Application of Newton’s second law to the free-body diagram gives (a) xmxckxF &&&=−−−

Noting that the centroidal mass moment of inertia of the thin disk is 2

21 mrI = and that

since the disk rolls without slip its angular acceleration is related to the acceleration of the mass center of the disk by αrx =&& , summation of moments about the mass center of the disk leads to

(b) 21)( 2 ⎟

⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛=+

=∑

rxmrFrtM

IM G

&&

α

Elimination of F between Equations (a) and (b) leads to

(c) )(23 tMkrxxcrxmr =++ &&&

2.27 The free-body diagram of the system at an arbitrary instant is shown below.

Application of Newton’s second law in the vertical direction to the free-body diagram leads to

(a) )()( xmxycxykkx

maF&&&& =−+−+−

=∑

Rearrangement of Equation (a) leads to (b) 2 kyyckxxcxm +=++ &&&&

52

Chapter 2

2.28 Let y(t) represent the displacement of the center of the disk, measured downward from the system’s equilibrium position. Let θ represent the clockwise angular rotation of the disk, measured from the system’s equilibrium position

When the center of the disk has moved distance of y downward, the cable has decreased in length by 2y. When the block has moved downward a distance x, the cable has increased in length by x. Since the cable is inextensible,

(a) 2

02xy

yx

=

=−

It is further assumed that there is no slip between the disk and the cable, thus

(b) 2

2

rx

rxx

ryx

=

=−

=−

θ

θ

θ

Free-body diagrams of the disk and the block, drawn at an arbitrary instant are shown below.

53

Chapter 2

Application of Newton’s second law to the free-body diagram of the block leads to

(c)

2

2

kxxmTxmkxT

maF

−−==−−

=∑

&&&&

Application of Newton’s second law to the free-body diagram of the disk leads to (d) 21 ymTTky &&=++−

Application of the moment equation to the free-body diagram of the disk gives

(e) 12 θ

α&&IrTrT

IM G

=−

=∑

Substitution of Equations (b) and (c) into Equation (e) leads to

( )

(f) 2

2

21

1

kxxrImT

rxIrTrkxxm

−⎟⎠⎞

⎜⎝⎛ +−=

⎟⎠⎞

⎜⎝⎛=−−−

&&

&&&&

Substitution of Equations (a),(c), and (f) into Equation (d) gives

(g) 025

225

222

2

2

=+⎟⎠⎞

⎜⎝⎛ +

⎟⎠⎞

⎜⎝⎛=−−−⎟

⎠⎞

⎜⎝⎛ +−⎟

⎠⎞

⎜⎝⎛−

kxxrIm

xmkxxmkxxrImxk

&&

&&&&&&

54

Chapter 2

2.29 A free-body diagram of the system at an arbitrary instant is shown below. Assuming small θ the system is linear and the static spring forces cancel with the gravity forces and neither are included on the free-body diagrams.

The bar rotates about a fixed axis through O, thus the moment equation may be applied as ∑ = (a) αOO IM

with clockwise moments taken as positive. Application of Equation (a) leads to

(b) 019

)()3)(3(22 =+

=−−

θθ

θθθ

λ&&

&&λλλλ

kI

Ikk

O

O

The mass moment of inertia of abut O is the sum of the moments of inertia of each of the particles. The moment of inertia of a particle about an axis through the particle is zero. The moment of inertia of each particle about O is then obtained using the parallel axis theorem. The moment of inertia of the particle of mass 2m about O is

(c) 18)3(2 221 λλ mmI P ==

The moment of inertia of the particle of mass m about an axis through O is (d) 2

2 λmI p = Thus

(e) 19 2

1

λm

III pwpO

=

+=

Substitution of Equation (e) into Equation (b) leads to

(f) 0

01919 22

=+

=+

θθ

θθ

km

km&&

λ&&λ

2.30 A free-body diagram of the system at an arbitrary instant is shown below. Assuming small θ the system is linear and the static spring forces cancel with the gravity forces and neither are included on the free-body diagrams.

55

Chapter 2

The bar rotates about a fixed axis through O, thus the moment equation may be applied as

∑ = (a) αOO IM with clockwise moments taken as positive. Application of Equation (a) leads to

(b) 019

)()3)(3(22 =+

=−−

θθ

θθθ

λ&&

&&λλλλ

kI

Ikk

O

O

The moment of inertia of the assembly about O is the sum of the moments of inertia of the two particles and the moment of inertia of the bar,

(c) 21 OpOpObO IIII ++= The moment of inertia of the bar about its cenroidal axis is obtained from Table 2.1 as

(d) 34

)4(121

2

2

λ

λ

m

mIb

=

=

The parallel axis theorem is used to determine the moment of inertia of the bar about O,

(e) 37

34

2

22

2

λ

λλ

m

mm

mdII bOb

=

+=

+=

The moment of inertia of a particle about an axis through the particle is zero. The moment of inertia of each particle about O is then obtained using the parallel axis theorem. The moment of inertia of the particle of mass 2m about O is

(f) 18)3(2 221 λλ mmIOP ==

The moment of inertia of the particle of mass m about an axis through O is (g) 2

2 λmIOp = Substitution of Equations (e), (f), and (g) into Equation (c) gives

(h) 3

64

1837

2

222

λ

λλλ

m

mmmIO

=

++=

Use of Equation (h) in Equation (b) leads to

(i) 0193

64 22 =+ θθ λ&&λ km

56

Chapter 2

2.31 For the system shown in Figure P2.31, let θ be the counterclockwise angular rotation of the disk, measured from the system’s equilibrium position, and let y be the displacement of the particle on the outer radius of the disk. Kinematics is used to give

(b) 333

(a)

xrxrry

rxrx

=⎟⎠⎞

⎜⎝⎛==

=⇒=

θ

θθ

Free-body diagrams of the disk and the block, drawn at an arbitrary instant are shown below

Application of Newton’s second law to the free-body diagram of the block leads to

(c) 2 xmkxT

maF&&=−−

=∑

Application of the moment equation, assuming moments are positive counterclockwise, to the free-body diagram of the disk gives

(d) )3(3 ⎟⎠⎞

⎜⎝⎛=−

=∑

rxIrkxTr

IM G

&&

α

Elimination of T between Equations (c) and (d) leads to

(e) 011 =+⎟⎠⎞

⎜⎝⎛ + krxx

rImr &&

2.32 Define θ as the clockwise angular rotation of the upper bar measured from the system’s equilibrium position and define φ as the counterclockwise rotation of the lower bar, measured from its equilibrium position. The link connecting the two bars is rigid.

57

Chapter 2

Thus the displacements of the particles on each bar where the link is attached must be the same. Assuming small displacements this leads to

(a) 43

32

2

θφ

φθ

=

=LL

Free-body diagrams of each bar, drawn at an arbitrary instant, are shown below.

Noting that the centroidal mass moment of inertia of a slender bar is 2

121 mLI = ,

application of the moment equation, assuming positive moments clockwise, to the upper bar gives

(b) 121

2:

222θθ

α

&&mLLFLLk

IM A

=⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛−

=∑

The parallel-axis theorem is used to determine the moment of inertia of the lower bar

about its pin support at B as 22

2

31

2121 mLLmmLI B =⎟

⎠⎞

⎜⎝⎛+= . Application of the moment

equation to the lower bar, assuming positive moments are clockwise, gives

( ) (c) 43

31

32

43 2 ⎟

⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛−

=∑θθ

α

&&& mLLFLcL

IM BB

Equation (a) is used to solve for the force in the rigid link as

(d) 21

61 2 θθ kLmLF += &&

Substitution of Equation (c) into Equation (d) leads to

(e) 43

41

32

21

61 222 θθθθ &&&&& cLmLLkLmL +=⎟

⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛ +−

Equation (e) is rearranged to give

58

Chapter 2

(f) 031

43

3613 222 =++ θθθ kLcLmL &&&

2.33 The free-body diagram of the bar at an arbitrary instant follows

Since the pin support is a fixed axis of rotation the appropriate form of the moment equation is

(a) ∑ = αOO IM where moments are taken positive counterclockwise to be consistent with the chosen positive direction for θ . Application of Equation (a) to the free-body diagram, assuming small θ , leads to

(b) 2

θθθ &&Ot ILmgk =−−

The moment of inertia of a slender bar about its centroidal axis is obtained from Table 2.1 as

(c) 121 2mLI =

Application of the parallel axis theorem leads to

(d) 31

2121

2

22

mL

LmmLIO

=

⎟⎠⎞

⎜⎝⎛+=

Use of Equation (d) in Equation (b) and subsequent rearrangement leads to

(e) 023

1 2 =⎟⎠⎞

⎜⎝⎛ ++ θθ LmgkmL t

&&

2.34 Free-body diagrams of the blocks, drawn at an arbitrary instant, are illustrated below

59

Chapter 2

Application of Newton’s second law to the left block gives

(a) 30)(10002003000 11211 xxxxx

maF&&& =−+−−

=∑

Application of Newton’s second law to the right block gives (b) 204000)(1000 2212 xxxx &&=−−−

Equations (a) and (b) are rearranged and written in a matrix form as

(c) 00

5000100010004000

200030

2

1

2

1⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡xx

xx&&&&

2.35 Free-body diagrams of the bar and the block, drawn at an arbitrary instant are shown below assuming small θ . Static spring forces and gravity cancel are not included as they will cancel with each other in the governing equations.

Summing forces on the free-body diagram of the block assuming positive forces act downward leads to

(a) 0)(

=−+=−−

=∑

θθkLkxxm

xmLxkmaF

&&&&

Since the bar rotates about a fixed axis through the pin support the appropriate moment equation to be applied to the free-body diagram of the bar is

60

Chapter 2

∑ = (b) αOO IM The moment of inertia of the slender bar about its own centroidal axis is obtained from

Table 2.1 as 2

121 mLI = . Application of the parallel axis theorem gives

(c) 31

2121

2

22

mL

LmmLIO

=

⎟⎠⎞

⎜⎝⎛+=

Application of Equation (b) to the free-body diagram of the bar, taking clockwise moments as positive, leads to

(d) 0231

31))(()(

22

2

=+−

=−+−

θθ

θθθ

kLkLxmL

mLLLxkLkL

&&

&&

Equations (a) and (d) provide the mathematical model for the system.

2.36 Free-body diagrams of each of the blocks at an arbitrary instant are illustrated.

Application of Newton’s second law to each of the blocks, assuming forces are positive to the right, leads to

(c) )()()(2(b) 2)()(2)((a) )(

32323

2232312

1121

xmtFxxcxxkxmxxcxxkxxk

xmxxkkx

&&&&&&&&

&&

=+−−−−=−+−+−−

=−+− Equations (a), (b), and (c) are rearranged as

(d) 02 211 =−+ kxkxxm&& (e) 022 321322 =−+−−+ kxkxkxxcxcxm &&&& (f) )(32323 tFkxkxxcxcxm =+−+− &&&&

Equations (d), (e), and (f) are summarized in matrix form as

00

00020

3

2

1

3

2

3

2

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣ −−−+

⎥⎥⎥

⎦⎢⎢⎢

⎣⎥⎥⎥

⎦⎢⎢⎢

⎣ −−+

⎥⎥⎥

⎦⎢⎢⎢

⎣⎥⎥⎥

⎦⎢⎢⎢

⎣ tFxxx

kkkkk

xx

cccc

xx

mm

&&

&&&&

2.37 Free-body diagrams of the link and the block are illustrated assuming small x.

(g) )(

00

02

0200000 11 ⎡ −⎤⎡⎤⎡⎤⎡⎤⎡ kkxxm &&&

61

Chapter 2

Since the link is massless, application of the moment equation to its free-body diagram leads to

( )

(a) )(31)(

03)()(

0

tFtG

tGtF

M O

−=

=+

=∑λλ

Application of Newton’s second law to the free-body diagram of the block gives

(b) )(31

)(

tFkxxcxm

xmxckxtG

−=++

=−−

&&&

&&&

2.38 Assuming small displacements and using similar triangles, as illustrated below, the displacement of the end of the spring attached to the bar of length is such that the spring is stretched.

λ3 )(2 ty

A free-body diagram of the bar at an arbitrary instant, assuming small θ , is illustrated. Note that the since θ is assumed positive clockwise rotation of the bar in the positive direction of θ leads to an increase in length of the spring. The bar rotates about a fixed axis at O, thus the appropriate form of the moment equation is

(a) αOO IM =∑ The moment of inertia of the slender bar about its own centroidal axis is obtained from

Table 2.1 as 2

121

λmI = . Application of the parallel axis theorem gives

62

Chapter 2

(b) 31

2121

2

22

λ

λλ

m

mmIO

=

⎟⎠⎞

⎜⎝⎛+=

Application of Equation (a) to the free-body diagram of the bar, taking clockwise moments as positive, leads to

( )

(c) )(223

131

2)(2

2222

2

tykmgkcm

mmgcyk

λλ

λ&λ&&λ

&&λλ

λ&λλλ

−=⎟⎠⎞

⎜⎝⎛ −++

=+−+−

θθθθ

θθθθ

2.39 Define )(tθ as the clockwise angular displacement of the crank from the aligned position, as illustrated below.

Define as the distance between the piston and the center of the crank at any instant. From the above diagram

)(tx p

(a) cos1.0sin3.0 θθ +=px The velocity of the piston is calculated as

(b) sin1.0cos3.0 θθθθ &&& −=px

Noting that , which is a constant, differentiation of Equation (b) leads to ωθ =&

(c) cos1.0sin3.0 22 θθθθ &&&& −−=px (a) Since the angular velocity is constant and the system is aligned at t=0,

(d) tωθ = Use of Equation (d) in Equations (a), (b), and (c) leads to

(g) )cos(1.0)sin(3.0

(f) )sin(1.0)cos(3.0

(e) )cos(1.0)sin(3.0

22 tta

ttv

ttx

p

p

p

ωωωω

ωωωω

ωω

−−=

−=

+=

(b)Assume the inertia effects of the crank and connecting rod are negligible. A free-body diagram of the engine, drawn at an arbitrary instant, is illustrated below

63

Chapter 2

Application of Newton’s second law, in the form of D’Alembert’s principle to the free-body diagram leads to

(h) 3.0900010003.10

)(3.01010009000

p

p

effext

xxxx

xxxxx

FF

&&&&&

&&&&&&&

=++

++=−−

= ∑∑

Use of Equation (g) in Equation (h) leads to ( ) (i) )cos(1.0)sin(3.03.0900010003.10 22 ttxxx ωωωω −−=++ &&&

Noting that srrpm /2.26250 ==ω , Equation (i) becomes (j) )cos(6.20)sin(8.61900010003.10 ttxxx ωω −−=++ &&&

64

Chapter 3

3. Electric Systems

3.1(a) The capacitance for a parallel plate capacitor is

(a) 0 dAKC ε=

Equation (a) is rearranged to solve for the separation distance as

(b) 0

CAK

dε

=

Substitution of given numerical values into Equation (b) gives

( )

(c) mm 410.0 F 105

m 5.0mN

C 1085.8)6.4(

8

22

212

=

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

= −

−

x

xd

(b) The charge on this parallel-plate capacitor when connected to a 12 V battery is

( )(d) C 100.6

V) 12(F 105 7

8

−

−

=

=

=

xx

Cvq

(c) The energy stored in the capacitor after being charged by a 12 V battery is

( )( )

(e) J 60.3

V 12F 10521

21

28

2

µ=

=

=

−x

CvE

3.2 The power delivered by the battery is

( )( )(a) W 30

A 5.1V 20 === viP

The energy stored in the battery is

( )( )

(b) J 1024.3 hr

s 3600hr 3 W30

5x

PtE

=

⎟⎠⎞

⎜⎝⎛=

=

3.3 (a) The power in the CD player is given by (a) viP =

Thus for a 200 W CD player run from a 120 V source,

65

Chapter 3

(b) A 67.1 V 120

W200

=

=

=vPi

(b) The resistance in the player is

(c) 72 A 1.67V 120

Ω=

=

=ivR

3.4 The power delivered by the heating coil is

(a) W 7.416 s 3600

hr 1hrJ101500 3

=

⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛=

=

x

tEP

The power is related to the resistance of the coil through

(b) 2

RvP =

The resistance is calculated as

( )

(c) 6.34 W7.416

V 120 2

2

Ω=

=

=PvR

3.5 The current in a series LRC circuit is (a) A )4.0100sin(10)( += tti

(a)The voltage across a 100 Ω resistor is [ ]( )

(b) V )4.0100sin(101 100A )4.0100sin(10

3 +=

Ω+==

txtiRvR

(b)The voltage change across the capacitor, assuming it is uncharged at t=0, is

66

Chapter 3

( )[ ][ ] (c) V )4.0100cos(921.0105

)4.0100cos()4.0cos(105

)4.0100cos()102.0(100

10

)4.0100sin(10102.01

1

5

5

06

06

0

+−=

+−=

+−=

+=

=

−

− ∫

∫

txtx

tx

dttx

idtC

v

t

t

t

C

(c)The voltage across the inductor is

( ) [ ]

(d) V 0.4)0t-250cos(10 0.4)00t(100)cos(1(0.25)(10)

)4.0100sin(100.25

+=+−=

+=

=

tdtd

dtdiLvL

(d) Application of KVL around the series LRC circuit shows that the voltage provided by the source is the sum of the voltage changes across the resistor, capacitor, and inductor. Thus from Equations (b), (c), and (d),

[ ][ ] (e) V )4.0100sin(101)4.0100cos(10003.51061.4

)4.0100cos(250)4.0100cos(921.0105)4.0100sin(101

)(

355

53

+++−=

+−+−++=

++=

txtxxttxtx

vvvtv LCR

3.6 (a) The energy stored in the capacitor is

(a) 21 2

cc CvE =

Substitution of Equation (c) from the solution of Problem 3.5 into Equation (a) leads to

( ) [ ]( )[ ]

(b) )]8.0200cos(5.0)4.0100cos(842.13482.1[105.2 )4.0100(cos)4.0100cos(842.18482.0105.2

)4.0100cos(921.0105 102.021

4

24

256

mNttxttx

VtxFxEc

⋅+++−=

+++−−=

+−= −

Equation (b) is plotted below

67

Chapter 3

(b) The energy stored in the inductor is

(c) 21 2LiEL =

Substituting given values into Equation (c) leads to

( )[ ]

(d) mN )4.0100(sin5.12

A 4.0100sin(10H 25.021

2

2

⋅+=

+=

t

tEL

Equation (d) is plotted below

68

Chapter 3

(c) The power dissipated by the resistor is

[ ][ ] (e) W )8.0200cos(1105

A )4.010sin(100) 100( 3

2

2

+−=

+Ω=

=

txt

RiP

The energy dissipated is

[ ]

[ ] (f) mN )8.0sin()8.0200sin(05.0105

)8.0200sin(2001105

)8.0200cos(1105

30

3

0

3

0

⋅++−=

⎟⎠⎞

⎜⎝⎛ +−=

+−=

=

∫

∫

ttx

ttx

dttx

PdtE

t

t

t

d

Equation (f) is plotted below.

69

Chapter 3

3.7 Currents are defined as in the diagram below.

Application of KCL at node B leads to

(a) 0321 =−− iii Application of KCL at node C gives

(b) 0543 =−− iii KVL is applied around each loop in a clockwise direction. For loop ABGH

(c) 01253 31 =+−− ii For loop BCFG

70

Chapter 3

(d) 0542 352 =+−− iii For loop CDEF

(e) 0443 544 =+−− iii Equations (a)-(e) are summarized in a matrix form as

(f)

00

1200

4700040520

0050311100

00111

5

4

3

2

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−

−−−−

iiiii

The equations represented by Equation (f) are solved simultaneously leading to (g) A 654.0 A, 373.0 A, 03.1 A, 26.1 A, 29.2 54321 ===== iiiii

(b) The voltage drops across the resistors are determined using Ohm’s law. The subscripts correspond to the labels in the figure

(h) V 49.14 V, 12.13V, 62.24 V, 52.22 V, 15.55 V, 87.63

4645

5423321

============

iviviviviviv i

(c) The power in each resistor is calculated as

(i) W 557.04 W,417.03

W,71.14 W,18.32 W,30.55P W,73.153246

245

254

223

232

211

====

========

iPiP

iPiPiiP

3.8 (a) The reduction of the circuit to a single resistor of equivalent resistance is illustrated below

71

Chapter 3

The reduction shows that the resistors may be replaced by a single resistor of equivalent resistance 1.3 Ω. (b) The power delivered to the circuit is

( )

(a) kW 111.0 3.1V 12

2

2

=Ω

=

=eqR

vP

3.9 The reduction of the resistive circuit is illustrated below.

The equivalent resistance of the combination of resistors is

(a) 83.6

23

157

Ω=

Ω=eqR

3.10 The circuit may be reduced as illustrated below

72

Chapter 3

Application of KCL at node B leads to (a) 0321 =−− iii

Application of KVL around loop ABEF gives

(b) 098

12710 31 =−− ii

Application pf KVL around loop BCDE leads to (c) 0921 32 =+− ii

Equations (a)-(c) are summarized in matrix form as

(d) 0

100

9210

908

127111

3

2

1

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−−

iii

The solution of Equation (d) is

(e) A 316.0A 135.0A 415.0

3

2

1

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

iii

The positive sign for indicates the current in loop BCDE flows in the direction assumed. Thus the current is flowing in the direction from the + to - terminals placed across the resistor, indicating a voltage decrease of

2i

(f) V 49.1 ) A)(11 135.0(

=Ω=v

3.11 From KVL the voltage change across each pair of series resistors is . Let be the current through and . Then

1v 1i

1R 2R

(a)

)(

21

11

2111

RRv

i

RRiv

+=

+=

1v . Let be the current through and . Then 2i 3R 4R

(b)

)(

43

12

4321

RRv

i

RRiv

+=

+=

73

Chapter 3

Define , , and as in the above diagram. Then Av Bv Cv

(e) (d) (c)

0

32

11

BC

AC

AB

vvvRivvRivv

−=−=−=

Substitution of Equations (c) and (d) into Equation (e) leads to

(f) 3211

11320

RiRiRivRivv AA

−=+−−=

(a) Use of Equations (a) and (b) in Equation (f) yields

(g) ))((

14321

3241

43

31

21

110

vRRRR

RRRR

RRRv

RRRvv

⎥⎦

⎤⎢⎣

⎡++

−=

+−

+=

(b) The output voltage is zero if (h) 3241 RRRR =

(c) The resistance in a wire is proportional to its length λ, λρ=R , where ρ is the resistivity of the wire. The strain of a wire initially of length aλ which has changed to length is bλ

(i) a

ab

a

ab

RRR −

=

−=

λλλ

ε

Suppose the strain gauge is calibrated such that 00 =v when the wire is unstretched

⎟⎟⎠

⎞⎜⎜⎝

⎛==

1

324 R

RRRRa . Then, when stretched the measured voltage is given by Equation (g)

with . bRR =4

3.12 Let be the current through the resistors. Since the resistors are in series Ohm’s law gives

i

(a) )( 211 RRiv += Equation (a) is rearranged to

(b) 21

1

RRvi+

=

74

Chapter 3

Using the definitions of voltages in the above diagram,

(e)

(d) (c)

21

12

2

2

1

iRvvviRv

vvviRvvvvv

AA

CB

AB

CA

−=+−−=

−=−=+=

Substitution of Equation (b) in Equation (e) leads to

(f) 121

1

21

2112

vRR

R

RRRv

vv

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

+−=

3.13 The relation between the voltage across a capacitor and the charge on the capacitor is

(a) Cvq = Application of conservation of charge at node B leads to

(b) 0321 =−− qqq Application of KVL around loop ABEF gives

(c) 06102105103 6

36

16

1 =+−−− −−− xq

xq

xq

Application of KVL around loop BCDE results in

(d) 0102104 6

36

2 =+− −− xq

xq

Equations (b)-(d) are summarized in matrix form as

75

Chapter 3

(e) 01060

21

410

210

158

1116

3

2

1

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−−−x

qqq

The solution of Equation (e) is

(f) C 0.189C 0.081C 271.0

10 6

3

2

1

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

qqq

The voltage across each capacitor is calculated using Equation (a). The voltage across the 3 µF capacitor is

(g) V 0902.0 F 3x10

C 0.271x10 6-

6-1

11

=

=

=Cqv

The voltage across the 5 µF capacitor is

(h) V 0541.0 F 5x10

C 0.271x10 6-

6-21

12

=

=

=Cqv

The voltage across the 4 µF capacitor is

(i) V 0203.0 F 4x10

C 0.081x10 6-

6-

31

23

=

=

=Cq

v

The voltage across the 2 µF capacitor is

(j) V 0947.0 F 2x10

C 0.189x10 6-

6-1

34

=

=

=Cq

v

76

Chapter 3

3.14 The reduction of the capacitive circuit is illustrated below

Thus the equivalent capacitance of the combination is

(a) F 52.1

F 79

120

µ

µ

=

=eqC

3.15 Define as the voltage drop across the resistor and as the voltage drop across the capacitor.

Rv Cv

77

Chapter 3

Application of KVL around loop BCDE leads to (b) 0=+− RC vv

Application of KCL at node be results in (c) 0)( 32 =−− iiti

The voltage drop across the resistor is calculated using Ohm’s law as

(d) 3

3

Rv

i

Riv

R

R

=

=

The voltage drop across the capacitor is obtained as

(e)

1

2

02

dtdv

C

dtdv

Ci

dtiC

v

R

C

t

C

=

=

= ∫

Substitution of Equations (d) and (e) in Equation (b) leads to

(f) )(

0)(

tiRv

dtdv

C

Rv

dtdv

Cti

RR

RR

=+

=−−

3.16 Application of KVL in a clockwise direction around the circuit leads to

(a) 0)(21 =+−−− tviRiRdtdiL

Equation (a) is rearranged to

(b) )()( 21 tviRRdtdiL =++

3.17 Let be the current in the circuit. Application of KVL over the loop, including the effect of mutual inductance, results in

)(ti

(a) 021 =−⎟⎠⎞

⎜⎝⎛ +−⎟

⎠⎞

⎜⎝⎛ +− Ri

dtdiM

dtdiL

dtdiM

dtdiLv

Equation (a) is rearranged as

( ) (b) 221 vRidtdiMLL =+++

78

Chapter 3

3.18 (a) When the switch is open the circuit is the resistive circuit shown below.

Application of KCL at node B gives (a) 0321 =−− iii

Application of KVL around loop ABEF leads to (b) 04610 31 =−− ii

Application of KVL around loop BCDE leads to

(c) 045 32 =+− ii Equations (a), (b), and (c) are summarized in matrix form as

(d) 0

100

540406111

3

2

1

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−−

iii

The solution of Equation (d) is

(e) A 676.0A 541.0

A 22.1

3

2

1

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

iii

(b) When the switch is closed the capacitor is connected as illustrated below

Application of KCL at node B and KVL around loops ABEF BCDE leads to

79

Chapter 3

(g) 0)(410515

(f) 0)(4610

210

262

211

=−+−−

=−−−

∫− iidtix

i

iiit

Equations (f) and (g) are rewritten as

(i) 010294

(h) 10410

02

521

21

=++−

=−

∫t

dtixii

ii

3.19 Currents through the circuit are defined below.

Application of KCL at node B gives

(a) 0321 =−− iii Application of KVL around loop ABEF leads to

(b) 0)(48 31 =+−− tvii Application of KVL around loop BCDE gives

(c) 04003.0 32 =+− i

dtdi

Equations (b) is rearranged to give

(d) 2)(41

3 iitvi −=

Use of Equation (d) in Equation (a) leads to

(e) )(121

31

02)(41

21

121

tvii

itvii

+=

⎥⎦⎤

⎢⎣⎡ =−−−

Equations (d) and (e) are used to give

(f) 32)(

121

23 itvi −=

Use of Equations (e) and (f) in Equation (c) leads to

(g) )(31

38003.0

032)(

1214003.0

22

22

tvidtdi

itvdtdi

=+

=⎥⎦⎤

⎢⎣⎡ −+−

80

Chapter 3

3.20 The circuit diagram is shown below.

Application of KVL around loop BCDE leads to

(a) 0)( 11

1 =−+−− iiRdtdiLRi

Equation (a) is rearranged as

(b) 21 RiRidtdiL i =+

3.21 The currents are defined in the circuit diagram shown below

Application of KCL at nodes B and C gives

(b) 0(a) 0

532

421

=−−=−−

iiiiii

Application of KVL around loop ABGH results in

(c) 0411

11 =−− iRdtdiLv

Application of KVL around loop BCFG gives

(d) 041522

2 =+−− iRiRdtdiL

Application of KVL around loop CDEF leads to

81

Chapter 3

(e) 01522

03 =++− ∫ iRvdti

C

t

Substituting Equations (a) and (b) into Equations (c), (d), and (e) and rearranging leads to

(h) 1

(g) 0)(

(f)

20

33222

32221112

2

121111

1

vdtiC

iRiR

iRiRRiRdtdi

L

viRiRdtdiL

t

=++−

=−++−

=−+

∫

3.22 Let i(t) be the current in the circuit after the switch is closed. Noting that the initial

voltage in the initially charged capacitor is 1

=)0(CQ

v application of KVL around the

circuit leads to

(a) 011

10102

=+−−− ∫∫ CQidt

Cidt

CRi

tt

The appropriate initial condition is . 0=)0(i

3.23 The currents are defined in the diagram below.

Application of KCL at node B gives

(a) 0321 =−− iii

Application of KVL around loop ABDE leads to

(b) 03111 =+−− v

dtdiLiR

Application of KVL around loop BCDE gives

(c) 031

2222 =+−−

dtdi

Ldtdi

LiR

Equation (a) is rearranged as (d) 213 iii −=

Use of Equation (d) in Equations (b) and (c) along with rearrangement gives

82

Chapter 3

3.24 The currents are defined in the circuit diagram shown below

( ) (f) 0

(e)

221

12

21

112

11

1

=+−+

=+−

iRdtdi

Ldtdi

LL

viRdtdi

Ldtdi

L

Application of KCL at node B leads to

(a) 0)( 21 =−− iiti Application of KVL around loop ABEF leads to

(b) 0102.013000

026 =−− ∫−

t

dtix

i

Application of KVL around loop BCDE results in

(c) 102.0160004.0

0261

1 ∫−+−−t

dtix

idtdi

3.25 Noting, from application of KCL, that the current through the inductor is , application of KVL around each loop leads to

21 ii −

(b) )(1

(a) )(

21220

2

2111

iidtdLiRdti

C

viidtdLiR

t

−+−−

=−−−

∫

Equations (a) and (b) are rearranged as

(d) 01

(c)

0222

21

1121

=+++−

=+−

∫t

dtiC

iRdtdi

Ldtdi

L

viRdtdiL

dtdiL

3.26 Currents are defined in the diagram below.

83

Chapter 3

Application of KCL at nodes B and C gives

(b) 0(a) 0

532

421

=−−=−−

iiiiii

Equations (a) and (b) are rearranged to give

(d) (c)

325

214

iiiiii

−=−=

Application of KVL around loop ABGH gives

(e) 0111

04

1

11 =+−−− ∫ viRdti

Cdtdi

Lt

Application of KVL around loop BCFG gives

(f) 011

04

1525

2

22 =+−−− ∫

t

dtiC

iRiCdt

diL

Application of KVL around loop CDEF gives

(g) 011

04

252

03

3

33 =++−− ∫∫

tt

dtiC

iRdtiCdt

diL

Substitution of Equations (c) and (d) into Equations (e)-(g) and rearranging leads to

(j) 0111

(i) 01111

(h) 11

03

3232

33

02

222

03

232

02

2122

22

01

1

02

101

111

11

=⎟⎟⎠

⎞⎜⎜⎝

⎛++++−−

=−−⎟⎟⎠

⎞⎜⎜⎝

⎛++++−

=−++

∫∫

∫∫∫

∫∫

tt

ttt

tt

dtiCC

iRdtdi

LdtiC

iR

dtiC

iRdtiCC

iRdtdi

LdtiC

vdtiC

dtiC

iRdtdi

L

3.27 Currents are defined in the diagram below

84

Chapter 3

Application of KCL at nodes B, C, and I leads to

(c) 0(b) 0(a) 0

765

532

421

=−−=−−=−−

iiiiiiiii

The resistor of resistance is in parallel with the capacitor of capacitance . Thus the potential change across each is the same

2R 2C

(d) 1

07

262 ∫=

t

dtiC

iR

Application of KVL around loops ABGJ, BCFG, and CDEF leads to

(g) 0

(f) 1

(e) 01

6223

3

04

14162

22

1410

41

11

=++−

−+−−

=+−−−

∫

∫

iRvdtdi

L

dtiC

iRiRdtdiL

viRdtiCdt

diL

t

t

Differentiation of Equation (d) with respect to time leads to

(h) 6227

726

2

dtdi

CRi

iCdtdi

R

=

=

Substitution of Equation (h) into Equation (c) gives

(i) 62265 dt

diCRii +=

Substitution of Equation (i) into Equation (b) gives

(j) 0622632 =−−−

dtdi

CRiii

Use of Equation (a) in Equations (g) and (h) leads to

(l) 011

(k) 11

620

21

212

20

11

11

10

21

210

11

111

1

=++++−−

=−−++

∫∫

∫∫

iRdtiC

iRdtdi

LdtiC

iR

vdtiC

iRdtiC

iRdtdi

L

tt

tt

Equations (g), (h), (k), and (l) form the mathematical model of the system.

85

Chapter 3

3.28 Currents are defined in the diagram below

Application of KCL at nodes B, C, and I leads to

(c) 0(b) 0(a) 0

765

532

421

=−−=−−=−−

iiiiiiiii

The resistor of resistance is in parallel with the capacitor of capacitance . Thus the potential change across each is the same

2R 2C

(d) 1

07

262 ∫=

t

dtiC

iR

Application of KVL around loops ABGJ, BCFG, and CDEF, including the mutual inductance between adjacent inductors leads to

(g) 0

(f) 1

(e) 01

62223

3

04

14162

3122

1410

41

211

=++−−

−+−−−−

=+−−−−

∫

∫

iRvdtdiM

dtdi

L

dtiC

iRiRdtdi

MdtdiM

dtdiL

viRdtiCdt

diM

dtdi

L

t

t

Differentiation of Equation (d) with respect to time leads to

(h) 6227

726

2

dtdi

CRi

iCdtdi

R

=

=

Substitution of Equation (h) into Equation (c) gives

86

Chapter 3

(i) 62265 dt

diCRii +=

Substitution of Equation (i) into Equation (b) gives

(j) 0622632 =−−−

dtdi

CRiii

Use of Equation (a) in Equations (g) and (h) leads to

(l) 011

(k) 11

620

21

2132

20

11

111

10

21

210

21

111

11

=+++++−−

=−−+++

∫∫

∫∫

iRdtiC

iRdtdiM

dtdiLdti

CiR

dtdiM

vdtiC

iRdtdiMdti

CiR

dtdiL

tt

tt

Equations (g), (h), (k), and (l) form the mathematical model of the system. 3.29 The analogous mechanical system is illustrated below

87

Chapter 3

3.30 The analogous mechanical system is illustrated below

3.31 The analogous mechanical system is illustrated below.

88

Chapter 3

3.32 The analogous mechanical system is illustrated below.

89

Chapter 3

3.33 The electric circuit analogy for the mechanical system is illustrated below.

3.34 The electric circuit analogy for the mechanical system is illustrated below

3.35 The circuit is the non-inverting circuit of Example 3.17(b).Use of Equation (g) of Example 3.17 leads to

(a) 3 2 41

1

1

1

11

22

v

v

vRRv

=

⎟⎠⎞

⎜⎝⎛

ΩΩ

+=

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

3.36 Assume the amplifier is an ideal amplifier. Thus referring to the diagram below,

90

Chapter 3

(a) 073 == ii

Applications of KCL at nodes A, B, and C give respectively

(d) 0(c) 0(b) 0

654

42

321

=−−=−

=−−

iiiii

iii

Application of Ohm’s law across the circuit components leads to

(h) 300

(g) 600

(f) 800

(e) 400

5

4

2

1

Ω−

=

Ω−

=

Ω−

=

Ω−=

CF

CB

BA

A

vvi

vvi

vvi

vi

Application of KVL around loop EFCD leads to (i) 0300100 156 =++− vii

Equation (h) is used in Equation (i), which is then rearranged to

(j) 100100

300100300

1001

16

CF

CF

vvv

vvvi

−+=

−+=

Substitution of Equations (e)-hi) and Equation (j) in Equations (b)-(d) leads to

(m) 0100100300600

(l) 0600800

(k) 0800400

1 =−

−−−

−−

=−

−−

=−

−−

CFCFCB

CBBA

BAA

vvvvvvv

vvvv

vvv

The amplifier is ideal, thus the potentials at each of its terminals is the same,

91

Chapter 3

(n) AF vv = Using Equation (n), Equations (k)-(l) are rearranged as

(q) 610(p) 0473(o) 03

1vvvvvv

vv

CB

CBA

BA

=−=+−

=+−

The solution of Equations (o)-(q) is

(t) 149

(s) 73

(r) 71

1

1

1

vv

vv

vv

C

B

A

−=

−=

−=

The output is 2v

(u) 143 1

2

v

vvv CB

=

−=

3.37 Currents are as defined below.

Assuming the amplifier is ideal,

(a) 043 == ii Since there is no current through the 400 Ω resistor, the voltage input to the amplifier is

which is also the voltage at its negative terminal. Application of KCL at A gives 1v(b) 021 =− ii

Use of Ohm’s law across the resistors gives

(d) 1800

(c) 600

32

11

Ω−

=

Ω−

=

vvi

vvi

A

A

Substitution of Equations (c) and (d) into Equation (b) and noting that leads to 2vvA =

92(e) 4

033

01800600

123

3221

3221

vvvvvvv

vvvv

−==+−−

=−

−−

Chapter 3

3.38 Currents are defined in the diagram below.

The first amplifier is an inverting amplifier, thus from Equation (c) of Example 3.17

(a) -2 300 600

1

1

11

2

v

v

vRR

vB

=ΩΩ

−=

−=

Application of KCL at node C assuming ideal amplifiers leads to (b) 021 =− ii

Applications of Ohm’s law across circuit components lead to

(d) 500

500

(c) 200

40002

400

2

22

1

1

1

v

vvi

v

v

vvi

C

CB

−=

−=

−=

−−=

−=

Substitution of Equations (c) and (d) in Equation (b) leads to

(e) 25

0500200

12

21

vv

vv

=

=+−

3.39 The circuit is that of the integrating circuit of Figure 3.28 whose output is given in Equation (3.66)

93

Chapter 3

( )( ) [ ]

[ ] (a) V 1)200cos(1017.4

)200sin(1033.8

V )200sin(100F 102 600

1

1

3

0

5

07

012

−=

−=

Ω−=

−=

∫

∫

∫

−

tx

dttx

dttx

dtvRC

v

t

t

t

3.40 Currents and potentials are defined below.

Since the amplifier is ideal

(b) (a) 0

2

4

vvi

A ==

Application of KCL at A gives (c) 0231 =−+ iii

Use of component laws leads to

( )

( ) (f) 1030

(e) 1050

(d) 16

63

26

2

11

A

A

A

vdtdxi

vvdtdxi

vvi

−=

−=

−=

−

−

Use of Equations (b), and (d)-(f) in Equation (c) results in

(g) 108.4

0103016

1224

2621

vvdt

dvx

dtdv

xvv

=+

=−−

−

−

94

Chapter 3

The output is the solution of the differential equation of Equation (g). 2v 3.41 The currents and potentials are as defined below.

Since the amplifier is ideal, and 03 =i 0=Av . Application of KCL at node A gives

(a) 021 =− ii Application of the component law for a capacitor gives

(b) 11 dtdv

Ci i=

Application of KCL at B gives (c) 0542 =−− iii

Noting that the parallel circuit components have the same voltage change, 22 vvvA −=− , application of component laws give

(e)

(d)

25

4

dtdv

Ci

Rv

i

o

o

−=

−=

Substitution of Equations (d) and (e) in Equation (c) leads to

(f) 22 dtdv

CRv

i oo −−=

Substitution of Equations (b) and (f) into Equation (a) leads to

(g)

0

12

21

dtdv

RCvdt

dvRC

dtdv

CRv

dtdv

C

io

o

ooi

−=+

=++

The output is the solution of Equation (g). )(tvo

95

Chapter 3

3.42 Currents and potentials are defined in the diagram below

Since the amplifier is ideal, and 04 =i oB vv = . Applications of KCL at nodes A and B give

(b) 0(a) 0

53

321

=+=−−

iiiii

Applications of component laws lead to

( )

(f)

(e)

(d)

(c)

15

3

22

1

dtdv

Ci

Rvv

i

vvdtdCi

Rvv

i

o

oA

oA

Ai

−=

−=

−=

−=

Substitution of Equations (c)-(f) in Equations (a) and (b) leads to

( )

(h) 0

(g) 0

1

2

=−−

=−

−−−−

dtdv

CR

vvR

vvvv

dtdC

Rvv

ooA

oAoA

Ai

Equation (h) is solved for resulting in Av

(i) 01 vdt

dvRCv o

A +=

Substitution of Equation (i) in Equation (g) gives

(j) 112

0

12

2

21

12

2

211

iooo

ooooi

vR

vRdt

dvC

dtvd

CRC

dtdv

Cdt

vdCRC

Rv

dtdv

CRv

=++

=−−−−

The output is the solution of Equation (j). )(tvo

96

Chapter 3

3.43 Application of KCL at node A

leads to

(a) 0321 =−− iii where

( )

(d)

(c)

(b)

3

22

11

Rvvi

vvdtdCi

Rvvi

BA

oA

A

−=

−=

−=

Assuming an ideal amplifier, oB vv = and substitution of Equations (b)-(d) in Equation (a) leads to

(e) 0)( 0

21 =

−−−−

−R

vvvvdtdC

Rvv A

oAA

Application of KCL at node B gives

(f) 0543 =−− iii Since the amplifier is ideal, . Additionally 05 =i

(g) 14 dt

dvCi o−= Substitution of Equations (d) and (g) in Equation (f) gives

(h)

0

1

1

dtdv

RCvv

dtdv

CR

vv

ooA

ooA

+=

=−−

Substituting Equation (h) into Equation (e) leads to

(i) 2

0

112

2

21

12

2

1211

Rv

Rv

dtdv

Cdt

vdRCC

dtdv

Cdt

vdRCC

dtdv

CRv

Rv

ooo

oooo

=++

=−−−−

97

Chapter 3

3.44 The differential equation governing the motion of the system shown is (a) )(12102 tFxxx =++ &&&

The differential Equation derived in Example 3.30 for the amplifier circuit of Figure 3.47 is

(a) 1

10

64

50

2

232

02

321 Rvv

RRR

dtdv

RCR

dtvd

RCC =++ which is of the form of Equation (a). The circuit is designed to simulate Equation (a) if

(e) 12

(d) 10

(c) 2(b) 1

64

5

2

23

321

1

=

=

==

RRRRCR

RCCR

There are many solutions of Equations (b)-(e). One solution is (f) 1 , 12 , 1R , 1 , 1.0 , 1 F, 1 ,F 2 6543211 Ω=Ω=Ω=Ω=Ω=Ω=== RRRRRCC

3.45 Consider the operational amplifier circuit shown below

Application of KCL at A leads to

(a) 021 =− ii However

(c)

(b) )(1

12

01

11

Rvvi

dtvvL

i

BA

t

A

−=

−= ∫

Assuming the amplifier is ideal, 0=Av , then substitution of Equations (b) and (c) into Equation (a) leads to

98

Chapter 3

(d)

01

01

1

1

101

1

∫

∫

−=

=+

t

B

Bt

dtvLR

v

Rv

dtvL

Application of KCL at node C leads to (e) 0543 =−− iii

The currents are given by

(h) )(1

(g)

(f)

02

25

3

24

23

∫ −=

−=

−=

t

C

C

CB

dtvvL

i

Rvv

i

Rvv

i

Assuming the amplifier is ideal, 0=Cv and substitution of Equations (d), (f),(g), and (h) in Equation (e) leads to

(i)

01

01

21

13

02

2

32

02

23

2

01

21

1

∫∫

∫∫

=+

=++−

tt

tt

dtvRLRR

dtvLR

v

dtvLR

vdtv

RLR

Differentiation of Equation (i) with respect to time leads to

(j) 121

132

2

32 v

RLRR

vLR

v =+&

If Ω===Ω= 3 H, 1 H, 1 , 10 1213 RLLR and Ω= 102R , Equation (j) becomes (k) 310 122 vvv =+&

3.46 Consider the operational amplifier circuit shown below

Referring to the circuit diagram application of KCL at node leads to 1Q

(a) 0321 =−− iii Assuming an ideal amplifier, and 03 =i 11

vvQ = . Thus

99

Chapter 3

(b) 1

11 R

vi −=

( ) (c) 112

12 A

A vvdtdC

Rvvi −+

−=

Substitution of Equations (b) and (c) into Equation (a) leads to

( )

(d) 111

0

11

211

21

112

1

1

1

dtdv

CRR

vvRdt

dvC

vvdtdC

Rvv

Rv

AA

AA

+⎟⎟⎠

⎞⎜⎜⎝

⎛+=+

=−−−

−−

Application of KCL at node gives 2Q(e) 0654 =−− iii

Assuming an ideal amplifier, and 06 =i 02=Qv . Thus

(f) 3

4 Rvi A=

(g) 24

5 dtdvC

Rvi BB −

−=

Substitution of Equations (f) and (g) in Equation (e) leads to

(h)

0)(

4

323

42

3

BB

A

BB

A

vRR

dtdv

CRv

Rvv

dtdC

Rv

−−=

=++

The final amplifier is simply an inverter, such that

(i) 5

6BC v

RR

v −=

Substitution of Equation (i) in Equation (h) leads to

(j) 6

53

6

325C

CA v

RRR

dtdv

RRCR

v +=

Substitution of Equation (j) in Equation (d) leads to

(k) 11 111

2162

53

2

325

6

5312

2

6

5321

dtdvCv

RRv

RRRR

dtdv

RRCR

RRRC

dtvd

RRRCC

AAC +⎟⎟

⎠

⎞⎜⎜⎝

⎛+=+⎟⎟

⎠

⎞⎜⎜⎝

⎛++

Equation (e) simulates a differential equation of the form (l) 23410 21 vvvvv CCC +=++ &&&&

when the circuit components are chosen with values such that

100

Chapter 3

(q) 3C

(p) 211

(o) 4

(n) 10

(m) 1

1

21

62

53

2

325

6

531

6

5321

=

=+

=

=+

=

RR

RRRR

RRCR

RRRC

RRRCC

3.47 The differential equations for a field controlled DC motor are

(b) 0

(a)

, =−+

=+

ffeqteq

fff

f

iKcdtdJ

viRdtdi

L

ωω

The kinetic energy of the system at an arbitrary instant is

( ) ( ) ( ) (c) 21

21

21 2

3442232

2111 ωωω rGGGGr JJJJJJT +++++=

where the gear ratios are used to determine the angular velocities of the shafts in terms of the angular velocity of the motor shaft 1ω

(d) 12

12

2211

ωω

ωω

nnnn

=

=

(e) 42

312

4

33

3423

ωωω

ωω

⎟⎟⎠

⎞⎜⎜⎝

⎛==

=

nnnn

nn

nn

Substitution of Equations (d) and (e) into Equation (c) and rearranging leads to

( ) ( ) (f) 21 2

1

2

42

3144

2

2

13211 ω

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟

⎟⎠

⎞⎜⎜⎝

⎛+++=

nnnn

JJnnJJJJT rGGGGr

The equivalent moment of inertia is determined from Equation (f) as

( ) ( ) (g) 2

42

3144

2

2

13211 ⎟⎟

⎠

⎞⎜⎜⎝

⎛++⎟

⎟⎠

⎞⎜⎜⎝

⎛+++=

nnnn

JJnnJJJJJ rGGGGreq

The work done by all torsional viscous damping moments is

(h) 11

2

42

313

2

2

121∫ ⎥

⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+−= ωω d

nnnn

cnnccW ttt

from which the equivalent torsional viscous damping coefficient is determined as

101

Chapter 3

(i) 2

42

313

2

2

121, ⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

nnnn

cnnccc ttteqt

3.48 The differential equations for an armature controlled dc servomotor are

(b) 0

(a)

, =−+

=++

aaeqteq

abaaa

a

iKcdtdJ

vKiRdtdi

L

ωω

ω

The kinetic energy of the system at an arbitrary instant is

( ) ( ) ( ) (c) 21

21

21 2

3442232

2111 ωωω rGGGGr JJJJJJT +++++=

where the gear ratios are used to determine the angular velocities of the shafts in terms of the angular velocity of the motor shaft 1ω

(d) 12

12

2211

ωω

ωω

nnnn

=

=

(e) 42

312

4

33

3423

ωωω

ωω

⎟⎟⎠

⎞⎜⎜⎝

⎛==

=

nnnn

nn

nn

Substitution of Equations (d) and (e) into Equation (c) and rearranging leads to

( ) ( ) (f) 21 2

1

2

42

3144

2

2

13211 ω

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟

⎟⎠

⎞⎜⎜⎝

⎛+++=

nnnn

JJnnJJJJT rGGGGr

The equivalent moment of inertia is determined from Equation (f) as

( ) ( ) (g) 2

42

3144

2

2

13211 ⎟⎟

⎠

⎞⎜⎜⎝

⎛++⎟

⎟⎠

⎞⎜⎜⎝

⎛+++=

nnnn

JJnnJJJJJ rGGGGreq

The work done by all torsional viscous damping moments is

(h) 11

2

42

313

2

2

121∫ ⎥

⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+−= ωω d

nnnn

cnnccW ttt

from which the equivalent torsional viscous damping coefficient is determined as

(i) 2

42

313

2

2

121, ⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

nnnn

cnn

ccc ttteqt

3.49 If the shafts are supported by elastic bearings then the appropriate dependent variables are the current in the field circuit and the angular rotation of the shaft θ. The differential equations are

fi

102

Chapter 3

(b) 0

(a)

,,2

2

=−++

=+

ffeqteqteq

fff

f

iKkcdtdJ

viRdtdi

L

θθθ

The kinetic energy of the system at an arbitrary instant is

( ) ( ) ( ) (c) 21

21

21 2

3442232

2111 ωωω rGGGGr JJJJJJT +++++=

where the gear ratios are used to determine the angular velocities of the shafts in terms of the angular velocity of the motor shaft 1ω

(d) 12

12

2211

ωω

ωω

nnnn

=

=

(e) 42

312

4

33

3423

ωωω

ωω

⎟⎟⎠

⎞⎜⎜⎝

⎛==

=

nnnn

nn

nn

Substitution of Equations (d) and (e) into Equation (c) and rearranging leads to

( ) ( ) (f) 21 2

1

2

42

3144

2

2

13211 ω

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟

⎟⎠

⎞⎜⎜⎝

⎛+++=

nnnn

JJnnJJJJT rGGGGr

The equivalent moment of inertia is determined from Equation (f) as

( ) ( ) (g) 2

42

3144

2

2

13211 ⎟⎟

⎠

⎞⎜⎜⎝

⎛++⎟

⎟⎠

⎞⎜⎜⎝

⎛+++=

nnnn

JJnn

JJJJJ rGGGGreq

The work done by all torsional viscous damping moments is

(h) 1 11

2

42

312

2

1∫ ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+−= ωω d

nnnn

nn

cW t

from which the equivalent torsional viscous damping coefficient is determined as

(i) 12

42

312

2

1,

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

nnnn

nn

cc teqt

The total potential energy stored in the bearings at an arbitrary instant is

(j) 121 2

42

312

2

1 θ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

nnnn

nn

kV t

The equivalent torsional stiffness is obtained from Equation (j) as

(k) 142

312

2

1,

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

nnnn

nn

kk teqt

3.50 The inductors in the circuit are in parallel and can be replaced by a single inductor. The motion of the mass induces a back emf, in loop BCDE of the circuit. be

103

Chapter 3

Application of KVL to loop ABEF gives

(a) )(11

02111 vdtii

CiR

t

=−+ ∫

Application of KVL to loop BCDE leads to

(b) 0)(1

012222 =−+++ ∫ dtii

CeiLiR

t

beq

Application of Newton’s second law to the free-body diagrams of the blocks, drawn at an arbitrary instant, leads to

( )(c) 2 21111

111211

FkxkxxcxmxmxxkxckxF

=−++=−+−−

&&&&&

(d) 0)(

2122

2212

=+−=−−kxkxxm

xmxxk&&

&&

3.51 Consider n resistors in parallel as illustrated below

KVL applied to each loop in the circuit shows that the voltage change across each resistor is the same

(a) 21 nvvvv ==== Κ The sum of the currents through each resistor is equal to the current entering the parallel combination

(b) 321 niiiii ++++= Κ The voltage change across each resistor is given by Ohm’s law

104

Chapter 3

(c) k

k

kkk

Rvi

Rivv

=

==

If the parallel combination were to be replaced by a single resistor of resistance then eqR

(d) eq

eq

Rvi

iRv

=

=

Substitution of Equations (c) and (d) into Equation (b) leads to

(e) 21 neq R

vRv

Rv

Rv

+++= Κ

Equation (e) is rearranged leading to

(f) 1

1

1111

1

21

∑=

=

+++=

n

k k

n

eq

R

RRR

RΚ

3.52 Consider n capacitors in parallel as illustrated below

KVL applied to each loop in the circuit shows that the voltage change across each capacitor is the same

(a) 21 nvvvv ==== Κ The sum of the currents through each capacitor is equal to the current entering the parallel combination

(b) 321 niiiii ++++= Κ The voltage change across each capacitor is given by

(c)

1

0

dtdvCi

iC

vv

kk

t

kk

k

=

== ∫

105

Chapter 3

If the parallel combination were to be replaced by a single capacitor of capacitance then eqC

(d)

1

0

dtdvCi

idtC

v

eq

t

eq

=

= ∫

Substitution of Equations (c) and (d) into Equation (b) leads to

(e) 21 dtdvC

dtdvC

dtdvC

dtdvC neq +++= Κ

Equation (e) is rearranged leading to

(f) 1

21

∑=

=

+++=n

kk

neq

C

CCCC Κ

106

Chapter 3

3.53 Consider n inductors in parallel as illustrated below

KVL applied to each loop in the circuit shows that the voltage change across each inductor is the same

(a) 21 nvvvv ==== Κ The sum of the currents through each inductor is equal to the current entering the parallel combination

(b) 321 niiiii ++++= Κ The voltage change across each inductor is given by

(c) 1

0∫=

==

t

kk

kkk

vdtL

i

dtdi

Lvv

If the parallel combination were to be replaced by a single inductor of inductance then eqL

(d) 1

0∫=

=

t

eq

eq

vdtL

i

dtdiLv

Substitution of Equations (c) and (d) into Equation (b) leads to

(e) 1111

002010∫∫∫∫ ++=t

n

ttt

eq

idtL

idtL

idtL

idtL

Κ

Equation (e) is rearranged leading to

(f) 1

1

1111

1

21

∑=

=

+++=

n

k k

n

eq

L

LLL

LΚ

107

Chapter 4

4. Fluid, Thermal, and Chemical Systems 4.1 The geometry of the proposed water clock is shown below.

Application of trigonometry to Figure (b) leads to

(a) 0015.0342.0 0015.0)20sin(

sin 21

+=+°=

+=

hh

rhr θ

Also the distance from the tip of the truncated cone to the hole is

(b) 00141.0 )20cos(0015.0

cos2

m

r

=°=

= θλ

The volume of a right cone of height z and base radius r is

(c) 31 2 zrV π=

Consider a control volume of the clock. Let z be the instantaneous height of the column of liquid in the control volume. Assume both ends of the control volume are open to the atmosphere and that the flow rate of air into the control volume is much smaller than the flow rate of the water leaving the control volume. At a given instant of time the volume of liquid in the clock is

( )[ ][ ] (d) 1017.31025.21003.1117.0

3

)00141.0()0015.0(0015.0342.03

31)(

31)(

96233

22

21

2

−−− −++=

−+=

−=

xzxzxz

zz

rzzrzV

π

π

ππ λ

The velocity of the liquid leaving the hole is approximated using Toricelli’s law as (e) 2gzv =

Thus the volumetric flow rate of the liquid leaving is

107

Chapter 4

( )[ ](f) 1013.3

0015.0 81.92

5

22

zx

zsm

vAQ

−=

⎟⎠⎞

⎜⎝⎛=

=

π

The application of the control volume form of Conservation of Mass to the clock is of the form: Rate of change of liquid leaving the control volume =Rate of change of volume of liquid in the control volume

[ ]

( ) ( ) (g) 1025.21003.2351.01013.33

1017.31025.21003.1117.03

1013.3

6325

962335

dtdzxzxzzx

xzxzxzdtdzx

dtdVQ

−−−

−−−−

++=⎟⎠⎞

⎜⎝⎛

⎭⎬⎫

⎩⎨⎧ −++−=

−=

π

π

Equation (g) is rearranged to

( ) (h) 1025.21003.2351.0103 5.065.035.15

dtdzzxzxzx −−−− ++=

Separation of variables in Equation (h) leads to ( ) (i) 1025.21003.2351.0103 5.065.035.15 dzzxzxzdtx −−−− ++=

Integration of Equation (i), requiring that the clock drains in 3600 s leads to

( )

(j) 1013.11035.1141.0108.0

1013.11035.1141.0103

1025.21003.2351.0103

5.065.135.20

5.065.135.23600

0

5

0

5.065.035.13600

0

5

zxzxz

zxzxztx

dzzxzxzdtx

hz

z

t

t

h

−−

=

=

−−=

=

−

−−−−

++=

++=

++= ∫∫

Equation (j) can be written as a 5th order polynomial in as 5.0z( ) ( ) (k) 0108.01013.11035.1141.0 5.0635.0355.0 =−++ −− zxzxz

The following MATLAB workspace is used to determine the roots of the polynomial >> clear >> p=[0.141 0 1.35e-3 0 1.13e-6 0 -0.108] p = 0.1410 0 0.0014 0 0.0000 0 -0.1080 >> aa=roots(p) aa = 0.9549

108

Chapter 4

0.4774 + 0.8298i 0.4774 - 0.8298i -0.4774 + 0.8298i -0.4774 - 0.8298i -0.9549 >> aa(1)^2 ans = 0.9118 >> Thus the water clock will drain in one hour if

(l) 912.0 mh = 4.2 Consider a control volume containing the balloon. Application of conservation of mass to the control volume leads to

(a) dtdmmin =&

where the total mass in the control volume is

(b) 34 3 ⎟

⎠⎞

⎜⎝⎛=

=

r

Vm

πρ

ρ

where r is the radius of the balloon. Substituting Equation (b) into Equation (a) leads to

( )

(c) 34

34

2

3

dtdrr

rdtdmin

πρ

πρ

=

=&

Rearrangement of Equation (c) gives

(d) 432

πρinm

dtdrr

&=

Integration of Equation (d) with respect to time is

109

Chapter 4

(e) 43

3

43

300

2

tmr

dtm

drr

in

tin

r

πρ

πρ

&

&

=

= ∫∫

Equation (e) is rearranged to solve for the time required to achieve a radius r as

(f) 9

4 3

inmrt&

πρ=

The time required to achieve a radius of 10 cm is

( )

(g) s 1036.2 s

kg 1.09

m 1.0mkg 169.04

3

33

−=

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

=

x

tπ

4.3 Application of Bernoulli’s equation from the surface of the reservoir to the exit of the reservoir assuming no losses

(a) 2 1

222

1

211

21

zg

vg

pzg

vg

p

hh

++=++

=

ρρρ

The pressure at the surface of the reservoir is given as kPa 2001 =p , the exit jet is open to the atmosphere, thus . Assuming the reservoir is large 01 =p 01 ≈v . Taking the datum at the tank exit leads to 0 and m 40 21 == zz . The density of water is taken

as 3mkg 1000=ρ . Substitution into Equation (a) leads to

110

Chapter 4

(b) sm 4.34

sm 81.92

m 40

sm 81.9

mkg 1000

mN 10200

2

2

22

23

23

=

⎟⎠⎞

⎜⎝⎛

=+⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛

v

vx

The water in the jet has a horizontal velocity of 34.4 m/s as it leaves the tank. The range of the jet is obtained by applying the projectile motion equations to the jet. Let x and y be the position of a fluid particle as it leaves the tank at t=0. The projectile motion equations are

(d)

(c) 21 2

vtx

gty

=

−=

The water hits the surface when m 10−=y . The time at which the fluid particle hits the surface is obtained using Equation (c) as

( )

(e) s 43.1 sm 81.9

m 102

2

21

2

21

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎟⎠⎞

⎜⎝⎛−

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

gyt

The range of the fluid jet is obtained using Equation (d) as

( )

(f) m 1.49

s 43.1sm 4.34

=

⎟⎠⎞

⎜⎝⎛=x

4.4 Application of extended Bernoulli’s equation between the surface of the reservoir (1) and the exit of the pipe (2) leads to

(a) 21 hhh += λ Choosing the datum at the free surface of the reservoir and assuming the velocity at the surface is small, the head at the surface of the reservoir is

(b) 0 2 1

211

1

=

++= zg

vg

phρ

The head at the exit of the pipe is

111

Chapter 4

( )

(d) 400510.0

m 40

sm81.92

Pa 0

2

22

2

22

2

222

2

−=

−+⎟⎠⎞

⎜⎝⎛

+=

++=

v

v

zg

vg

ph

ρ

The head loss through the pipe is

( )( )(d) 0199.0

m 0.2m 260510.0003.0

2

22

22

22

v

v

DL

gvfh

=

⎟⎠⎞

⎜⎝⎛=

⎟⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=λ

Substitution of Equations (b)-(d) into Equation (a) leads to

(e) sm 8.23

0199.0400510.00

2

22

22

=

+−=

v

vv

4.5 Application of extended Bernoulli’s equation between the surface of the reservoir (1) and the exit of the pipe (2) leads to

(a) 21 hhh += λ Choosing the datum at the free surface of the reservoir and assuming the velocity at the surface is small, the head at the surface of the reservoir is

(b) m 10.5 sm 81.9

mkg 1000

mN 105

2

23

24

1

211

1

=

⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛

=

++=

x

zg

vg

phρ

The head at the exit of the pipe is

( )

(d) 400510.0

m 40

sm81.92

Pa 0

2

22

2

22

2

222

2

−=

−+⎟⎠⎞

⎜⎝⎛

+=

++=

v

v

zg

vg

ph

ρ

The head loss through the pipe is

112

Chapter 4

( )( )(d) 0199.0

m 0.2m 260510.0003.0

2

22

22

22

v

v

DL

gvfh

=

⎟⎠⎞

⎜⎝⎛=

⎟⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=λ

Substitution of Equations (b)-(d) into Equation (a) leads to

(e) sm 2.25

0199.0400510.010.5

2

22

22

=

+−=

v

vv

4.6 Application of extended Bernoulli’s equation between the surface of the reservoir (1) and the exit of the pipe (2) leads to

(a) 21 hhh += λ Choosing the datum at the free surface of the reservoir and assuming the velocity at the surface is small, the head at the surface of the reservoir is

(b) 0 2 1

211

1

=

++= zg

vg

phρ

The head at the exit of the pipe is

( )

(d) 400510.0

m 40

sm81.92

Pa 0

2

22

2

22

2

222

2

−=

−+⎟⎠⎞

⎜⎝⎛

+=

++=

v

v

zg

vg

ph

ρ

The head loss through the pipe is

( )(d) 63.6

m 0.2m 260510.0

2

22

22

22

fv

vf

DL

gvfh

=

⎟⎠⎞

⎜⎝⎛=

⎟⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=λ

Substitution of Equations (b)-(d) into Equation (a) leads to

(e) 0510.063.6

40

63.6400510.00

21

2

22

22

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

+−=

fv

fvv

113

Chapter 4

The friction factor is unknown as is the velocity. They are related from the Colebrook formula on which the Moody diagram is based. The relative roughness ratio is given as

002.0=Dε . An iterative solution using the Moody diagram is required. The procedure is

1. Guess a value of the velocity

2. Calculate the Reynolds number µρvD

=Re , assuming 3mkg 1000=ρ and

23

msN 1031.1 ⋅

= −xµ . In terms of the velocity the Reynolds number is calculated as

( )

(f) 1053.1

sN 1031.1

m 2.0mkg 1000

Re

5

23

3

vxm

x

v

=

⋅

⎟⎠⎞

⎜⎝⎛

=−

3. Determine a friction factor from the Moody diagram (Fig. 4.5) 4. Calculate the velocity using Equation (e) 5. Compare the calculated velocity to the guessed value. Use the calculated value as

the new initial guess, continuing the procedure until convergence is achieved.

The procedure, using an initial guess of sm 202 =v is summarized in the Table below.

Iteration Guess for (m/s) 2v Re (Eq. f) f (Moody diagram)

2v (m/s) (Eq.e)

1 20 61006.3 x 0.023 14.02 2 14.02 61015.2 x 0.023 14.02

Since the friction factor corresponding to 002.0=Dε is flat for Re>1 , it only takes

two iterations to converge to

610x

sm 02.142 =v .

4.7 The velocity of the flow through the pipe is calculated in terms of the diameter of the pipe as

(a) 255.0

4

sm 2.0

2

2

3

D

D

AQv

=

=

=

π

114

Chapter 4

Application of extended Bernoulli’s equation from the pressure head to the flow exit, assuming they are at the same elevation and the exit pressure is atmospheric gives

(b) 1031.31031.318

sm 81.92

255.0

Dm 10

sm 81.92

255.0

m 18

5

2

4

3

2

2

2

2

2

2

21

Dfx

Dx

DfD

hhh

−−

+=

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛+

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

=

+= λ

Since the friction factor is a function of the Reynolds number, which is a function of the diameter,

( )

(c) 1095.1

msN 1031.1

D255.0mkg 1000

Re

5

23

23

Dx

x

D

vD

=

⋅

⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛

=

=

−

µρ

an iterative process using the Moody diagram is necessary to solve for the required pipe diameter. The process is to guess a diameter, calculate the Reynolds number using Equation (c), use the Moody diagram to determine the friction factor for a smooth pipe, and use Equation (b) to calculate a new diameter. Use this new diameter as an initial guess for the subsequent iteration. The process is summarized in the table below using an initial guess of 10 cm. Iteration D (m) Re (Eq. c) f (Moody

diagram) D (Eq. b)

1 0.1 61095.1 x 0.0105 0.125

2 0.125 61056.1 x 0.0109 0.125 Thus the required diameter is 12.5 cm.

4.8 Before the rocket is filled with water it contains air at atmospheric pressure. The total mass of the air is

( ) (a) 2 LD

Vm

a

aaa

πρ

ρ

=

=

The ideal gas law implies

(b) 0a

0

TRp

TRp

a

aa

=

=

ρ

ρ

115

Chapter 4

where is the atmospheric pressure. Substitution of Equation (b) into Equation (a) leads to

0p

(c) 2

0

TRLDp

ma

aπ

=

As the rocket is being filled with water, the mass of the air remains constant. Its density is calculated as

(d) )(

)()(

2 hLDmzV

mh

a

a

a

−=

=

π

ρ

(a) Assuming the process is isothermal use of the ideal gas law and Equation (c) in Equation (d) leads to

(e) )(

)()(

0

2

20

phL

Lhp

hLDTRLDp

TRhp

aa

−=

−=

ππ

(b) The pressure on the surface when the water level is z is obtained by replacing h by z in Equation (e) leading to

(f) )( 0pzL

Lzp−

=

(c) Let 1 represent the free surface of the water and 2 represent the exit of the rocket. Assume acceleration of the particles on the surface of the water is negligible. Application of Bernoulli’s equation leads to

(g) 22

1

22220

20

2

22

2

21

21

1

1

gv

gp

zdtdz

ggp

zLL

zg

vg

pz

gv

gp

+=+⎟⎠⎞

⎜⎝⎛+

−

++=++

ρρ

ρρ

Equation (g) is rearranged as

(h) 2221

20

2⎥⎥⎦

⎤

⎢⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛

−= gz

dtdzp

zLzv

ρ

Equation (h) is a nonlinear first-order differential equation whose solution is z(t). The solution must satisfy

(i) )0( hz = (d) Consider a control volume consisting of the rocket. The principle of conservation

of mass, assuming the water is incompressible, is written as: rate at which water leaver the control volume through its exit=rate at which water accumulates in rocket

116

Chapter 4

( )

(h) 2221

20

222

dtdzgz

dtdzp

zLz

Dd

zDdtddv

−=⎥⎥⎦

⎤

⎢⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛

−⎟⎠⎞

⎜⎝⎛

−=

ρ

ππ

4.9 Application of extended Bernoulli’s equation from the free surface of the tank to the exit of the pipe assuming the velocity of the fluid on the surface of the reservoir is small leads to

(a) 21 λhhh += where

(c) 510.0

0

sm 81.92

0

2

(b) m 2000 2

22

2

22

2

222

2

1

211

1

v

v

zg

vg

ph

zg

vg

ph

=

+⎟⎠⎞

⎜⎝⎛

+=

++=

++=

++=

ρ

ρ

The head loss through the pipe is

( )(d) 3.15

m 0.3m 600510.0

2

22

22

22

fv

vf

DL

gvfh

=

⎟⎠⎞

⎜⎝⎛=

⎟⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=λ

Substitution of Equations (b)-(d) into Equation (a) leads to

(e) 0510.03.15

20

63.6400510.00

21

2

22

22

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

+−=

fv

fvv

The friction factor is unknown as is the velocity. They are related from the Colebrook formula on which the Moody diagram is based. The relative roughness ratio is calculated as

(f) 105

m 0.3m 105.1

4

4

−

−

=

=

x

xDε

An iterative solution using the Moody diagram is required. The procedure is 1. Guess a value of the velocity

117

Chapter 4

2. Calculate the Reynolds number µρvD

=Re , assuming 3mkg 1000=ρ and

23

msN 1031.1 ⋅

= −xµ . In terms of the velocity the Reynolds number is calculated as

( )

(g) 1029.2

sN 1031.1

m 3.0mkg 1000

Re

25

23

23

vxm

x

v

=

⋅

⎟⎠⎞

⎜⎝⎛

=−

3. Determine a friction factor from the Moody diagram (Fig. 4.5) 4. Calculate the velocity using Equation (e) 5. Compare the calculated velocity to the guessed value. Use the calculated value as

the new initial guess, continuing the procedure until convergence is achieved.

The procedure, using an initial guess of sm 202 =v is summarized in the Table below.

Iteration Guess for (m/s) 2v Re (Eq. g) f (Moody diagram)

2v (m/s) (Eq.e)

1 20 61058.4 x 0.0165 8.11 2 8.11 61086.1 x 0.017 8.02

Since the friction factor corresponding to 002.0=Dε is flat for Re>1 , it only takes

two iterations to converge to

610x

sm 02.82 =v . The flow rate is calculated as

( )

(h) m 567.0

m 3.04s

m 02.8

3

4

s

vAQ

=

⎟⎠⎞

⎜⎝⎛=

=

π

(b) Since the Reynolds number is much greater than 2300, the flow is turbulent and the resistance in the pipe is

(i) ms 6.70

m 0.567

m 202

2

2

3

=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

=

=

s

QhR

118

Chapter 4

4.10 The pipe’s resistance is calculated as

(a) 2QhR =

The head loss across the pipe is

( )

(b) 74.5

m 0.3m 30

sm 81.92

m 3.04

m 075.0

2

2

2

2

3

2

f

s

f

DL

gvfh

=

⎟⎠⎞

⎜⎝⎛

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

⎟⎠⎞

⎜⎝⎛

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

=

⎟⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

π

The Reynolds number for the flow is

( )

(c) 1001.1 m

sN 0.290

m) 3.0(m 3.0

4

m 075.0

mkg 917

Re

3

2

2

3

3

x

s

vD

=

⋅

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎟⎠⎞

⎜⎝⎛

=

=

π

µρ

The friction factor is determined using the smooth pipe curve on the Moody diagram as . Thus from Equation (a) 07.0=f

(d) ms 10.71

m 0.075

m )07.0)(74.5(2

2

3

=

=

s

R

4.11 The tanks are operating at steady state. Assuming turbulent flow, the resistance in a pipe is

(a) 2=qh

R λ

119

Chapter 4

The head loss in the first pipe is the difference in the levels between the two tanks, and the flow rate through the pipe is the flow rate into the first

tank,

( ) ( ) m 7m 8m 151 =−=λh

sm

8.1=3

1q . Thus

(b) ms 78.7

sm 1.8

m 72 231 ==R

The head loss in the second pipe is equal to the level of the liquid in the second tank, . The flow rate through the second pipe is the sum of the flow rates into the

two tanks

m 8=2λh

sssq

333

2m 2.3m 4.1m 8.1 =⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛= . Hence

(c) ms

0.5=m

3.2

m 82= 232

s

R

4.12 The three pipes are in parallel, thus their equivalent resistance is the sum of the reciprocals of the individual resistances

(a) ms 69.1

ms 0.5

1

ms 0.4

1

ms 0.7

11

111

1

2

222

321

=

++=

++=

RRR

Req

The head loss across each pipe is the same for all pipes in a parallel combination while the total flow rate is the sum of the individual flow rates,

(c) s

m 9.4

(b) 21

21

21

321

3

332211

QQQQ

QRQRQRh

++==

===

The flow rate is related to the equivalent resistance through

120

Chapter 4

(d) m 6.16

m 9.4ms 69.1

21

21

2

3

2

=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛=

=

=

s

QRh

QhR

eq

eq

Substituting Equation (d) into Equations (b) leads to

(e) m 74.4

ms 7.0

m) 6.16(2

2

3

2

11

s

RhQ

=

=

=

(f) m 35.8

ms 4.0

m) 6.16(2

2

3

2

22

s

RhQ

=

=

=

(g) m 68.6

ms 5.0

m) 6.16(2

2

3

2

33

s

RhQ

=

=

=

4.13 The appropriate differential equation is

(a) )(1 tqhRdt

dhA p=+

The steady-state resistance is obtained as

(b) 5.12 4.2

1522 23 ms

smm

qh

Rs

s ===

From the given information the perturbation in flow rate is . Thus Equation (1) becomes

)(2.0)( tutq p −=

121

Chapter 4

(c) )(2.008.0650

)(2.05.12

1650

tuhdtdh

tuhdtdh

−=+

−=+

4.14 The mathematical model for the two-tank system is

(b) )(111

(a) )(11

2221

11

22

121

11

11

tqhRR

hRdt

dhA

tqhR

hRdt

dhA

p

p

=⎟⎟⎠

⎞⎜⎜⎝

⎛++−

=−+

The resistances are calculated using the steady-state values

(c) ms 45.1

sm 4.0

m 8.6m 7.92

2

2

3

1

211

=

−=

−=

s

ss

qhh

R

(d) ms 4.3

sm 0

sm 4.0

m 8.62

2

2

33

21

22

=

+=

+=

ss

s

qqh

R

The perturbation in flow rates are

(f) m )(2.0)(

(e) 0)(3

2

1

stutq

tq

p

p

=

=

Substitution of Equations (c)-(f) in Equations (a) and (b) leads to

(g) 045.11

45.11450 21

1 =−+ hhdtdh

(h) )(2.040.31

45.11

45.11790 21

2 tuhhdt

dh=⎟

⎠⎞

⎜⎝⎛ ++−

Equations (g) and (h) are rearranged as

122

Chapter 4

(j) )(2.0984.0690.0790

(i) 0690.0690.0450

212

211

tuhhdt

dh

hhdtdh

=+−

=−+

4.15 The mathematical model for the two-tank system is

(b) )(111

(a) )(11

2221

11

22

121

11

11

tqhRR

hRdt

dhA

tqhR

hRdt

dhA

p

p

=⎟⎟⎠

⎞⎜⎜⎝

⎛++−

=−+

The resistances are calculated by

(d) 2

(c) 2

12

11

2

1

s

s

qh

R

qh

R

λ

λ

=

=

The head loss in the first pipe is obtained by application of extended Bernoulli’s equation between the free surface of the reservoir and the exit of the pipe

(e) m 12.4

m 15m 13

sm 81.9

mkg 1000

mN 106

1

1

1

23

24

21

=

+=+⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛

+=

λ

λ

λ

h

hx

hhh

Substitution of Equation (e) into Equation (c) leads to

(f) ms 06.2

sm 4.0

m 12.42

2

1

=

=R

(g) s

m 5.7

ms 4.0

m 152

2

2

2

=

=R

Substitution in Equations (a) and (b) leads to

123

Chapter 4

( )

( ) (i) 05.7

106.21

06.21m 8.1

4

(h) )(6.006.21

06.21m 2

4

2122

2112

=⎟⎠⎞

⎜⎝⎛ ++−

=−+

hhdt

dh

tuhhdtdh

π

π

Equations (h) and (i) are simplified to

(k) 0619.0485.054.2

(j) )(6.0485.0485.014.3

212

211

=+−

=−+

hhdt

dh

tuhhdtdh

4.16 Application of conservation of mass to each tank leads to an equation of the form

(a) ,, outiinii

i QQdt

dHA −=

With the exception of the first tank both the inlet and outlet flow rates are related to the pipe resistances. It is shown in the text that linearization of the nonlinear equations obtained through application of Equation (a) is consistent with approximating the change in head across a pipe and the flow rate through the pipe by

(b) QhR ∆

=

where R is the resistance of the pipe at steady state. Consider the piping system connecting the three tanks. Let be the flow rates through the pipes. Then from conservation of mass

321 and , QQQ

(c) 321 QQQ += Defining as the head at the junction of the three pipes, the resistances are related to the flow rates and perturbations in level by

mh

(f)

(e)

(d)

3

33

2

22

1

11

Qhh

R

Qhh

R

Qhh

R

m

m

m

−=

−=

−=

Solving Equations (d)-(f) for the flow rates and substituting into Equation (c) leads to

124

Chapter 4

(g) 111

1

111

3

3

2

2

1

1

321

3

3

2

2

1

1

321

3

3

2

2

1

1

⎟⎟⎠

⎞⎜⎜⎝

⎛++

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++=

++=⎟⎟⎠

⎞⎜⎜⎝

⎛++

−+

−=

−

Rh

Rh

Rh

RRR

h

Rh

Rh

Rh

RRRh

Rhh

Rhh

Rhh

m

m

mmm

Then

(h) 1

11111

2

31

3

21

133221

3

3

2

2

1

1

321

11

1

11

⎟⎟⎠

⎞⎜⎜⎝

⎛ −+

−++

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛++

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++−=

−=

Rhh

Rhh

RRRRRR

Rh

Rh

Rh

RRR

hR

Rhh

Q m

In a similar fashion

(j) 1

(i) 1

1

32

2

31

1332213

1

23

3

21

1332212

⎟⎟⎠

⎞⎜⎜⎝

⎛ −+

−++

=

⎟⎟⎠

⎞⎜⎜⎝

⎛ −+

−++

=

Rhh

Rhh

RRRRRRQ

Rhh

Rhh

RRRRRRQ

Using Equations (h)-(i) the appropriate linear equations for the liquid levels in the tanks are

(k) 1

2

31

3

21

133221

11 iq

Rhh

Rhh

RRRRRRdtdhA =⎟⎟

⎠

⎞⎜⎜⎝

⎛ −+

−++

+

(l) 01

1

23

3

21

1332215

222 =⎟⎟

⎠

⎞⎜⎜⎝

⎛ −+

−++

−+R

hhR

hhRRRRRRR

hdt

dhA

(m) 01

1

32

2

31

1332214

333 =⎟⎟

⎠

⎞⎜⎜⎝

⎛ −+

−++

−+R

hhR

hhRRRRRRR

hdtdh

A

4.17 If the process is isentropic then the pressure and density are related by

(a) γρCp = where C is a constant and γ is the ratio of specific heats. The constant is evaluated from the initial state

125

Chapter 4

(b) 0

0

00

γ

γ

ρ

ρp

C

Cp

=

=

The ideal gas law relates pressure, temperature, and density by (c) TRp aρ=

Substitution of Equation (c) into Equation (a) leads to

(d) 11

0

01

0

0

−

−

=

=

=

γ

γγ

γγ

ρ

ρρ

ρρ

ρ

BT

Tp

R

pTR

a

a

where

(e) 1

1

0

0−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

γγρp

RB a

Substitution of Equation (e) into Equation (a) leads to

(f)

1

11

0

0

11

0

0

−

−−

−

=

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

γγ

γγ

γγ

γ

γ

γγ

ρ

ρ

DT

TBp

BTp

p

The mass of the air in the pressure vessel is as given by Equation (d) in the solution of Example 4.10,

(g) Vma ρ= Use of Equation (d) in Equation (g) leads to

(h) 11−= γBVTma

As in Example 4.10 the mass flow rate into the vessel is related to the resistance, pressure, and upstream pressure by

(i) 0

Rpp

m−

=&

Substitution of Equation (f) into Equation (h) leads to

(j) 1

0

RDTp

m−−

=γγ

&

The control volume form of conservation of mass for the pressure vessel is

126

Chapter 4

(k) 1

10

12

101

1

RDTp

dtdTBVT

RDTp

BVTdtd

mdt

dma

−−−

−−

−=

−

−=⎟

⎟⎠

⎞⎜⎜⎝

⎛

=

γγ

γγ

γγ

γ

γ

&

4.18 The mass of the air in the pressure vessel is as given by Equation (d) in the solution of Example 4.10,

(a) Vma ρ= As in Example 4.10 the mass flow rate into the vessel is related to the resistance, pressure, and upstream pressure by

(b) 0

Rpp

m−

=&

The control volume form of conservation of mass for the pressure vessel is

( )

(c) 0

0

RVp

RVTR

dtd

RTRp

Vdtd

mdt

dm

a

a

a

=+

−=

=

ρρ

ρρ

&

4.19 The appropriate mathematical model is that of Equation (h) of Example 4.10

(a) sg

s

g

ppdtdp

TRRV

Rp

Rp

dtdp

TRV

=+

=+

The volume of the tank is

( )

(b) 5.33

234

34

3

3

3

m

m

rV

=

=

=

π

π

The resistance is determined from the given information by

127

Chapter 4

(c) 1 1025.1

6.0

105.2 1010

4

23

23

smx

skg

mNx

mNx

mpp

R s

⋅=

−=

−=

&

Thus

( )

( )

(d) 1044.6

313 077.2

5.33 1 1025.1

2

2

2

24

sx

KKs

m

msm

x

TRRV

g

=

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

⎟⎠⎞

⎜⎝⎛

⋅=

Substituting Equation (d) into Equation (a) leads to

(e) 10101044.6 32 xpdtdpx =+

4.20 Application to conservation of mass to the control volume containing the tank

(a) outin mmdtdm && −=

where the mass of the gas in the chamber is Vm ρ= and from the ideal gas law,

TRp

g

=ρ . Since the process is isothermal T is constant and

(b) dtdp

TRV

dtdm

g

=

Application of extended Bernoulli’s equation between the exit of the tank and the end of the pipe leads to

(c) le hg

pgp

+=ρρ

Equation (c) is rearranged to (d) le ghpp ρ=−

Assuming laminar flow

128

Chapter 4

(e) 32

264

2Re64

2

2

2

vgD

LDL

gv

vD

DL

gvhl

ρµ

ρµ

=

=

=

Substitution of Equation (e) into Equation (d) leads to

( )

( ) (f) 128

128

4

432

4

4

2

2

2

eout

out

e

ppL

Dm

mD

L

D

DV

DLpp

−=

=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

=−

νππ

ν

π

π

ρν

&

&

Substitution of Equations (b) and (f) into Equation (a) leads to

( )

(g) 128128R

V

128

44

g

4

ein

eing

pL

DmpL

Ddtdp

T

ppL

Dmdtdp

TRV

νπ

νπ

νπ

+=+

−−=

&

&

4.21 A free-body diagram of the bar is illustrated below

129

Chapter 4

Summation of moments about the pin support, noting that the bar is massless leads to

(a)

0)()(0

1

1

FabF

aFbFM O

=

=−

=∑

Summation of forces on the free-body diagram of the piston of mass leads to 1m

(b) 11

11

1

Aym

AF

abp

ympAFab

amF

&&

&&

−⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛−=

=−

=∑

where y is the displacement of the piston. Summation of forces on the free-body diagram of the piston of mass leads to 2m

(c) 22 xmxckxpA &&&=−− Substitution of Equation (b) into Equation (c) leads to

(d) 1

2

1

212 aA

FbAkxxcyAAmxm −=+++ &&&&&

Assuming the hydraulic liquid is incompressible the volume of the air displaced by each piston must be equivalent. Thus

(e) 1

2

21

xAA

y

xAyA

=

=

Use of Equation (e) in Equation (d) leads to

(f) )(1

21

2

1

22 tF

aAbAkxxcxm

AAm ⎟⎟

⎠

⎞⎜⎜⎝

⎛−=++

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+ &&&

130

Chapter 4

4.22 Free-body diagrams of the pistons are illustrated below.

Noting that the pressure in the hydraulic chamber is constant and the pistons are massless, summing forces on each of the pistons leads to

(a) )(

00)(

0

1

21

12

1

tFAAF

FpApAtF

F

−=

=+=−

=∑

Summation of moments about the pin support applied to the free-body diagram of the bar leads to

(b) 21 α

α

O

OO

IbFaFIM=+−

=∑

The mass moment of inertia of the bar about O, assuming a>b is obtained using Table 2.1 and the parallel axis theorem

( ) ( )

( ) (c) 6

21212

41

2

22)(

2121

22

2222

22

babam

babababam

baambamIO

+−⎟⎠⎞

⎜⎝⎛=

⎥⎦⎤

⎢⎣⎡ ++++−⎟⎠⎞

⎜⎝⎛=

⎟⎠⎞

⎜⎝⎛ +

−⎟⎠⎞

⎜⎝⎛++⎟

⎠⎞

⎜⎝⎛=

Assuming small displacements the angular acceleration of the bar is related to the acceleration of the block by

131

Chapter 4

(d) bxxb&&&&

=

=

α

α

Substitution of Equations (a), (c), and (d) in Equation (b) leads to

( )

( ) (e) )(6

6)(

1

22222

222

1

2

+⎟⎟⎠

⎞⎜⎜⎝

⎛−+−⎟

⎠⎞

⎜⎝⎛=

⎟⎠⎞

⎜⎝⎛+−⎟

⎠⎞

⎜⎝⎛=+⎟⎟

⎠

⎞⎜⎜⎝

⎛

tFbAaA

xbababmF

bxbabambFatF

AA

&&

&&

Application of Newton’s second law to the free-body diagram of the block, drawn at an arbitrary instant leads to

(f) 2 xmkxxcF &&& =−−− Substituting Equation (e) in Equation (f) and rearranging leads to

(g) )(76 1

22

tFbAaAkxxcx

ba

bam

⎟⎟⎠

⎞⎜⎜⎝

⎛=++

⎥⎥⎦

⎤

⎢⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛ &&&

4.23 The potential energy of the system at an arbitrary instant is simply

(a) 21 2kxV =

The kinetic energy of the system is the sum of the kinetic energies of the block, the bar and the two pistons. Let θ represent the angular rotation of the bar, y the displacement of the piston of mass and z the displacement of the piston of mass . The kinetic energy of the system is

1m 2m

(b) 21

21

21

21 2

22

122 zmymIxmT o &&&& +++= θ

The mass moment of inertia of the bar about O, assuming a>b is obtained using Table 2.1 and the parallel axis theorem

( ) ( )

( ) (c) 6

21212

41

2

22)(

2121

22

2222

22

babam

babababam

baambamIO

+−⎟⎠⎞

⎜⎝⎛=

⎥⎦⎤

⎢⎣⎡ ++++−⎟⎠⎞

⎜⎝⎛=

⎟⎠⎞

⎜⎝⎛ +

−⎟⎠⎞

⎜⎝⎛++⎟

⎠⎞

⎜⎝⎛=

The displacement of the end of the bar which is a distance b from the support is equal to the displacement of the block of mass m. Assuming small displacements

(d) bxxb

=

=

θ

θ

132

Chapter 4

The displacement of the end of the bar which is a distance a from the pin support is equal to the displacement of the piston of mass . Assuming small displacements and using Equation (d)

2m

(e) ba x

az

=

= θ

Conservation of mass applied to the fluid reservoir implies that the volumetric flow rate is constant in the reservoir. To this end,

(f) 1

2

1

2

21

xba

AA

zAAy

zAyA

&

&&

&&

⎟⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

=

=

Use of Equations (c)-(f) in Equation (b) leads to

( ) (g) 21

21

621

21 2

2

2

1

21

2222 ⎟

⎠⎞

⎜⎝⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟

⎠⎞

⎜⎝⎛+−⎟

⎠⎞

⎜⎝⎛+= x

bamx

bAaA

mbxbabamxmT &&&

&

Equation (g) is simplified to

(h) 162

1 22

2

2

1

212

2

xbam

bAaAm

ba

bammT &

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+−+=

The work done by the viscous damping force is

(i) 2

1

∫−=x

x

dxxcW &

The work done by the external force is

(j) )(

)(

2

1

2

1

1

2∫

∫

⎟⎟⎠

⎞⎜⎜⎝

⎛=

=

x

x

z

z

dxbAaA

tF

dztFW

Application of the energy method leads to a differential equation of the form (k) )(tFxkxcxm eqeqeqeq =++ &&&

where the equivalent mass is determined from the kinetic energy, Equation (g), as

(l) 16

2

2

2

1

212

2

⎟⎠⎞

⎜⎝⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+−+=

bam

bAaA

mba

bammmeq

The equivalent stiffness is determined using the potential energy, Equation (a), as (m) xkeq =

The equivalent damping coefficient is determined using the work done by the viscous damping force as

(n) cceq =

133

Chapter 4

The equivalent force is determined using Equation (j) as

(o) )()(1

2⎟⎟⎠

⎞⎜⎜⎝

⎛=

bAaAtFtFeq

4.24 Let w(t) represent the displacement of the particle to which the spring and viscous damper are attached, as illustrated in the figure below.

Using similar triangles,

(a) ya

baabzw

bwy

bawz

+−=

+=

++

Summing forces to zero at the end of the walking beam leads to (b) 0)( =−+ xwkwc&

The motor equation is (c) ˆ xAyC p &=

Substitution of Equation (c) into Equation (a) leads to

(d) ˆ xC

Aa

baabzw p &+

−=

Use of Equation (d) in Equation (b) leads to

(e) ˆˆ

0ˆˆ

zakbz

acbkxx

C

Aa

bakxC

Aa

bac

kxxC

Aa

baabzkx

C

Aa

baazbc

pp

pp

+=+⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ +

+⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ +

=−⎟⎟⎠

⎞⎜⎜⎝

⎛ +−+⎟⎟

⎠

⎞⎜⎜⎝

⎛ +−

&&&&

&&&&

134

Chapter 4

4.25 Let z(t) represent the displacement of the point on the walking beam connected to the valve, w(t) represent the displacement of the end of the walking beam, and q(t) represent the displacement of the rod. Assume small displacements.

Using similar triangles on the walking beam

(a) za

ayaw

awz

awy

+−=

+=

++

λλ

λ

Using similar triangles on the bar leads to

(b) xa

aq

ax

aq

λ

λ

+=

=+

The servomotor equation is

(c) ˆ

ˆ

xC

Az

xAzC

p

p

&

&

=

=

Use of Equation (c) in Equation (a) leads to

(d) ˆ xCA

aayaw p &λ

λ+

−=

Summing forces to zero at the end of the walking beam leads to (e) 0)( =+− kwqwc &&

Substitution of Equations (b) and (d) in Equation (e) leads to

(f) ˆ

0ˆ

yakyacxakxacax

C

Aa

ac

xayakxa

axC

Aa

ayac

p

p

λ&

λλλ&

λ&&λ

λλ

λ&

λ&&λ&

λ

+=+

++

+⎟⎟⎠

⎞⎜⎜⎝

⎛ +

=⎥⎦⎤

⎢⎣⎡ +

−+⎥⎦

⎤⎢⎣

⎡+

−+

−

135

Chapter 4

4.26 Consider a control volume consisting of the plate. The control volume form of conservation of energy applied to the plate is Rate at which energy accumulates in the control volume=Rate at which energy enters the control volume through its boundary –rate at which energy leaves the control volume through its boundary. The rate of accumulation of energy is the rate of change of internal energy,

(a) dtdTdcw

dtdU

pλρ=

Energy is transferred into the control volume due to the friction between the plate and the surface. The rate of energy dissipation due to friction is

( )

(b) 0

dgVw

dxdgxwdtdW

x

λ

λ&

µρ

µρ

=

= ∫

Energy is transferred from the control volume due to heat transfer with the ambient. Using Newton’s law of cooling

(c) )( 0TThwdQ −=& Thus application of Conservation of Energy to the control volume leads to

(d)

)(

0

0

hTgVhTdtdTc

TThwddgVwdtdTdcw

p

p

−=+

−−=

λλ

λλ

µρρ

µρρ

4.27 A control volume consisting of the aquarium is illustrated below.

The control volume form of conservation of energy applied to this control volume is Rate at which energy enters the control volume- rate at which energy leaves the control volume +rate at which heat is transferred to the control volume=rate at which energy accumulates in control volume Rate at which energy enters control volume = inpin Tcm&

136

Chapter 4

Rate at which energy leaves control volume= outpout Tcm& Rate at which heat is transferred to the control volume =Q&

Rate at which energy accumulates in control volume = dtdTVc pρ where T is the

temperature of the water in the aquarium under the lumped property assumption Application of the control volume form of conservation of mass leads to outin mm && =It is assumed that the water in the aquarium is well mixed such that TTout = Thus the energy equation is written as

(a) inpp TcmQTcmdtdTVc &&& +=+ρ

(a) In the steady state 0=dtdT , leading to

(b) 0=+− QTcmTcm pinp&&&

The mass flow rate is calculated as

(c) s

kg 1.85x10

m 8.1mkg 1003.1

3

3

33

=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛=

sxm&

Substitution into Equation (a) leads to

( )

(d) s

MJ 2.74

C 20C 03Ckg

J 100.4s

kg 1085.1

)(

33

=

−⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

⎟⎠⎞

⎜⎝⎛=

−=

xx

TTcmQ inp&&

(b) If the heater stops working, 0=Q& and Equation (a) becomes

(e) inppp TcmTcmdtdTVc && =+ρ

Substitution of given values leads to

137

Chapter 4

( )

( )

(f) 1017.2104.71047.2

K 293Kkg

J 100.4s

kg 1085.1

KkgJ 100.4

skg 1085.1

KkgJ 100.4m 600

mkg 1003.1

969

33

33323

3

xTxdtdTx

xx

TxxdtdTxx

=+

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

⎟⎠⎞

⎜⎝⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

⎟⎠⎞

⎜⎝⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅

⎟⎠⎞

⎜⎝⎛

Equation (f) is supplemented by the initial condition (g) K 303)0( =T

4.28 A control volume consisting of the aquarium is illustrated below.

The control volume form of conservation of energy applied to this control volume is Rate at which energy enters the control volume- rate at which energy leaves the control volume +rate at which heat is transferred to the control volume=rate at which energy accumulates in control volume Rate at which energy enters control volume = inpin Tcm& Rate at which energy leaves control volume= outpout Tcm& Rate at which heat is transferred to the control volume =Q&

Rate at which energy accumulates in control volume = dtdTVc pρ where T is the

temperature of the water in the aquarium under the lumped property assumption Application of the control volume form of conservation of mass leads to outin mm && =It is assumed that the water in the aquarium is well mixed such that TTout = Thus the energy equation is written as

(a) inpp TcmQTcmdtdTVc &&& +=+ρ

138

Chapter 4

When the cooling and filtration stop working the flow is recirculated such that the temperature of the stream entering is the temperature of the tank 15 s before. Thus the rate at energy enters the control volume is

(b) )15( −tTcm p& Equation (a) is modified as

(c) )15( QtTcmTcmdtdTVc pp

&&& =−−+ρ

The rate of heat transfer is equal to the rate such that the tank is in steady state when the system is working properly. This steady-state equation is

(d) 0=+− QTcmTcm pinp&&&

The mass flow rate is calculated as

(e) s

kg 1.85x10

m 8.1mkg 1003.1

3

3

33

=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛=

sxm&

Substitution into Equation (a) leads to

( )

(f) s

MJ 2.74

C 20C 03Ckg

J 100.4s

kg 1085.1

)(

33

=

−⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

⎟⎠⎞

⎜⎝⎛=

−=

xx

TTcmQ inp&&

Substitution of given and calculated values in Equation (c) leads to

( )

(g) 1042.7)]15([104.71047.2

sJ 1042.7)15(

KkgJ 100.4

skg 1085.1

KkgJ 100.4

skg 1085.1

KkgJ 100.4m 600

mkg 1003.1

769

733

33323

3

xtTTxdtdTx

xtTxx

TxxdtdTxx

=−−+

=−⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

⎟⎠⎞

⎜⎝⎛−

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

⎟⎠⎞

⎜⎝⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅

⎟⎠⎞

⎜⎝⎛

4.29 A control volume consisting of the aquarium is illustrated below.

139

Chapter 4

The control volume form of conservation of energy applied to this control volume is Rate at which energy enters the control volume- rate at which energy leaves the control volume +rate at which heat is transferred to the control volume=rate at which energy accumulates in control volume Rate at which energy enters control volume = inpin Tcm& Rate at which energy leaves control volume= outpout Tcm& Rate at which heat is transferred to the control volume =Q&

Rate at which energy accumulates in control volume = dtdTVc pρ where T is the

temperature of the water in the aquarium under the lumped property assumption Application of the control volume form of conservation of mass leads to outin mm && =It is assumed that the water in the aquarium is well mixed such that TTout = Thus the energy equation is written as

(a) inpp TcmQTcmdtdTVc &&& +=+ρ

The rate of heat transfer is equal to the rate such that the tank is in steady state when the system is working properly. This steady-state equation is

(b) 0=+− QTcmTcm pinp&&&

The mass flow rate is calculated as

(c) s

kg 1.85x10

m 8.1mkg 1003.1

3

3

33

=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛=

sxm&

Substitution into Equation (b) leads to

( )

(d) s

MJ 2.74

C 20C 03Ckg

J 100.4s

kg 1085.1

)(

33

=

−⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

⎟⎠⎞

⎜⎝⎛=

−=

xx

TTcmQ inp&&

If the flow rate suddenly changes to s

3m 6.1 a transient perturbation in temperature is

induced in the aquarium. Defining

140

Chapter 4

(e) m )(2.0s

m 8.1 33

stu

QQQ ps

−=

+=

and (f) )()( tTTtT ps +=

where . Noting that K 303C 30 ==sT Qm ρ=& . Assuming no resistance in the system . Substitution of Equations (e) and (f) into Equation (a) leads to outin mm && =

( ) ( ) ( ) (g) )(Qs QTcQQTTcQTTdtdVc inppspspppsp

&++=++++ ρρρ

Noting that 0=dt

dTs and using Equation (b), Equation (g) is reduced to

( ) (h) inpppppsp

p TcQTcQQdt

dTVc ρρρ =++

Substitution of known values in Equation (h) leads to

( )

( ) ( )

(i) 1041.21059.61047.2

K 293Kkg

J 100.4)(2.0mkg 1003.1

KkgJ 100.4m 6.1

mkg 1003.1

KkgJ 100.4m 600

mkg 1003.1

869

33

3

33

3332

33

xTxdtdTx

xtux

Txs

xdtdTxx

−=+

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

−⎟⎠⎞

⎜⎝⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅

⎟⎠⎞

⎜⎝⎛

4.30 During the reaction 2 moles of reactant A are consumed to make 1 mole of reactant B. The rate of change of the number of moles of component A due to the reaction is

2 BAk→

(a) 2A

reaction

A kVCdt

dn−=⎟

⎠⎞

⎜⎝⎛

while the rate of change of number of moles of reactant B made during the reaction is

(b) 2A

reaction

B kVCdt

dn=⎟

⎠⎞

⎜⎝⎛

Application of the conservation of species equation to reactant A gives

(c) 2

2

AiAAA

AAAiA

qCkVCqCdt

dCV

kVCqCqCdt

dCV

=++

−−=

Application of the conservation of species equation to reactant B gives

(d) 2

2

BiBAB

ABBiB

qCqCkVCdt

dCV

kVCqCqCdt

dCV

=+−

+−=

141

Chapter 4

Equations (c) and (d) form the mathematical model for the system.

4.31 If the reaction is of order 1.5 then the rate of consumption of reactant A is BAk→

(a) 5.1A

A kVCdt

dC−=

(a) Application of the conservation of species equation to a CSTR with this reaction to species A leads to

(b)

5.1

5.1

AiAAA

AAAiA

qCkVCqCdt

dCV

kVCqCqCdt

dCV

=++

−−=

(b) The concentration of the reactant A in the inlet stream is suddenly perturbed according to

(c) ,,, ipAisAiA CCC += which leads to a perturbation in the concentration of species A in the tank according to

(d) ApAsA CCC += The steady-state condition before the perturbation occurs is obtained using Equation (b) as

(e) 5.1AisAsAs qCkVCqC =+

Substitution of Equations (c) and (d) into Equation (b) leads to

( ) (f) )()( 5.1AipAisApAsApAs

Ap CCqCCkVCCqdt

dCV +=++++

The nonlinear term is rewritten as

(g) 1)(5.1

5.15.1⎟⎟⎠

⎞⎜⎜⎝

⎛+=+

As

ApAsApAs C

CkVCCCkV

Since the binomial expansion is used and truncated after the linear term to provide a linearization of Equation (g) of

1/ <<AsAp CC

(h) 5.1

5.11)(

5.05.1

5.15.1

ApAsAs

As

ApAsApAs

CkVCkVC

CC

kVCCCkV

+=

⎟⎟⎠

⎞⎜⎜⎝

⎛+≈+

Use of Equation (h) in Equation (f) gives

( ) (i) 5.1)( 5.05.1AipAisApAsAsApAs

Ap CCqCkVCkVCCCqdt

dCV +=++++

Subtracting Equation (e) from Equation (i) leads to

(j) 5.1 5.0AipApAsAp

Ap qCCkVCqCdt

dCV =++

4.32 When the reaction occurs in a CSTR the rate at which the reactants are absorbed are

CBAk→+

142

Chapter 4

(c)

(b)

(a)

2

2

2

BAC

BAB

BAA

CCkVdt

dn

CCkVdt

dn

CCkVdt

dn

=

−=

−=

Application of conservation of species to each component in the CSTR leads to

(f)

(e)

(d)

2

2

2

CiBACC

BiBABB

AiBAAA

qCCCkVqCdt

dCV

qCCCkVqCdt

dCV

qCCCkVqCdt

dCV

=−+

=++

=++

4.33 The equations for the species are unaffected by the presence of the jacket, except of course that the jacket affects the temperature. The equations for the species concentrations are the same as those of Equations (a) and (b) of Example 4.17,

( )

(b)

(a)

)/(

)/(

BiBARTEB

AiARTEA

qCqCCVedt

dCV

qCCVeqdt

dCV

=+−

=++

−

−

α

α

The equation for the temperature in the reactor is modified to take into account the heat transfer between the mixture and the jacket,

(c) )(1w

c

TTR

Q −=&

Thus Equation (c) of Example (4.17) is modified as

(d) )(1 )/(

dtdTVcCeVQTqcTqcTT

R pARTE

pipwc

ραλρρ =+−−+−− −&

Consider a control volume encasing the jacket. Application of conservation of energy to the jacket leads to

(e) 1)(1)(21

ww

wc

wwiwww

ww TRRR

TTR

TTmcdt

dTcm

++−−+−= &

In deriving Equation (e) it is assumed that the resistances are in series, is the total mass of the water in the jacket and

wm

ww qm ρ=& is the mass flow rate of the water into the jacket. 4.34 Let be the mass of the drug that has yet to be absorbed and let is the rate of absorption. The rate at which the mass of the drug to be absorbed changes is

)(tma ak

(a) aaa mk

dtdm

−=

Equation (a) is rearranged to

143

Chapter 4

(b) 0=+ aaa mk

dtdm

(a) Considering the one-compartment model, application of conservation of mass to the drug in the control volume leads to

Rate of accumulation =rate of absorption –rate of elimination

(c) 0=−+

−=

aae

eaa

mkVCkdtdCV

VCkmkdtdCV

Equations (b) and (c) constitute the mathematical model for the one-compartment model. (b) The two-compartment model is changed from that of Example 4.30 only by replacing the infusion term by the absorption term. The three equations are

(f) 0

(e) 0)(

(d) 0

21

21

=+−

=−−++

=+

ttppt

t

aattppep

p

aaa

CVkCVkdt

dCV

mkCVkCVkkdt

dCV

mkdt

dm

144

Chapter 5

5. Laplace Transforms

5.1 Recall ( )tt eet ωωω −−=21)sinh( . Thus

( )

(a) 1121

21

21)sinh(L

0

)(

0

)(

0

)(

0

)(

0

⎥⎦⎤

⎢⎣⎡

++

−−=

⎥⎦

⎤⎢⎣

⎡−=

⎥⎦⎤

⎢⎣⎡ −=

∞+−∞−−

∞+−∞ −−

∞−−

∫∫

∫

tsts

tsts

ttst

es

es

dtedte

dteeet

ωω

ωω

ωω

ωω

ω

The exponential functions in Equation (a) approach zero as ∞→t if ω>)Re(s . Thus,

( )

(b)

))(()(

21

1121)sinh(L

22 ωω

ωωωωωω

ω

−=

−+−−+

=

⎥⎦⎤

⎢⎣⎡

+−

−=

s

ssssss

t

5.2 By definition

)a()sin(sin0

L ∫∞

−= dtetttt stωω

Application of integration by parts to Equation (a) with

(c) )( )sin((b)

tgvdttedvdtdutu

st ==

==− ω

where

[ ] )d()cos()sin()( 22 ttss

etgst

ωωωω

++

−=−

Equation (d) is obtained through integration by parts. Substitution into Equation (a) gives

[ ] [ ] )e()cos()sin()cos()sin()(0

220

22 ∫∞ −∞=

=

−

⎭⎬⎫

⎩⎨⎧

++

−−⎭⎬⎫

⎩⎨⎧

++

−= dtttss

ettss

tesFstt

t

st

ωωωω

ωωωω

Noting that , Equation (e) reduces to 0for0lim >→−

∞→ste st

t

[ ] )f()cos()sin()(0

22 ∫∞ −

⎭⎬⎫

⎩⎨⎧

++

= dtttss

esFst

ωωωω

Equation (f) is rewritten as

)g()cos()sin()( 2222 LL ts

ts

ssF ωω

ωωω +

++

=

145

Chapter 5

The definition of the transform, applying integration by parts, is used in Example 5.3 leading to

)h()sin( 22 Lω

ωω+

=s

t

A similar procedure is used to show that

)i()cos( 22 Lω

ω+

=s

st

Substitution of Equations (h) and (i) in Equation (g) leads to

( ))j(2

)(

222

22222222

ω

ωωω

ωω

ωω

+=

⎟⎠⎞

⎜⎝⎛

+⎟⎠⎞

⎜⎝⎛

++⎟

⎠⎞

⎜⎝⎛

+⎟⎠⎞

⎜⎝⎛

+=

ss

ss

sssssF

5.3 Recall that 32 2

st =L . Thus application of the first shifting theorem leads to

)a()3(

2

2

3

33

3

232

LL

−=

=

=

−→

−→

s

s

tet

ss

ss

t

5.4 It is noted that

)b(4

2)2sin(

)a(4

)2cos(

2

2

L

L

+=

+=

st

sst

In order to apply the second shifting theorem the transform must be of the form `)()( atuatf −−L . That is wherever t appears in the function to be transformed it must

appear as t-a. This is not the case in the problem at hand. However algebraic manipulation may be used such that

)c(]2)1(2cos[)]11(2cos[)2cos(

+−=+−=

ttt

Use of a trigonometric identity for the cosine of the sum of two angles allows Equation (c) to be rewritten as

)d()2sin()]1(2sin[)2cos()]1(2cos[)2cos( −−−= ttt Thus

[ ] )e()1()2sin()]1(2sin[)2cos()]1(2cos[)1()2cos( LL −−−−=− tutttut Application of the linearity of the transform leads to

)f()1()]1(2sin[)2sin()1()1(2cos[)2cos()1()2cos( LLL −−−−−=− tuttuttut Application of the Second Shifting Theorem to Equation (f) leads to

146

Chapter 5

( ) )g(49093.04161.0

42)2sin(

4)2cos()1()2cos(

2

22

L

++

−=

+−

+=−

−

−−

sess

essetut

s

ss

5.5 Noting that 2

1s

t =L the First Shifting Theorem is used to determine

(a) )3(

1L 23

+=−

ste t

Linearity and the Second Shifting Theorem are used leading to

(b) 3

2)3()3(

1

L2 )2()2(L)(L

)2()22(L)(L )2(L)(L)]2()([L

26

2

26

2

)2()2(36

)2(363

)22(33

333

+−

+−

+=

−

−−−=

−+−−=

−−=−−

−−

−−

−−−−

−−−−

+−−−

−−−

see

see

s

eetuetetute

tuettutetutetutetutute

ss

tut

tt

tt

ttt

5.6 Table 5.1 is used to determine that

)a(16

)4cos( 2 L+

=s

st

Application of linearity, the property of differentiation of the transform and Equation (a) leads to

( )( )

( ))b(

16162

16)2()1(162

162

)4cos(2)4cos(2

22

2

22

2

2

LL

+

−=

+

−+−=

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

+−=

=

ss

ssss

ss

dsd

tttt

5.7 The function is written using unit step functions as

[ ] )a()4()1()1()(20)]4()1([20)]1()([20)(

−−−−−=−−−+−−=

tututttutututututtf

Application of linearity of the transform leads to [ ]

[ ] )b()4()1()1()(20)4()1()1()(20)(

LLL LL

−−−−−=−−−−−=

tututttutututttutf

Application of the Second-Shifting Theorem to each of the transforms lead to

)c(120)(4

22 L ⎥⎦

⎤⎢⎣

⎡−−=

−−

se

se

stf

ss

147

Chapter 5

5.8 The function of Figure P5.8 has a mathematical formulation in terms of unit step functions as

[ ] [ ](a) )4(20)1()1(20)(20

)4()1(20)1()(20)(−+−−−=

−−−+−−=tututttu

tututututtF

Thus, using Equation (a) and linearity of the transform, (b) )4(L20)1()1(L20)(L20)(L −+−−−= tututttutF

Recall s

eatus

tas−

=−= )(12 L ,L and from the Second Shifting Theorem

2)1()1(setut

s−

=−−L . Thus, Equation (b) becomes

(c) 202020)(L4

22 se

se

stF

ss −−

+−=

5.9 Application of transform pair 1 from Table 5.1 leads to

)a(2

12 L+

=−

se t

Linearity of the transform shows that [ ] )b()3(2)(2)3()(2 222 LLL −−=−− −−− tuetuetutue ttt

The first term on the right-hand side of Equation (c) is evaluated directly using Equation (a). Determination of the transform of the second term on the right-hand side of Equation (b) requires application of the Second Shifting Theorem. However it is not in the form for direct application of the shifting theorem. Properties of the exponential function and linearity of the transform are used to rewrite Equation (b) as

[ ] )c()3(2)(2

)3(2)(2)3(2)(2)3()(2

)2(262

)2(262

)33(222

LL

LL

LLL

−−=

−−=

−−=−−

−−−−

−−−−

+−−−−

tueetuetueetue

tuetuetutue

tt

tt

ttt

The Second Shifting Theorem is applied to the second-term on the right-hand side of Equation (c) leading to

[ ] )d(2

22

2)3()(23

62 L+

−+

=−−−

−−

see

stutue

st

5.10 The transform is evaluated using the convolution property. Noting that the convolution between two functions, f(t) and g(t) is defined as

)a()()()()(0

∫ −=×t

ftgftgtf τττ

and that )b()()()()( L sGsFtgtf =×

then

148

Chapter 5

( ) )c(204

416)2(

41

)4sin()](2sin[

22

22

2

0

)(

LLL

++=

⎟⎟⎠

⎞⎜⎜⎝

⎛++

⎟⎠⎞

⎜⎝⎛=

=⎭⎬⎫

⎩⎨⎧

− −−−∫

sss

ss

tetdte tt

t τττ τ

5.11 It is given that

)a(21

Ls

t π=

⎭⎬⎫

⎩⎨⎧ −

(a) Note that ( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛=

−21

21

ttt . Thus from the property of differentiation of the transform

using Equation (a)

)b(2

21

23

23

21

21

21

LL

s

s

sdsd

sdsd

tdsdt

π

π

π

π

=

⎟⎠⎞

⎜⎝⎛−−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎭⎬⎫

⎩⎨⎧

−=⎭⎬⎫

⎩⎨⎧

−

−

−

(b) Another application of the property of differentiation of the transform, using Equation (b) leads to

149

Chapter 5

)c(4

3

23

2

2

2

25

25

23

23

21

23

LL

−

−

−

−

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎭⎬⎫

⎩⎨⎧

−=⎭⎬⎫

⎩⎨⎧

s

s

sdsd

sdsd

tdsdt

π

π

π

π

(c) Use of the First Shifting Property and Equation (b) leads to

( ))d(

32

2

23

323

3

21

321

LL

+=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

+→

+→

−

s

s

tet

ss

ss

t

π

π

5.12 The final value theorem is applied to the transform leading to

)a(241

)15)(4)(2(5

)15)(4)(3(5lim

)15)(4)(3(5lim

)(lim)(lim

2

2

0

2

2

0

0

=

=

⎥⎦

⎤⎢⎣

⎡+++

+=

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+++

+=

=

→

→

→∞→

ssss

ssssss

ssXtx

s

s

st

150

Chapter 5

5.13 The initial value theorem is applied to the transform leading to

)a(41

17643lim

17643lim

)(lim)0(

2

2

2

=

⎟⎟⎠

⎞⎜⎜⎝

⎛++

+=

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

+++

=

=

∞→

∞→

∞→

ssss

ssss

ssXx

s

s

s

5.14 It is given that 1

1)()(20+

==s

tJsF L

(a) Using the property of derivative of the transform

[ ]

( )

( ))a(

1

)2(121

11

)()(

23

2

23

2

2

0

L

+=

+⎟⎠⎞

⎜⎝⎛−−=

⎥⎦

⎤⎢⎣

⎡

+−=

−=

−

s

s

ss

sdsd

sFdsdttJ

(b) The change of sale property with 21

=a is used to evaluate

)b(4

1

121

121

21

21)2(

2

2

0

L

+=

+⎟⎠⎞

⎜⎝⎛

=

⎟⎠⎞

⎜⎝⎛=

s

s

sFtJ

(c) Knowing that ( ))()( 01 tJdtdtJ −= and 1)0(0 =J , use of the property of transform of

the first derivative leads to

151

Chapter 5

[ ]

)c(1

1

11

)1()(

))(()(

2

2

0

01

LL

+−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡−

+−=

−−=

⎟⎠⎞

⎜⎝⎛−=

s

ss

s

JssF

tJdtdtJ

5.15 The definitions of the hyperbolic trigonometric functions are

( ) ( )

( ) )b(21)sinh(

)a(21cosh

tt

tt

eet

eet

ωω

ωω

ω

ω

−

−

−=

+=

Substitution of Equations (a) and (b) into the transform and using linearity of the transform leads to

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) [ ]

( ) ( ) [ ] )c(coscos21

sinsin21

cos21sin

21sinhcoscoshsin

LL

LL

LL

tete

tete

teeteetttt

tt

tt

tttt

ωω

ωω

ωωωωωω

ωω

ωω

ωωωω

−

−

−−

−−

+=

⎭⎬⎫

⎩⎨⎧ −−+=−

Applications of transform pairs from Table 5.1 and the First Shifting Theorem give

( ) ( )

( ) ( )

( ) ( )

( ) ( )

(g) cosL

(f) cosL

(e) sinL

(d) sinL

22

22

22

22

ωωωω

ωωωω

ωωωω

ωωωω

ω

ω

ω

ω

+++

=

+−−

=

++=

+−=

−

−

sste

sste

ste

ste

t

t

t

t

Substituting Equations (d)-(g) in Equation (c) leads to

152

Chapter 5

( ) ( ) ( ) ( ) ( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )( )[ ] ( )[ ]( )[ ]( )[ ] (h) )2()2(

21

2221

21

21

21

21sinhcoscoshsinL

2222

2222

2222

2222

2222

2222

2222

ωωωωωωωωωω

ωωω

ωωω

ωωω

ωωω

ωωω

ωωω

ωωω

ωωω

ωωω

ωωωωωωω

+−+++−++++−−

=

⎥⎦

⎤⎢⎣

⎡

+++

++−

−−=

⎥⎦

⎤⎢⎣

⎡

+++

+++

+

⎥⎦

⎤⎢⎣

⎡

+−−

−+−

=

⎥⎦

⎤⎢⎣

⎡

+++

−+−

−−

⎥⎦

⎤⎢⎣

⎡

+++

+−=−

ssssss

ss

ss

ss

s

ss

s

ss

ss

sstttt

After a little more algebra, Equation (h) reduces to

( ) ( ) ( ) ( ) )i(4

4sinhcoscoshsin 22

3

Lω

ωωωωω+

=−s

tttt

5.16 The definition of the Laplace transform of a function is

(a) )()(0∫∞

−= dtetfsF st

Note that, from the definition of the Laplace transform

(b) )()(0∫ ∫∫∞ ∞

−∞

=s

st

s

dsdtetfdssF

Interchanging the order of integration in Equation (b) leads to

(c) )(1

1)(

)(

)()(

0

0

0

0

∫

∫

∫ ∫

∫ ∫∫

∞−

∞ ∞=

=

−

∞ ∞−

∞ ∞−

∞

⎥⎦⎤

⎢⎣⎡=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

=

dtetft

dtet

tf

dtdsetf

dtdsetfdssF

st

s

ss

st

s

st

s

st

s

Equation (c) shows that

153

Chapter 5

(d) )()(1L ∫∞

=⎭⎬⎫

⎩⎨⎧

s

dssFtft

Application of Equation (d) with )]cos(1[2)( ttf ω−= gives

( )(e) ln

2

22)cos1(2L

2

22

22

2

22

⎟⎟⎠

⎞⎜⎜⎝

⎛ +=

+=

⎥⎦⎤

⎢⎣⎡

+−=

⎭⎬⎫

⎩⎨⎧ −

∫

∫∞

∞

ss

dsss

dss

ss

tt

s

s

ω

ωω

ωω

5.17 The definition of the Laplace transform of a function is

(a) )()(0∫∞

−= dtetfsF st

Note that, from the definition of the Laplace transform

(b) )()(0∫ ∫∫∞ ∞

−∞

=s

st

s

dsdtetfdssF

Interchanging the order of integration in Equation (b) leads to

(c) )(1

1)(

)(

)()(

0

0

0

0

∫

∫

∫ ∫

∫ ∫∫

∞−

∞ ∞=

=

−

∞ ∞−

∞ ∞−

∞

⎥⎦⎤

⎢⎣⎡=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

=

dtetft

dtet

tf

dtdsetf

dtdsetfdssF

st

s

ss

st

s

st

s

st

s

Equation (c) shows that

(d) )()(1L ∫∞

=⎭⎬⎫

⎩⎨⎧

s

dssFtft

Application of Equation (d) with )sin()( ttf ω= gives

(e) tan

sin1L

1

22

⎟⎠⎞

⎜⎝⎛=

+=

⎭⎬⎫

⎩⎨⎧

−

∞

∫

s

dss

tt s

ω

ωωω

154

Chapter 5

5.18 Let . Then by definition )()( tfsF L=

(a) L0∫∞

−⎟⎠⎞

⎜⎝⎛=

⎭⎬⎫

⎩⎨⎧

⎟⎠⎞

⎜⎝⎛ dte

atf

atf st

Let

(d) (c)

(b)

advdtavtatv

==

=

Changing the variable of integration in Equation (a) through introduction of Equations (b)-(d) leads to

(e) )(

)(

)()(L

0

)(

0

asaF

dvevfa

advevfatf

vas

asv

=

=

=⎭⎬⎫

⎩⎨⎧

⎟⎠⎞

⎜⎝⎛

∫

∫∞

−

∞−

Equation (e) is a statement of the change of scale property. 5.19 The property of differentiation of the transform states

(a) )0()(L xssXdtdx

−=⎭⎬⎫

⎩⎨⎧

Rearrangement of Equation (a) leads to

(b) )0(L)( xdtdxssX +⎭⎬⎫

⎩⎨⎧=

Taking the limit of Equation (b) for large s leads to

(c) )0(lim

)0(Llim)(lim

0

xdtedtdx

xdtdxssX

st

s

ss

+⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎥⎦

⎤⎢⎣

⎡+

⎭⎬⎫

⎩⎨⎧=

∫∞

−

∞→

∞→∞→

Since the integral on the right-hand side of Equation (c) exists the limiting process may be interchanged with the integration,

(d) )0(lim)(lim0

xdtedtdxssX st

ss+⎟

⎠⎞

⎜⎝⎛= ∫

∞−

∞→∞→

Since the exponential functions approaches zero for large s, the limit in Equation (d) is zero leading to the statement of the Initial Value Theorem

(e) )0()(lim xssXs

=∞→

155

Chapter 5

5.20 Let . Then )()( tfsF L=

(a) )()(

0⎥⎦

⎤⎢⎣

⎡= ∫

∞− dtetf

dsd

dssdF st

If the integral exists, the order of differentiation and integration in Equation (a) may be interchanged

[ ]

( )

(b) )(

)(

)()(

0

0

0

∫

∫

∫

∞−

∞−

∞−

−=

−=

=

dtettf

dttetf

dtetfdsd

dssdF

st

st

st

Equation (b) is rewritten as

(c) )()(Lds

sdFttf −=

Equation (c) is the statement of differentiation of the transform for n=1. The general property is proved by induction. Assume the property holds for some value of n, that is

( ) (d) )()1()(L sFdsdtft n

nnn −=

and consider

( ) ( )

(e) )()()1(

)1(

⎥⎦

⎤⎢⎣

⎡=+

+

sFdsd

dsdsF

dsd

n

n

n

n

Use of Equation (d) and the definition of the transform in Equation (e) leads to

( )

( )

(g) )(L)1(

)()1(

))(()1(

)()1(

)()1()(

11

0

11

0

0

0)1(

)1(

tft

dtetft

dtettft

dtetftdsd

dtetftdsdsF

dsd

nn

stnn

stnn

stnn

stnnn

n

++

∞−++

∞−

−∞

∞−

+

+

−=

−=

−−=

−=

⎥⎦

⎤⎢⎣

⎡−=

∫

∫

∫

∫

Equation (g) is rearranged as

( ) (h) )()1()(L )1(

)1(11 sF

dsdtft n

nnn

+

+++ −=

156

Chapter 5

Equation (c) shows that the property of differentiation of the transform holds for n=1. Equation (h) shows that if the property holds for any value of n, then it also holds for n+1. Thus the property is proved by induction. 5.21 The Laplace transform of a periodic function of period T is calculated by

(a) )(1

1)(0∫ −

−−=

Tst

Ts dtetfe

sF

The function for this problem is periodic of period 0tT = whose definition over one period is

(b)

2

32

20

)(0

00

00

⎪⎩

⎪⎨

⎧

<<−

<<=

ttt

F

ttF

tF

Use of Equation (b) in Equation (a) leads to

(c) 32

351

11

32

321

11

32

11

32

11)(

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

20

220

2

020

0

2

0

2

00

⎥⎦

⎤⎢⎣

⎡+−⎟

⎠⎞

⎜⎝⎛

−=

⎥⎦

⎤⎢⎣

⎡−+−⎟

⎠⎞

⎜⎝⎛

−=

⎥⎦

⎤⎢⎣

⎡+−

−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎠⎞

⎜⎝⎛−+

−=

−−

−

−−−

−

=

=

−=

=

−−

−−− ∫∫

stst

st

st

stst

st

ttt

t

stt

t

t

stst

t

t

st

t

stst

ees

Fe

eees

Fe

esF

es

Fe

dteFdteFe

sF

5.22 The transform may be written as

)a(9

331

93

913)(

22

2

⎟⎠⎞

⎜⎝⎛

++⎟

⎠⎞

⎜⎝⎛

+=

++

=

sss

sssX

The inverse transform is obtained by applying linearity and use of transform pairs from Table 5.1

)b()3sin(31)3cos(3

93

31

93)( 2

12

1

LL

tt

ssstx

+=

⎭⎬⎫

⎩⎨⎧

++

⎭⎬⎫

⎩⎨⎧

+= −−

5.23 A partial fraction decomposition is assumed as

(a) 9)9(

1322 ++

+=++

sCBs

sA

sss

157

Chapter 5

The residue, A, is determined as

(b) 91

)9()13(

02 =++

==sss

ssA

Thus

(c) 9

91

)9(13

22 ++

+=++

sCBs

ssss

Evaluation of Equation (c) at s=1 leads to

( )

(d) 2699101

91

104

=+

++=

CB

CB

Evaluation of Equation (c) at s=-1 leads to

(e) 2899

)(101

91

102

=+−

+−+−=−−

CB

CB

Equations (d) and (e) are solved simultaneously yielding 391

=−= CB . Thus

)f(9

391

91)( 2

+

+−+=

s

s

ssX

Linearity and transform pairs from Table 5.1 are used to invert Equation (f) leading to

)g()3sin()3cos(91)(

91)( tttutx +−=

5.24 The denominator of the transform has poles at s=3, -3. Thus the transform has a partial fraction decomposition of the form

)a(339

13)( 212

++

−=

−+

=sA

sA

sssX

The residues are determined as

)b(35

331)3(3

9)13)(3(

321

=++

=

−+−

==ss

ssA

)c(34

331)3(39

)13)(3(

322

=−−+−

=

−++

=−=ss

ssA

Thus

)d(3

43

531)( ⎟

⎠⎞

⎜⎝⎛

++

−=

sssX

158

Chapter 5

The inverse transform of Equation (d) is obtained using the transform as

eat

−=

1L

( ) )e(4531)( 33 tt eetx −+=

5.25 The poles of the transform are 3,3,0 −=s . Thus a partial fraction decomposition is assumed as

)a(33)9(

13 3212

−+

++=

−+

sA

sA

sA

sss

The residues are determined as

( ))b(

91

901)0(39)13(

021

−=−+

=

−+

==sss

ssA

( ))c(

94

)33)(3(1)3(3

9)13)(3(

322

−

=−−−+−

=

−++

=−=sss

ssA

( ))d(

95

)33)(3(1)3(3

9)13)(3(

323

=++

=

−+−

==sss

ssA

Substituting Equations (b)-(d) into Equation (a) leads to

)e(3

53

4191)( ⎟

⎠⎞

⎜⎝⎛

−+

+−−=

ssssX

The inverse transform of Equation (d) is obtained using the transform as

eat

−=

1L

[ ] )f(54)(91)( 33 tt eetutx +−−= −

5.26 A partial fraction decomposition is assumed as

)a(9)9(

1322 ++

+=++

sCBs

sA

sss

The residue, A, is determined as

)b(91

)9()13(

02 =++

==sss

ssA

Thus

(c) 9

91

)9(13

22 ++

+=++

sCBs

ssss

159

Chapter 5

Evaluation of Equation (c) at s=1 leads to

( )

(d) 2699101

91

104

=+

++=

CB

CB

Evaluation of Equation (c) at s=-1 leads to

(e) 2899

)(101

91

102

=+−

+−+−=−−

CB

CB

Equations (d) and (e) are solved simultaneously yielding 391

=−= CB . Thus

( ) )f(19

391

91)( 2

2 ses

s

ssX −−

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+

+−+=

It is noted that )3sin(9

3)3cos(9

)(12

12

11 ts

ts

stus

=⎭⎬⎫

⎩⎨⎧

+=

⎭⎬⎫

⎩⎨⎧

+=

⎭⎬⎫

⎩⎨⎧ −−− L L L . These

transform pairs and the second shifting theorem are used to invert the transform of Equation (f) leading to

( ) (g) )2())2(3sin()2(3cos91

91)()3sin()3cos(

91

91)( −⎥⎦

⎤⎢⎣⎡ −+−−−⎥⎦

⎤⎢⎣⎡ +−= tutttutttx

5.27 The appropriate partial fraction decomposition is

(a) 1613)16)(1)(3(

222

2

++

++

++

=+++

+s

DCssB

sA

ssss

Residues are used to obtain

(b) 5013

]16)3)[(13(2)3(

)16)(1(2

2

23

2

2

−=

+−+−+−

=

+++

=−=sss

sA

(c) 343

]16)1)[(31(2)1(

)16)(3(2

2

21

2

2

=

+−+−+−

=

+++

=−=sss

sB

160

Chapter 5

Thus Equation (a) becomes

(d) 161

1343

31

5013

)16)(3)(1(2

22

2

++

+⎟⎠⎞

⎜⎝⎛

++⎟

⎠⎞

⎜⎝⎛

+−=

++++

sDCs

ssssss

Evaluation of Equation (d) at s=0 gives

(e) 283.1161

1343

31

5013

)16)(3)(1(2

=

+⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛−=

D

D

Evaluation of Equation (d) at s=2, using Equation (e) leads to

(f) 2621.020

283.1231

343

51

5013

)20)(5)(3(6

=

++⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛−=

C

C

Thus

(g) 16

283.12621.01

1343

31

5013)( 2 +

++⎟

⎠⎞

⎜⎝⎛

++⎟

⎠⎞

⎜⎝⎛

+−=

ss

sssX

Inversion of Equation (g) gives (h) )4sin(3208.0)4cos(2621.00882.026.0)( 3 tteetx tt +++−= −−

5.28 Completing the square of the denominator leads to

)a(4)3(

32136

32)(

2

2

++

+=

+++

=

ss

ssssX

For purposes of application of the First Shifting Theorem to invert the transform, Equation (a) is rewritten as

)b(4)3(

223

4)3(32

4)3(3

4)3(32

4)3(3)33(2)(

22

22

2

++−

+++

=

++−

+++

=

+++−+

=

sss

sss

sssX

The First Shifting Theorem, linearity of the inverse transform and transform pairs from Table 5.1 are used to invert Equation (b) to obtain

)c()2cos(23)2sin(2)( 23 tetetx tt −− −=

5.29 The appropriate partial fraction decomposition is

(a) 262

)262(45)(

2

2

2

+++

+=

+++

=

ssCBs

sA

sssssX

161

Chapter 5

The residue theorem is applied leading to

(b) 132

26245

02

2

=

+++

==sss

sA

Use of Equation (b) in Equation (a) leads to

( )

(c) 4134

132

)(26213245

2621

132

26245

2

22

22

2

+⎟⎠⎞

⎜⎝⎛ ++⎟

⎠⎞

⎜⎝⎛ +=

++++=+

+++

+⎟⎠⎞

⎜⎝⎛=

+++

sCsB

sCBssss

ssCBs

ssss

Equating coefficients of like powers of s in Equation (c) leads to

(e) 134

1340

(d) 1363

1325

−=

+=

=

+=

C

C

B

B

Use of Equations (c), (d), and (e) in Equation (a) leads to

(f) 25)1(

6725)1()1(631

131

25)1(4631

131

2624632

131)(

22

2

2

⎥⎦

⎤⎢⎣

⎡++

−+++

+=

⎟⎟⎠

⎞⎜⎜⎝

⎛++−

+=

⎟⎠⎞

⎜⎝⎛

++−

+=

sss

s

ss

s

sss

ssX

Inversion of Equation (f) leads to

(g) )5sin(5

67)5cos(63)(131)( ⎟

⎠⎞

⎜⎝⎛ −+= −− tetetutx tt

5.30 The appropriate partial fraction decomposition for is )(sX

(a) )4(42)4)(2(

7222 +

++

++

=++

+s

Cs

Bs

Ass

s

The residue A is calculated as

162

Chapter 5

(b) 43

)42(7)2(2

)4(72

2

22

=

+−+−

=

++

=−=ss

sA

Use of Equation (b) in Equation (a) leads to

( ) (c) )2()4)(2(44372

)4(421

43

)4)(2(72

2

22

++++++=+

++

++⎟

⎠⎞

⎜⎝⎛

+=

+++

sCssBss

sC

sB

ssss

Evaluating Equation (c) at s=-4 leads to

(d) 21

)24(7)4(2

=

+−=+−

C

C

Evaluating Equation (c) for s=0, using Equation (d) leads to

( )

(e) 43

)2(21)4)(2(4

437)0(2 2

−=

++=+

B

B

Thus

(f) )4(

24

32

341)( 2 ⎥

⎦

⎤⎢⎣

⎡+

++

−+

=sss

sX

Inversion of Equation (f) leads to

[ ] (g) 23341)( 442 ttt teeetx −−− +−=

5.31 The appropriate partial fraction decomposition is

( )

( )(a)

1142

)1(421)(

22

22

++

++

+++

=

+++=

sD

sC

ssBAs

ssssX

Multiplication of Equation (a) by the denominator leads to ( )

( )(b) )44()262()32()(

)42(463)12)(( )42()1)(42(1)(1

23

2232

222

DCBsDCBAsDCBAsCAssDsssCssBAs

ssDsssCsBAs

++++++++++++=

++++++++++=

+++++++++=

Equating coefficients of like powers of s in Equation (b) leads to the following set of equations to solve for the unknown coefficients

163

Chapter 5

(c)

0100

4410262113120101

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

DCBA

The solution of Equation (c) is 31,

31,0,

31

−===−= DCBA . Thus the transform of the

response is

(d) )1(

11

13)1(

13)1(

131

)1(

11

13)1(3

1

)1(1

11

4231)(

222

22

22

⎥⎦

⎤⎢⎣

⎡+

−+

+++

−++

+=

⎥⎦

⎤⎢⎣

⎡+

−+

+++

=

⎥⎦

⎤⎢⎣

⎡+

−+

+++

=

sssss

ssss

ssssssX

Inversion of Equation (d) leads to

( ) ( ) (e) 13sin3

13cos31)( ⎥

⎦

⎤⎢⎣

⎡−+−= − tttetx t

5.32 (a) The MATLAB worksheet to determine the partial fraction decomposition of

(a) )9(

13)( 2 ++

=ssssX

is provided below >> clear >> N=[3 1] N = 3 1 >> D=[1 0 9 0] D = 1 0 9 0 >> [r,p,k]=residue(N,D) r = -0.0556 - 0.5000i -0.0556 + 0.5000i

164

Chapter 5

0.1111 p = 0 + 3.0000i 0 - 3.0000i 0 k = [] >> syms s >> Num=r(2)*(s-p(1))+r(1)*(s-p(2)) Num = (-1/18+1/2*i)*(s-3*i)-(1/18+1/2*i)*(s+3*i) >> Num1=vpa(simplify(Num),3) Num1 = -.111*s+3. >> Thus the appropriate partial fraction decomposition is

(b) 9

3111.0111.0)( 2 ++−

+=s

ss

sX

(b) The MATLAB workspace to help determine the partial fraction decomposition for

(c) 33)9(

13 3212 −

++

+=−+

sA

sA

sA

sss

is provided below >> clear >> N=[3 1] N = 3 1 >> D=[1 0 -9 0] D =

165

Chapter 5

1 0 -9 0 >> [r,p,k]=residue(N,D) r = 0.5556 -0.4444 -0.1111 p = 3 -3 0 k = [] >> The partial fraction decomposition determined from the above is

(d) 3

4444.03

5556.01111.0)(+

−−

+−

=sss

sX

(c) The MATLAB worksheet to determine the partial fraction decomposition of

(e) )262(

45)( 2

2

+++

=sss

ssX

is provided below >> clear >> N=[5 0 4] N = 5 0 4 >> D=[1 2 26 0] D = 1 2 26 0 >> [r,p,k]=residue(N,D)

166

Chapter 5

r = 2.4231 + 0.5154i 2.4231 - 0.5154i 0.1538 p = -1.0000 + 5.0000i -1.0000 - 5.0000i 0 k = [] >> syms s >> Num=r(1)*(s-p(2))+r(2)*(s-p(1)) Num = (63/26+67/130*i)*(s+1+5*i)+(63/26-67/130*i)*(s+1-5*i) >> Num1=vpa(simplify(Num),3) Num1 = 4.85*s-.308 >> The partial fraction decomposition is determined from the MATLAB work as

(f) 262308.085.41538.0)( 2 ++

−+=

sss

ssX

(d) The MATLAB worksheet to determine the partial fraction decomposition for

(g) )4)(2(

72)( 2+++

=ss

ssX

is provided below >> clear >> N=[2 7] N = 2 7

167

Chapter 5

>> syms s >> Ds=(s+2)*(s+4)^2 Ds = (s+2)*(s+4)^2 >> expand(Ds) ans = s^3+10*s^2+32*s+32 >> D=[1 10 32 32] D = 1 10 32 32 >> [r,p,k]=residue(N,D) r = -0.7500 0.5000 0.7500 p = -4.0000 -4.0000 -2.0000 k = [] >> The partial fraction decomposition is determined from the MATLAB workspace as

(h) 2

75.0)4(

5.0475.0)( 2 +

++

++

−=

ssssX

168

Chapter 5

5.33 The mathematical problem is

)b(0)0()a(23 3

==+ −

xexx t&

Taking the Laplace transform of both sides of Equation (a), defining using the linearity property leads to )()( txsX L=

)c(23 3 LL texx −=+&

Noting from Table 5.1 that 3

13

+=−

se tL and using the property of transform of

derivatives Equation (c) is rewritten as

)d(3

2)(3)0()( +

=+−s

sXxssX

Substitution of the initial conditions, Equation (b) into Equation (d) and solving for X(s) results in

)e()3(

2)( 2 +

=s

sX

Equation (e) is inverted using the first shifting theorem and the known transform pair

2

1s

t =L leading to

)f(2)( 3 ttetx −=

5.34 The mathematical problem is

(c) 1)0((b) 0)0((a) 0172

==

=++

xx

xxx

&

&&&

Taking the Laplace transform of both sides of Equation (a), defining using the linearity property leads to )()( txsX L=

)d(0172 LLL =++ xxx &&& Using the property of transform of derivatives Equation (d) is rewritten as

[ ] )e(0)(17)0()(2)0()0()(2 =+−+−− sXsxsXxsxsXs & Substitution of the initial conditions, Equations (b) and (c) into Equation (e) and solving for X(s) results in

)f(172

1)( 2 ++

=ss

sX

Equation (f) is rewritten by completing the square in the denominator of the fraction on its right-hand side

( ))g(

1611)( 22

++

=s

sX

The inverse of the right-hand side of Equation (g) is obtained by using the First Shifting Theorem and known transform pairs leading to

)h()4sin(41)( tetx t−=

169

Chapter 5

5.35 The mathematical problem is

)c(0)0()b(0)0()a()5sin(325

==

=+

xx

txx

&

&&

Taking the Laplace transform of both sides of Equation (a), defining using the linearity property leads to )()( txsX L=

)d()5sin(325 LLL txx =+&&

Noting from Table 5.1 that 25

5)5sin( 2 +=

stL , using the property of transform of

derivatives Equation (d) is rewritten as

)e(25

15)(25)0()0()( 22

+=+−−

ssXxsxsXs &

Substitution of the initial conditions, Equations (b) and (c) into Equation (e) and solving for X(s) results in

( ))f(

2515)( 22

+

=s

sX

There are several ways to invert the transform in Equation (f). Note that

)g(25

525

553)( 22 ⎟

⎠⎞

⎜⎝⎛

+⎟⎠⎞

⎜⎝⎛

+=

sssX

Application of the convolution property leads to

[ ]

[ ]

)h()5(sin)5cos()5cos()5sin()5sin(53

)5sin()5cos()5cos()5sin()5sin(53

)(5sin)5sin(53

)5sin()5sin(53)(

0

2

0

0

0

⎥⎦

⎤⎢⎣

⎡−=

−=

−=

×=

∫∫

∫

∫

tt

t

t

dtdt

dtt

dt

tttx

τττττ

ττττ

τττ

Use of trig identities on the integrands leads to

[ ]

)i()10sin()5cos(101)5cos()5sin(

101)10cos()5sin(

101

103

)10sin()5cos(101)5cos()10cos()5sin(

101

103

)10cos(1)5cos(21)10sin()5sin(

21

53)(

000

00

⎭⎬⎫

⎩⎨⎧ +−+−=

⎭⎬⎫

⎩⎨⎧ +−−=

⎭⎬⎫

⎩⎨⎧

−−=

=

=

=

=

=

=

∫∫

ttttttt

ttt

dtdttx

ttt

tt

τ

τ

τ

τ

τ

ττττ

ττττ

Further algebra gives

170

Chapter 5

( )

)j()5sin(100

2)5cos(103

)105sin(101)5sin(

101)5cos(

103

)10sin()5cos()10cos()5sin(101)5sin(

101)5cos(

103)(

ttt

ttttt

ttttttttx

+=

⎥⎦⎤

⎢⎣⎡ −−+=

⎥⎦⎤

⎢⎣⎡ −−+=

5.36 The mathematical problem is

(c) 0)0((b) )((a) 749 2

==

=+ −

xx

exx t

&

&&

00

Taking the Laplace transform of both sides of Equation (a), defining using the linearity property leads to )()( txsX L=

)d(749 2 LLL texx −=+&&

Noting from Table 5.1 that 2

12

+=−

se tL and using the property of transform of

derivatives Equation (d) is rewritten as

)e(2

7)(49)0()0()(2 +

=++−s

sXxsxsXs &

Substitution of the initial conditions, Equations (a) and (b) into Equation (e) and solving for X(s) results in

)f()49)(2(

7)( 2 ++

=ss

sX

The appropriate partial fraction decomposition of the right-hand side of Equation (f) is

)g(492)49)(2(

722 ++

++

=++ s

CBss

Ass

The residue A is determined as

)h(537

49)2(7

297

2

22

=

+−=

+=

−=ssA

Substitution of Equation (h) in Equation (g) gives

( ) ( ) )i()2()495377

492537

)49)(2(7

2

22

++++=

++

++

=++

sCBss

sCBs

sss

171

Chapter 5

Evaluation of Equation (i) at s=0 leads to

)j(5314

53491

27

)2()49(5377

=

⎟⎠⎞

⎜⎝⎛ −=

+=

C

C

Evaluation of Equation (i) at s=2 leads to

)k(5314

)4(5314)53(

5377

−=

⎟⎠⎞

⎜⎝⎛ ++=

B

Bs

Use of Equations (g), (h), (j), and (k) in Equation (f) leads to

)l(4922

21

537)( 2 ⎟

⎠⎞

⎜⎝⎛

+−

++

=s

ss

sX

Inversion of Equation (l) using transform pairs found in Table 5.1 leads to

)m()7sin(72)7cos(2

537)( 2 ⎥⎦

⎤⎢⎣⎡ +−= − ttetx t

5.37 The mathematical problem is

(d) 5.0)0((c) 0)0((b) )((a) 050252

===

=+++

xxx

xxxx

&&&

&&&&&&

00

Taking the Laplace transform of both sides of Equation (a), defining using the linearity property leads to )()( txsX L=

)e(0502525 LLLL =+++ xxxx &&&&&& Using the property of transform of derivatives Equation (d) is rewritten as

[ ][ ] )f(0)(50)0()(25

)0()0()(2)0()0()0()( 223

=+−+−−+−−−

sXxssXxsxsXsxsxssxsXs &&&&

Substitution of the initial conditions, Equations (a), (b), and (c) into Equation (f) and solving for X(s) results in

)g(50252

5.0)( 23 +++

=sss

sX

The poles of the transform (the roots of the polynomial in the denominator of Equation (g) are . Thus Equation (g) can be written as js 0.5,2 ±−=

)h()25)(2(

5.0)( 2 ++

=ss

sX

A partial fraction decomposition leads to

172

Chapter 5

)i(252

)( 2 ++

++

=s

CBss

AsX

Application of the residue theorem leads to

)j(581

295.0

)25)(2()2(5.0lim 22

=

=

+++

=−→ ss

sAs

Thus

( ) )k()2)((255815.0

2521

581

)25)(2(5.0

2

22

++++=

++

++

=++

sCBss

sCBs

sss

Setting s=0 in Equation (k) leads to

( )

)l(291

225581

21

=

+=

C

C

Setting s=-1 in Equation (l) leads to

( )

)m(581

)1(29126

581

21

−=

⎟⎠⎞

⎜⎝⎛ +−+=

B

B

Substitution of Equations (j), (l), and (m) into Equation (h) gives

)n(252

21

581)( 2 ⎟

⎠⎞

⎜⎝⎛

++−

++

=s

ss

sX

Inversion of Equation (n) leads to

)o()5sin(52)5cos(

581)( 2 ⎥⎦

⎤⎢⎣⎡ +−= − ttetx t

5.38 The mathematical problem is

(d) 0)0((c) 0)0((b) 03(a) 22

2

1

212

211

==

=+−=−+ −

xx

xxxexxx t

&&

Taking the Laplace transforms of Equations (a) and (b) and using linearity of the transform leads to

173

Chapter 5

(f) 0)(3)(L

(e) 3

2)()(2L

212

211

=+−+

=−+

sXsXxs

sXsXx

&

&

Using the property of the transform of the first derivative and application of Equations (c) and (d) leads to

(h) 0)()3()(

(g) 3

2)()()2(

21

21

=++−+

=−+

sXssXs

sXsXs

Equations (g) and (h) are summarized in a matrix form as

(i) 0

32

)()(

3112

2

1

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡+−−+

ssXsX

ss

Cramer’s rule is applied to determine the solution of Equation (i) as

( )( )

(k) )55)(3(

2

5501

322

)(

(j) 553

)1(2

311230

11

2

)(

2

22

2

1

+++=

++−

++

=

++++

=

+−−++

−+

=

sss

ss

ss

sX

sssss

ss

s

sX

MATLAB is used to help determine the partial fraction decompositions for Equations (j) and (k). The MATLAB workspace is >> clear >> syms s >> Ds=(s+3)*(s^2+5*s+5) Ds = (s+3)*(s^2+5*s+5) >> Ds1=expand(Ds) Ds1 = s^3+8*s^2+20*s+15

174

Chapter 5

>> D=[1 8 20 15] D = 1 8 20 15 >> N=[2 2] N = 2 2 >> [r,p,k]=residue(N,D) r = -3.7889 4.0000 -0.2111 p = -3.6180 -3.0000 -1.3820 k = [] >> N=[2] N = 2 >> [r,p,k]=residue(N,D) r = 1.4472 -2.0000 0.5528

175

Chapter 5

p = -3.6180 -3.0000 -1.3820 k = [] >> The partial fraction decompositions determined using MATLAB are

(m) 382.1

5538.0618.3

447.13

2)(

(l) 382.1

2111.0618.3

789.33

4)(

2

1

++

++

+−

=

+−

+−

+=

ssssX

ssssX

Inversion of Equations (l) and (m) leads to

(o) 5538.0447.12)(

(n) 2111.0789.34)(382.1618.33

2

382.1618.331

ttt

ttt

eeetx

eeetx−−−

−−−

++−=

−−=

5.39 The mathematical problem is

(c) 0)0((b) 001.)0((a) 02502010

==

=++

xx

xxx

&

&&&

Defining )()( txsx L= , taking the Laplace transform of Equation (a), and using linearity of the transform leads to

(d) 0L250L20L10 =++ xxx &&& Using the properties of transforms of first and second derivatives and use of the initial conditions of Equations (b) and (c) leads to

[ ] [ ]

(e) 252002.0001.0

250201002.001.0)(

02.01.)()2502010(0)(250]001.)([20]001.)([10

0)(250)0()(20)0()0()(10

2

2

2

2

2

+++

=

+++

=

+=++

=+−+−

=+−+−−

sss

ssssx

ssxsssxsxsssxs

sxxsxsxsxsxs &

The transform of Equation (e) has only complex poles. The technique of completing the square is used to write the denominator as . Thus 24)1( 2 ++s

176

Chapter 5

(f) 24)1(

001.024)1(

)1(001.0

24)1(002.0001.0)(

22

2

+++

+++

=

+++

=

sss

sssx

Inversion of Equation (f) leads to

( ) ( ) (g) 24sin24124cos001.0)( ⎥

⎦

⎤⎢⎣

⎡+= −− tetetx tt

5.40 The mathematical problem is

(c) 0)0((b) 0)0((a) )400sin(2002502010

==

=++

xx

txxx

&

&&&

Defining , taking the Laplace transform of Equation (a), and using linearity of the transform leads to

)()( txsX L=

(d) )400sin(L200L250L20L10 txxx =++ &&& Using the properties of transforms of first and second derivatives and use of the initial conditions of Equations (b) and (c) leads to

[ ] [ ]

( )( )

( )( ) )e(106.1252

108106.12502010

108)(

106.1108.)()2502010(

106.1108)(250)(20)(10

)400()400(200)(250)0()(20)0()0()(10

522

3

522

4

52

42

52

42

222

xsss

xxsss

xsX

xsxsXss

xsxsXssXsXs

ssXxssXxsxsXs

+++=

+++=

+=++

+=++

+=+−+−− &

The appropriate partial fraction decomposition is of the form

)f(106.1252

)( 522 xsDCs

ssBAssX

++

+++

+=

Multiplying through by the denominator of the right-hand side of Equation (e) leads to

( ) ( ) ( )( ) )g(25106.1

225106.12108)252)(()106.1)((108

5

5233

2523

DBxsDCAxsDCBsCAx

ssDCsxsBAsx

++

+++++++=

++++++=

Equating coefficients of like powers of s leads to

177

Chapter 5

)h(

108000

250106.102250106.112100101

35

5

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

xDCBA

xx

Then solution to the set of equations specified in Equation (h) is )i(05.0,1025.6,05.0,1025.6 77 −=−=== −− DxCBxA

Substitution of Equation (i) into Equation (f) leads to

)j(106.1

05.01025.6252

05.01025.6)( 52

7

2

7

xssx

sssxsX

++

−+++

=−−

Completing the square in the denominator of the first term leads to

(k) 106.1

05.01025.624)1(

05.01025.6)( 52

7

2

7

xssx

ssxsX

++

−+++

=−−

Inversion of Equation (k) leads to

)l()400sin(40005.)400cos(1026.6

)24sin(2505.)24cos(1025.6)(

7

7

ttx

tetextx tt

++

+=

−

−−−

5.41 The mathematical problem is

(c) 0)0((b) 0)0((a) )]2()([2002502010

==

−−=++

xx

tutuxxx

&

&&&

Defining , taking the Laplace transform of Equation (a), and using linearity of the transform leads to

)()( txsX L=

(d) )2()(L200L250L20L10 −−=++ tutuxxx &&& Using the properties of transforms of first and second derivatives and use of the initial conditions of Equations (b) and (c) leads to

[ ]

( )

( )(e) )1(

)252(200

)1(2502010

200)(

)1(200)(2502010

)1(200)(250)]0()([20)0()0()(10

22

22

22

22

s

s

s

s

esss

esss

sX

es

sXss

es

sXxsXxsxsXs

−

−

−

−

−++

=

−++

=

−=++

−=+−+−− &

The appropriate partial fraction decomposition is

178

Chapter 5

( )

( )

( )

(g) 124)1(

824)1(

)1(88

124)1(

1688

1252

1688)(

222

22

22

s

s

s

ess

ss

es

ss

ess

ss

sX

−

−

−

−⎥⎦

⎤⎢⎣

⎡++

−++

+−=

−⎥⎦

⎤⎢⎣

⎡++

+−=

−⎟⎠⎞

⎜⎝⎛

+++

−=

Equation (g) is inverted using the first shifting theorem, the second shifting theorem and known transforms leading to

(h) )2()]2(90.4sin[63.1)2()]2(90.4cos[8 )2(8)90.4sin(63.1)90.4cos(8)(8)(

)2()2( −−−−−−

−−−−=−−−−

−−

tutetutetutetetutx

tt

tt

5.42 The mathematical problem is

(b) 0)0(

(a) )500sin(100010410001.00

5

=

=++ ∫i

tidtxidtdi t

Define )()( tisI L= . Taking the Laplace transform of Equation (a) and using linearity of the transform gives

(c) )500sin(L1000L104L1000L1.00

5 tidtxidtdi t

=⎭⎬⎫

⎩⎨⎧

++⎭⎬⎫

⎩⎨⎧ ∫

Use of properties of transform of the first derivative and transform of an integral and application of the initial condition, Equation (b), leads to

[ ]

( )

( ) (d) )500(10410000

105)(

)500(10410001.0105)(

)500(105)(10410001.0

)500()500(1000)(104)(1000)0()(1.0

2262

6

2252

5

22

55

22

5

+++=

+++=

+=⎟⎟

⎠

⎞⎜⎜⎝

⎛++

+=++−

sxsssxsI

sxsssxsI

sxsI

sxs

ssI

sxsIissI

The roots of are determined as 62 10410000 xss ++ 4.4171 −=s and . Thus Equation (d) can be rewritten as

32 1058.9 xs −=

(e) )500)(1058.9)(4.417(

105)( 223

5

+++=

sxsssxsI

which has a partial fraction decomposition of

(f) 5001058.94.417

)( 22321

++

++

++

=s

CBsxs

As

AsI

179

Chapter 5

The residues are determined as

(g) 0537.0)500)(1058.9(

105

4.417223

5

1 −=++

=−=ssxs

sxA

(h) 1068.5.0)500)(4.417(

105 3

1058.922

5

23

−

−=

=++

= xss

sxAxs

Thus

(i) 5001058.9

1068.54.417

0537.0)500)(1058.9)(4.417(

105223

3

223

5

++

++

++−

=+++

−

sCBs

xsx

ssxsssx

Evaluation of Equation (i) at s=0 leads to 01.32=C . Further evaluation of Equation (i) at s=500 leads to B=0.1221. Thus

(j) 500

01.321221.01058.9

1068.54.417

0537.0)( 223

3

++

++

++

−=

−

ss

xsx

ssI

Inversion of Equation (j) leads to (k) )500(0640.0)500cos(1221.061068.50537.0)(

31058.934.417 ttexeti txt +++−= −−− 5.43 The mathematical problem is

(b) 0)0(

(a) 100010410001.0 02.0

0

5

=

=++ −∫i

eidtxidtdi t

t

Define )()( tisI L= . Taking the Laplace transform of Equation (a) and using linearity of the transform gives

(c) L1000L104L1000L1.0 02.0

0

5 tt

eidtxidtdi −=

⎭⎬⎫

⎩⎨⎧

++⎭⎬⎫

⎩⎨⎧ ∫

Use of properties of transform of the first derivative and transform of an integral and application of the initial condition, Equation (b), leads to

[ ]

( )

( ) (d) )02.0(10410000

101)(

)02.(10410001.01000)(

02.1000)(10410001.0

02.01000)(104)(1000)0()(1.0

62

4

52

5

5

+++=

+++=

+=⎟⎟

⎠

⎞⎜⎜⎝

⎛++

+=++−

sxsssxsI

sxssssI

ssI

sxs

ssI

sxsIissI

The roots of are determined as 62 10410000 xss ++ 4.4171 −=s and . Thus Equation (d) can be rewritten as

32 1058.9 xs −=

(e) )02.0)(1058.9)(4.417(

101)( 3

4

+++=

sxsssxsI

which has a partial fraction decomposition of

180

Chapter 5

(f) 02.0

1051058.9

091.14.417

091.1)(5

3 +−

+−

+=

−

sx

xsssI

Inversion of Equation (f) leads to (g) 105091.1091.1)( 02.051058.94.417 3 ttxt exeeti −−−− −−=

5.44 The mathematical problem is

(b) 0)0(

(a) )]25.0()([100010410001.00

5

=

−−=++ ∫i

tutuidtxidtdi t

Define . Taking the Laplace transform of Equation (a) and using linearity of the transform gives

)()( tisI L=

(c) )25.0()(L1000L104L1000L1.00

5 −−=⎭⎬⎫

⎩⎨⎧

++⎭⎬⎫

⎩⎨⎧ ∫ tutuidtxi

dtdi t

Use of properties of transform of the first derivative and transform of an integral and application of the initial condition, Equation (b), leads to

[ ] ( )

( )

( )( )

( )( ) (d)

10410000110000)(

10410001.011000)(

11000)(10410001.0

11000)(104)(1000)0()(1.0

62

25.0

52

25.0

25.05

25.05

xssesI

xssesI

es

sIs

xs

es

sIs

xsIissI

s

s

s

s

++−

=

++−

=

−=⎟⎟⎠

⎞⎜⎜⎝

⎛++

−=++−

−

−

−

−

The roots of are determined as 62 10410000 xss ++ 4.4171 −=s and . Thus Equation (d) can be rewritten as

32 1058.9 xs −=

( )

( )

( ) (e) 11058.9

091.14.417

091.1

11058.94.417

)1058.9)(4.417(

110000)(

25.03

25.03

3

25.0

s

s

s

exss

exs

Bs

Axss

esI

−

−

−

−⎟⎠⎞

⎜⎝⎛

+−

+=

−⎟⎠⎞

⎜⎝⎛

++

+=

++−

=

Inversion of Equation (e) leads to ( )

( ) (f) )25.0(0916.1091.1

)(0916.1091.1)()25.0(1058.9)25.0(4.417

1058.94.417

3

3

−−

−−=−−−−

−−

tuee

tueetitxt

txt

181

Chapter 5

5.45 The mathematical model is

(d) 0)0((c) 0)0(

(b) 011

(a) )(11

2

1

022

2

01

02

011

==

=+++−

=−+

∫∫

∫∫

ii

dtiC

Ridtdi

LdtiC

tvdtiC

dtiC

Ri

tt

tt

Substitution of given values into Equations (a) and (b) leads to

(f) 01055002.0105

(e) )(500105105500

02

42

2

01

4

02

4

01

41

=+++−

=−+

∫∫

∫∫tt

tt

dtixidtdi

dtix

tudtixdtixi

Taking the Laplace transforms of Equations (e) and (f) and using linearity leads to

(h) 0L105)(500L2.0L105

(g) 500L105L105)(500

01

42

2

01

4

02

4

01

41

=⎭⎬⎫

⎩⎨⎧

++⎭⎬⎫

⎩⎨⎧+

⎭⎬⎫

⎩⎨⎧

−

=⎭⎬⎫

⎩⎨⎧

−⎭⎬⎫

⎩⎨⎧

+

∫∫

∫∫tt

tt

dtixsIdtdidtix

sdtixdtixsI

Application of the property of transform of the first derivative, transform of an integral, and the initial conditions, Equations (c) and (d), to Equations (g) and (h) leads to

(j) 0)(105)(500)(2.0105

(i) 500)(105)(105)(500

2

4

22

4

2

4

1

4

1

=+++−

=−+

sIs

xsIssIs

xs

sIs

xsIs

xsI

Equations (i) and (j) are summarized in matrix form as

(k) 0

500

)()(

1055002.0105

105105500

2

144

44

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡=⎥

⎦

⎤⎢⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

++−

−+ssI

sI

sxs

sx

sx

sx

Solution of Equation (k) using Cramer’s rule gives

182

Chapter 5

( )( )

( ) (l) 105106.2105.2105.2

105106.21001055002.0500

1051055002.0105500

1055002.0500

1055002.0105

105105500

1055002.00

105500

)(

532

532

752

42

2444

4

44

44

4

4

1

xxssxsxs

xsxssxss

sx

sxs

sx

sxs

s

sxs

sx

sx

sx

sxs

sx

s

sI

++++

=

++++

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−⎟⎟

⎠

⎞⎜⎜⎝

⎛++⎟⎟

⎠

⎞⎜⎜⎝

⎛+

⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟

⎠⎞

⎜⎝⎛

=

+++−

−+

++

−

=

and

( )

( ) (m) 105106.2

105.2

105106.2100105.2

1051055002.0105500

105500

1055002.0105

105105500

0105

500105500

)(

532

5

752

7

2444

4

44

44

4

4

2

xxssx

xsxssx

sx

sxs

sx

sx

s

sxs

sx

sx

sx

sx

ssx

sI

++=

++=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−⎟⎟

⎠

⎞⎜⎜⎝

⎛++⎟⎟

⎠

⎞⎜⎜⎝

⎛+

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

=

+++−

−+

−

+

=

The quadratic polynomial in the denominators of Equations (l) and (m) have roots of

and . Partial fraction decompositions are performed with the aid of MATLAB

31 1039.2 xs −= 2

2 1009.2 xs −=

>> clear

183

Chapter 5

>> N=[1 2.5e3 2.5e5] N = 1 2500 250000 >> D=[1 2.6e3 5e5 0] D = 1 2600 500000 0 >> [r,p,k]=residue(N,D) r = -0.0021 0.5021 0.5000 p = 1.0e+003 * -2.3909 -0.2091 0 k = [] >> r(1)*10 ans = -0.0209 >> N=[2.5e5] N = 250000

184

Chapter 5

>> [r,p,k]=residue(N,D) r = 0.0479 -0.5479 0.5000 p = 1.0e+003 * -2.3909 -0.2091 0 k = [] >> The partial fraction decompositions are

)(o 1009.2

5479.01039.2

1079.45.0)(

(n) 1009.2

502.01039.2

1009.25.0)(

23

2

2

23

3

1

xsxsx

sti

xsxsx

sti

+−

++=

++

+−=

−

−

Equations (n) and (o) are inverted leading to

(q) 5479.01079.4)(5.0)(

(p) 502.01009.2)(5.0)(23

23

1009.21039.222

1009.21039.231

txtx

txtx

eextuti

eextuti−−−

−−−

−+=

+−=

The following MATLAB m file is used to plot Equations (p) and (q) t=0:.0001:0.02 i1=0.5-2.09e-3*exp(-2.39e3.*t)+0.502*exp(-2.09e2.*t) i2=0.5+4.79e-2*exp(-2.39e3.*t)-0.5479*exp(-2.09e2.*t) plot(t,i1,'-',t,i2,'.') xlabel('t (s)') ylabel('i (A)') legend('i_1(t)','i_2(t)') title('Solution of Problem 5.45') Execution of the above file leads to the following plot

185

Chapter 5

5.46 The differential equations obtained from the system modeling are

(b) 0)((a) 0)(

2321222

2212111

=++−=−++

xkkxkxmxkxkkxm

&&&&

Substitution of given values into Equations (a) and (b) leads to

(d) 0105.2105.120

(c) 0105.1105.210

24

14

2

24

14

1

=+−

=−+

xxxxx

xxxxx&&

&&

The given initial conditions are

(h) 0)0((g) 0)0((f) 0)0((e) 001.0)0(

2

2

1

1

====

xxxx

&

&

Taking the Laplace transforms of Equations (c) and (d) using the initial conditions of Equations (e)-(h) leads to

( )(j) 0)(105.2)(105.1)(20

)(i 0)(105.1)(105.2001.0)(10

24

14

22

24

14

12

=+−

=−+−

sXxsXxsXs

sXxsXxssXs

Equations (i) and (j) are rearranged and written in matrix form as

186

Chapter 5

(k) 001.0

)()(

105.220105.1105.1105.210

2

1424

442

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

+−−+ s

sXsX

xsxxxs

Cramer’s rule is used to solve the simultaneous equations of Equation (j) leading to

( )( )( ) ( )

( )( )(l)

1011.31054.7

9.6431046.2

1011.39.64325.10010.0

1023750

25.10010.0

104105.72002502.0

105.1105.220105.210105.22001.0

105.220105.1105.1105.210

105.2200105.101.0

)(

32

3

2

3

322

3

624

3

8254

3

244242

42

424

442

42

4

1

xssx

ssx

xssss

xssss

xsxsss

xxsxsxss

xsxxxs

xsxs

sX

+−

+=

+++

=

+++

=

+++

=

−−++

+=

+−−+

+−

=

−−

Inversion of Equation (l) leads to (m) )74.55cos(1054.7)38.25cos(1046.2)( 33

1 txtxtx −− +=

( )( )( ) ( )

( )( )(n)

1011.30305.0

9.6430305.0

1011.39.64375.0

1023750

75.0

104105.7200150

105.1105.220105.210105.101.0

105.220105.1105.1105.210

0105.101.0105.210

)(

322

322

624

8254

244242

4

424

442

4

42

2

xss

ss

xsssxss

sxsxs

sxxsxs

xs

xsxxxs

xsxs

sX

+−

+=

++=

++=

++=

−−++

−−=

+−−+

−+

=

Inversion of Equation (l) leads to

187

Chapter 5

(o) )74.55cos(0305.0)38.25cos(0305.0)(1 tttx −= The following MATLAB code is written to generate a plot of the system response t=0:.001:0.5 x1=2.46e-3*cos(25.38.^t)+7.54e-3*cos(55.74.*t); x2=0.0305*cos(25.38.*t)-0.0305*cos(55.74.*t); plot(t,x1,'-',t,x2,'.') xlabel('t (s)') ylabel('x (m)') legend('x_1(t)','x_2(t)') title('Solution of Problem 5.46')

The following MATLAB plot is generated

5.47 Substitution of numerical values into Equations (a) and (b) of the problem statement leads to

)(b )(5.0225.01.0150

(a) 0125.01.0100

212

211

tuhhdt

dh

hhdtdh

−=+−

=−+

Since the dependent variables represent perturbations from a steady state, . Let 0)0()0( 21 == hh )()( 11 thsH L= and )()( 22 thsH L= . Taking the Laplace

188

Chapter 5

transform of Equation(a), using linearity or the transform and the property of transform of the first derivative leads to

[ ](c) 0125.0)1.0100(

0125.01.0)0(100

0125.01.0L100

21

2111

211

=−+

=−+−

=−+⎭⎬⎫

⎩⎨⎧

HHsHHhHs

HHdtdh

Repeating the procedure with Equation (b) leads to

[ ]

( ) (d) 5.01.0225.0150

5.0225.01.0)0(150

5.0225.01.0L150

12

2122

212

sHHs

sHHhHs

sHH

dtdh

−=−+

−=+−−

−=+−

⎭⎬⎫

⎩⎨⎧

Equations (c) and (d) are summarized in matrix form as

(e) 5.00

225.01501.0125.01.0100

2

1

⎥⎥⎦

⎤

⎢⎢⎣

⎡−=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡+−

−+

sHH

ss

Cramer’s rule is used to solve the equations represented by Equation (e) as

( )72

6

241

1067.60025.1017.4

(f) 01.05.37105.1

0625.0

225.01501.0125.01.0100

225.01505.0125.00

−

−

++−

=

++

−=

+−−+

+−

−

=

xsssx

ssxs

ss

ssH

( )72

63

242

1067.60025.1033.31033.3

(g) 01.05.37105.1

05.50

225.01501.0125.01.0100

5.01.001.0100

−

−−

+++−

=

++

+−=

+−−+

−−

+

=

xsssxsx

ssxs

ss

s

s

H

Partial fraction decompositions lead to

(i) 000304.

55.7002196.056.20.5)(

(h) 000304.

26.7002196.000.125.6)(

2

1

+−

++−=

++

+−−=

ssssH

ssssH

Inversion of the transforms in Equations (h) and (i) leads to

189

Chapter 5

(k) 55.756.2)(0.5)(

(j) 26.700.1)(25.6)(000304.0002196.0

2

000304.002196.01

tt

tt

eetuth

eetuth−−

−−

−+−=

+−−=

5.48 The equations obtained for the perturbation in temperatures of oil and water in the double pipe heat exchanger of Example 4.25 are

( ) (a) 22 02

oiowoooooo

ioo mcULrmcLUrdt

dLrc θθπθπ

θπρ && =−++

( ) ( ) (b) 22222 wiwwwwwooo

woww mcmcLUrLUr

dtd

Lrrc θθπθπθ

πρ && =++−−

It is desired to determine the perturbations in temperatures of the oil and water when 0)( =toiθ , )(3)( tutwi =θ and

32

o32

mkg

1050.7= ,mkg

1060.9= xxw ρρC kg

J 104.2 ,Ckg

J 1019.4 33

⋅=

⋅= xcxc ow

m 2.5L cm, 2.4 cm, 2.2 cm 2 s

kg 85.4 s

kg 51.2 m

W 1065.8 222 ======

⋅= rrrmm

CxU oiow &&

The appropriate initial conditions are 0)0()0( == wo θθ . Substitution of given values into Equations (a) and (b) leads to

(d) )(1016.31041.21036.11004.4

(c) 01036.11019.11065.5

4444

443

tuxxxdt

dx

xxdt

dx

wow

woo

=+−

=−+

θθθ

θθθ

Taking the Laplace transforms of Equations (c) and (d), applying initial conditions, and summarizing the resulting equations in a matrix form leads to

(e) 1016.30

)()(

1041.21004.41036.11036.11019.11065.5 4

444

443

⎥⎥⎦

⎤

⎢⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡ΘΘ

⎥⎦

⎤⎢⎣

⎡

+−−+

sx

ss

xsxxxxsx

w

o

Cramer’r rule is used to determine a solution to the simultaneous equations of Equation (e) leading to

190

Chapter 5

( )

(f) )411.049.2(

73.1

)02.117.648.2(

4.30

)1036.1()104.21004.4(1019.11065.5

1016.3)1036.1(

1041.21004.41036.11036.11019.11065.5

1041.21004.41016.31036.10

)(

2

2

244443

44

444

443

444

4

++=

++=

−−++

⎟⎟⎠

⎞⎜⎜⎝

⎛

=

⎥⎦

⎤⎢⎣

⎡

+−−+

+

−

=Θ

sss

sss

xxsxxsxsxx

xsxxxxsx

xsxsx

x

so

( )

(g) )4113.049.2(

52.1718.0

)02.117.648.2(

3.761.78s

)1036.1()104.21004.4(1019.11065.5

1016.3)1019.11065.5(

1041.21004.41036.11036.11019.11065.5

1016.31036.1

01019.11065.5

)(

2

2

244443

443

444

443

44

43

+++

=

+++

=

−−++

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

⎥⎦

⎤⎢⎣

⎡

+−−+

−

+

=Θ

ssss

sss

xxsxxsxsxxsx

xsxxxxsx

sxx

xsx

sw

Partial fraction decompositions lead to

(i) 517.0

53.187.1

0360.049.1)(

(h) 517.0

30.297.1

602.070.1)(0

+−

++=Θ

+−

++=Θ

ssss

ssss

w

Inversion of Equations (h) and (i) leads to

(k) 53.10360.049.1)(

(j) 30.2602.070.1)(517.097.1

w

517.097.1

tt

tto

eet

eet−−

−−

−+=

−+=

θ

θ

5.49 Then differential equations obtained for the concentrations of the reactant and product in the constant volume CSTR of Example 4.16 are

191

Chapter 5

(b)

(a) )(

BiBAB

AiAA

qCqCkVCdt

dCV

qCCkVqdt

dCV

=+−

=++

It is desired to determine the concentrations of the reactant A and the product B when

3m 31000.1 −= xV , L

molsm 13

)(2.0)(100.2100.2 36 tutCxks

xq Ai === −−− and

. Substitution of these values into Equations (a) and (b) leads to 0)( =tCBi

(d) 0102102101

(c) )(104104101

663

763

=+−

=+

−−−

−−−

BAB

AA

CxCxdt

dCx

tuxCxdt

dCx

Taking the Laplace transforms of Equations (c) and (d), taking initial conditions to be zero, simplifying, and writing the resulting equations in a matrix form leads to

(e) 0

104

)()(

1021020104

4

33

3

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

−++

−

−−

−

sx

sCsC

xxsxs

B

A

Equation (e) is used to determine

(g) )104)(102(

108

)(102

2x10)(

(f) )104(

104)(

33

7

3

3-

3

4

−−

−

−

−

−

++=

+=

+=

xsxssx

sCxs

sC

xssxsC

AB

A

Partial fraction decompositions of Equations (f) and (g) leads to

(i) 104

1102

211.0)(

(h) 104

111.0)(

33

3

⎟⎠⎞

⎜⎝⎛

++

+−=

⎟⎠⎞

⎜⎝⎛

+−=

−−

−

xsxsssC

xsssC

B

A

Inversion of Equations (h) and (i) leads to ( )( ) (k) 211.0)(

(j) 11.0)(33

3

104102

104

txtxB

txA

eetC

etC−−

−

−−

−

+−=

−=

The following MATLAB program is used to plot the time dependence of the concentrations % Problem 5.49 dt=.1/4e-3; for i=1:101 t(i)=(i-1)*dt; CA(i)=0.1*(1-exp(-4e-3*t(i)));

192

Chapter 5

CB(i)=0.1*(1-2*exp(-2e-3*t(i))+exp(-4e-3*t(i))); end plot(t,CA,'-',t,CB,'.') xlabel('t (s)') ylabel('C (mol/L') title('Response of system of Problem 5.49') legend('C_A(t)','C_B(t)')

The resulting MATLAB generated plot is

193

Chapter 5

194

Chapter 6

6. Transient Analysis and Time Domain Response

6.1 The differential equations derived in Example 2.26 are

(a)

00

222111222111

22112211

222

2111122

112221222

2111122

112221

⎥⎦

⎤⎢⎣

⎡+−+−

+++

=⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡+−−+

+⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡+−−+

+⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

ycycykykycycykyk

xkkkkkkkkx

ccccccccx

Im

&λ&λλλ

&&

λλλλλλ

&&

λλλλλλ

&&&&

θθθ

Substituting given values and taking 02 =y leads to

(b) 7201004.1

900103.1

10367.3101.9101.9108.2

19283203201700

250001500

115

115

54

45

⎥⎦

⎤⎢⎣

⎡

−−+

=⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

yyxyyx

xxx

xxxx

&&

&&

&&&&

θθθ

Define )()( txsX L= and )()( ts θL=Θ . Taking the Laplace transforms of Equations (b) and using the properties of linearity of the transforms and transforms of derivatives and assuming all initial conditions are zero leads to

(c)

)(1004.1720

103.1900)()(

10367.31928250101.9320101.9320108.217001500

15

5

524

452

sYxs

xsssX

xssxsxsxss

⎥⎦

⎤⎢⎣

⎡

−−+

=⎥⎦

⎤⎢⎣

⎡Θ⎥

⎦

⎤⎢⎣

⎡

++++++

Cramer’s rule is used to determine the solution of Equation (c) as ( )

( )

( ) (d) 1060.81005.11078.51032.31075.3

)(1032.51052.61045.31025.2

10367.31928250101.9320101.9320108.217001500

10367.31928250)(1004.1720101.9320)(103.1900

)(

1110283645

1082735

524

452

525

45

xsxsxsxsxsYxsxsxsx

xssxsxsxss

xsssYxsxssYxs

sX

+++++++

=

++++++

++−−++

=

( )

( )

( ) (e) 1060.81005.11078.51032.31075.3)(1010.41002.51058.11008.1

10367.31928250101.9320101.9320108.217001500

)(1004.1720101.9320)(103.1900108.217001500

)(

1110283645

1082836

524

452

54

552

xsxsxsxsxsYxsxsxsx

xssxsxsxss

sYxsxssYxsxss

s

++++−−−−

=

++++++

−−++++

=Θ

The transfer functions are determined from Equations (d) and (e) as

195

Chapter 6

( ) )(1060.81005.11078.51032.31075.3

)(1032.51052.61045.31025.2)()()( 1110283645

1082735

1 f xsxsxsxsx

sYxsxsxsxsYsXsG

+++++++

==

( ) (g)

1060.81005.11078.51032.31075.3)(1010.41002.51058.11008.1

)()()( 1110283645

1082836

2 xsxsxsxsxsYxsxsxsx

sYssG

++++−−−−

=Θ

=

6.2 The integro-differential equations of derived in Example 3.28 are

(b) 011

(a) 11

22

02

01

02

011

=+++−

=−+

∫∫

∫∫

Ridtdi

LdtiC

dtiC

vdtiC

dtiC

Ri

tt

tt

Taking the Laplace transform of Equations (a) and (b), assuming all initial conditions are zero leads to

(d) 0)()()(1)(1

(c) )()(1)(1)(

2221

211

=+++−

=−+

sRIsLsIsICs

sICs

sVsICs

sICs

sRI

Equations (c) and (d) are summarized in matrix form by

(e) 0

)()()(

11

11

2

1⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++−

−+ sVsIsI

CsRLs

Cs

CsCsR

Cramer’s rule is used to solve for as )(1 sI

196

Chapter 6

( )(f)

2

)(1

2

)(1

1111

)(1

11

11

10

1)(

)(

22

2

2

1

CRsLCRRLs

sVC

RsLs

CsRLCRRLs

sVCs

RLs

CsCsCsRLs

CsR

sVCs

RLs

CsRLs

Cs

CsCsR

CsRLsCs

sV

sI

+++

⎟⎠⎞

⎜⎝⎛ ++

=

+++

⎟⎠⎞

⎜⎝⎛ ++

=

⎟⎠⎞

⎜⎝⎛−⎟⎠⎞

⎜⎝⎛−−⎟

⎠⎞

⎜⎝⎛ ++⎟⎠⎞

⎜⎝⎛ +

⎟⎠⎞

⎜⎝⎛ ++

=

++−

−+

++

−

=

Similarly solving for leads to )(2 sI

( )(g)

2

)(1

2

)(1

2

01

)(1

)(

22

2

22

CRsLCRRLs

sVC

CRLCRRLs

sVCs

CRLCRRLs

Cs

sVCs

R

sI

+++=

+++

⎟⎠⎞

⎜⎝⎛

=

+++

−

+

=

The transfer functions are defined as

197

Chapter 6

( )

( )(i)

2

1

)()()(

(h) 2

1

)()(

)(

22

22

22

2

11

CRsLCRRLs

C

sVsIsG

CRsLCRRLs

CRsLs

sVsI

sG

+++=

=

+++

++=

=

6.3 The differential equations derived in Example 3.22 governing the current in the field circuit and angular velocity in a field controlled dc servomotor are

(b)

(a)

TiKcdtdJ

viRdtdi

L

fft

ffff

f

=−+

=+

ωω

Taking the Laplace transform of Equations (a) and (b), setting T=0 and assuming all initial conditions are zero leads to

(d) 0)()()(

(c) )()()(

=−Ω+Ω

=+

sIKscsJs

sVsIRssIL

fft

fffff

Equation (c) is rewritten as

(e) )(

)(ff

ff RsL

sVsI

+=

The first transfer function is determined as

(f) 1

)()(

)(1

ff

f

f

RsL

sVsI

sG

+=

=

Substitution of Equation (e) in Equation (d) leads to

(g) ))((

)()(

0)(

)()(

fff

f

ff

ft

cJsRsLsVK

s

RsLsVK

scJs

++=Ω

=+

−Ω+

The second transfer function is obtained from Equation (g) as

198

Chapter 6

(h) ))((

)()()(2

tff

f

f

cJsRsLK

sVssG

++=

Ω=

6.4 Equations (s), (t), and (u) of Example 4.17 are

(c)

(b) 0

(a) 0

12)/(

2)/(

2)/(

QQCTRT

ECVeTqcdt

dTVc

qCCTRT

ECVedt

dCV

CTRT

ECVeqCdt

dCV

Asps

ApRTE

pvp

v

BpAsps

ApRTEBp

Asps

ApRTE

ApAp

s

s

s

&&−=⎟⎟⎠

⎞⎜⎜⎝

⎛+−+

=+⎟⎟⎠

⎞⎜⎜⎝

⎛+−

=⎟⎟⎠

⎞⎜⎜⎝

⎛+++

−

−

−

λαρρ

α

α

Taking the Laplace transforms of Equations (a)-(c) assuming all initial conditions are zero leads to

(d) 0)()()()( 2)/( =⎥

⎦

⎤⎢⎣

⎡+++ − sTC

RTEsCVesqCsVsC pAs

sAp

RTEapap

sα

(e) 0)()()()( 2)/( =+⎥

⎦

⎤⎢⎣

⎡+− − sqCsTC

RTEsCVesVsC BppAs

sAp

RTEBp

sα

(f) )()()()()( 2)/( sQsTC

RTEsCVesTqcsVsTc pAs

sAp

RTEpvpV

s =⎥⎥⎦

⎤

⎢⎢⎣

⎡+−+ −λαρρ

Equations (d)-(f) are summarized in matrix form as

(g)

)(

00

)()()(

0

0

2)/()/(

2)/()/(

2)/()/(

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛−+−

⎟⎟⎠

⎞⎜⎜⎝

⎛−−+

⎟⎟⎠

⎞⎜⎜⎝

⎛++

−−

−−

−−

sQsTsCsC

RTEC

VeqcVscVe

RTEC

VeVeqVs

RTEC

VeVeqVs

p

Bp

Ap

s

AsRTEvv

RTE

s

AsRTERTE

s

AsRTERTE

ss

ss

ss

λαρρλα

αα

αα

Cramer’s rule is used to solve Equation (g) leading to

199

Chapter 6

[ ](h)

)(

)(

0

0

0)(

0

00

)(

2

2)/(

2)/()/(

2)/()/(

2)/()/(

2)/(

2)/()/(

2)/(

sD

sQRTEC

Ve

RTEC

VeqcVscVe

RTEC

VeVeqVs

RTEC

VeVeqVs

RTEC

VeqcVscsQ

RTEC

VeVe

RTEC

Ve

sC

s

AsRTE

s

AsRTEvv

RTE

s

AsRTERTE

s

AsRTERTE

s

AsRTEvv

s

AsRTERTE

s

AsRTE

Ap

s

ss

ss

ss

s

ss

s

⎟⎟⎠

⎞⎜⎜⎝

⎛

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−+−

⎟⎟⎠

⎞⎜⎜⎝

⎛−−+

⎟⎟⎠

⎞⎜⎜⎝

⎛++

⎟⎟⎠

⎞⎜⎜⎝

⎛−+

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

⎟⎟⎠

⎞⎜⎜⎝

⎛

=

−

−−

−−

−−

−

−−

−

α

λαρρλα

αα

αα

λαρρ

αα

α

where

[ ]

[ ] (i)

)(

2

2)/(

2)/()/()/(

⎪⎭

⎪⎬⎫⎟⎟⎠

⎞⎜⎜⎝

⎛+

⎪⎩

⎪⎨⎧

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+++−=

−

−−−

s

AsRTE

s

AsRTEvv

RTERTE

RTEC

Ve

RTEC

VeqcVscVeqVsVesD

s

sss

αλ

λαρραα

[ ](j)

)(

)(

)(

)(

0

0

)(

2

2)/(

2)/()/(

2)/(

2)/()/(

sD

sQRTEC

Ve

sDRTEC

VeqcVscsQVe

RTEC

VeqVs

RTEC

VeVeqVs

sC

s

AsRTE

s

AsRTEvv

RTE

s

AsRTE

s

AsRTERTE

Bp

s

ss

s

ss

⎟⎟⎠

⎞⎜⎜⎝

⎛

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−+−

⎟⎟⎠

⎞⎜⎜⎝

⎛−+

⎟⎟⎠

⎞⎜⎜⎝

⎛++

=

−

−−

−

−−

α

λαρρλα

α

αα

200

Chapter 6

[ ] (k) )(

)(

)(

)(0000

)(

)/()/(

)/(

)/(

)/(

sDsQVeqVsVe

sD

sQVeVeqVs

VeqVs

sT

ss

s

s

s

RTERTE

RTE

RTE

RTE

p

−−

−

−

−

++−=

−−+

++

=

αα

λαα

α

The transfer functions are obtained from Equations (i), (j), and (k) as

[ ](l)

)(

)()(

)(

2

2)/(

1

sDRTEC

Ve

sQsC

sG

s

AsRTE

Ap

s

⎟⎟⎠

⎞⎜⎜⎝

⎛

=

=

−α

[ ](m)

)(

)()(

)(

2

2)/(

2

sDRTEC

Ve

sQsC

sG

s

AsRTE

Bp

s

⎟⎟⎠

⎞⎜⎜⎝

⎛

=

=

−α

[ ] (n) )(

)()(

)(

)/()/(

3

sDVeqVsVe

sQsT

sG

ss RTERTE

p

−− ++−=

=

αα

6.5 Equations (k) and (l) of Example 4.25 are

( ) (a) 22 oi2 θθπθπ

θπρ oowooooo

oioo mcULrmcLUr

dtd

Lrc && =−++

( ) ( ) (b) 22222 wiwwwwwooo

woww mcmcLUrLUr

dtd

Lrrc θθπθπθ

πρ && =++−−

The desired transfer functions are obtained by setting 0=oiθ and taking the Laplace transforms of Equations (a) and (b) assuming all initial conditions are zero,

( ) (c) 0)(2)(2)(2 =Θ−Θ++Θ sULrsmcLUrsLsrc wooooooioo πππρ &

( ) ( ) (d) (s))(2)(2)(222 wiwwwwwooowoww mcsmcLUrsLUrsLsrrc Θ=Θ++Θ−Θ− &&πππρ

Equations (c) and (d) are summarized in matrix form as

201

Chapter 6

( )( ) ( )

(e) )(

0

)()(

2222

222

2

⎥⎦

⎤⎢⎣

⎡Θ

=⎥⎦

⎤⎢⎣

⎡ΘΘ

⎥⎦

⎤⎢⎣

⎡

++−−−++

smc

ss

mcLUrLsrrcULrULrmcLUrLsrc

wiww

w

o

wwoowwo

ooooioo

&

&&

ππρππππρ

Application of Cramer’s rule to solve Equation (e) leads to

( ) ( )( )

( ) ( )

( )[ ] ( ) ( )[ ] ( )

(f) )(

)(2

222)(2

2222

2)(20

)(

2222

2

222

2

222

sDsmULcr

ULrmcLUrLsrrcmcLUrLsrcsmULcr

mcLUrLsrrcULrULrmcLUrLsrc

mcLUrLsrrcsmcULr

s

wiwwo

owwoowwoooioo

wiwwo

wwoowwo

ooooioo

wwoowwwiww

o

o

Θ=

−++−++

Θ=

++−−−++

++−Θ−

=Θ

&&&

&

&&

&&

ππππρππρ

πππρπ

πππρ

ππρπ

where ( ) ( )( ) ( )( )[

( )( )( )]( )( ) (g) 2

2

2)(

022

2

22222

2

wwooo

wwoooooww

wwoiooowwioo

mcmcLUrmmccsmcLUrLrrc

mcLUrLrcsLrrcLrcsD

&&&&&

&

++++−

+++−=

πππρ

ππρπρπρ

Also,

The required transfer functions are

(i) )(

2

)()(

)(1

sDmULcr

ss

sG

wwo

wi

o

&π=

ΘΘ

=

( )[ ](j)

)(2

)()(

)(

2

2

sDmcmcLUrLsrc

ss

sG

wwoooioo

wi

w

&&++=

ΘΘ

=

ππρ

6.6 The transfer function derived in Problem 6.1 is ( ) (a)

4510.91011.11024.61032.31075.31010.41002.51058.11008.1

)()()( 109283645

1082836

2 ++++−−−−

=Θ

=sxsxsxsxxsxsxsx

sYssG

The system is stable if all roots of the polynomial in the denominator have negative real parts. To this end consider

202

Chapter 6

(b) 01052.21096.21017.38533.804510.91011.11024.61032.31075.3

532334

109283645

=++++

=++++

xsxsxsssxsxsxsx

Equation (6.29) is applied to Equation (b) leading to a Routh’s matrix of the form

(c)

002.52x10002.17x1002.52x102.84x1001096.28533.8

1052.21017.31

5

3

53

3

53

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=x

xx

B

Calculations in obtaining elements of B are

(d) 10841.2 853.8

)1096.2)(1()1017.3)(8533.8(

3

33

1,2

2,21,12,11,21,3

x

xx

bbbbb

b

=

−=

−=

(e) 1052.2 853.8

)0)(1()1052.2)(8533.8(

5

5

1,2

3,21,13,11,22,3

x

x

bbbbb

b

=

−=

−=

(f) 1017.2 1084.2

)1052.2)(853.8()1096.2)(1084.2(

3

3

533

1,3

2,31,22,21,31,4

xx

xxx

bbbbb

b

=

−=

−=

(g) 0 1,3

3,32,23,21,32,4

=

−=

bbbbb

b

(h) 1052.2 1017.2

)0)(1084.2()1052.2)(1017.2(

5

3

353

1,4

2,41,32.31.41,5

xx

xxx

bbbbb

b

=

−=

−=

Since there are no sign changes in the first column of B the system is stable. 6.7 The transfer function of a third-order system is

(a) 20158

4)( 23 ++++

=sss

ssG

203

Chapter 6

Routh’s method is applied in the solution of Example 6.9 to derive a stability criteria for a third-order system. If the denominator of the transfer function is of the form

(b) )( 322

13 asasassD +++=

the system is stable if (c) 321 aaa >

Equation (c) is checked for the transfer function of Equation (a) by

(d) 20120? 20)15)(8(

>>

Equation (d) shows that the system is stable. 6.8 The transfer function of a second-order system is

(a) 25)410(5

)254()( 23

ii

i

KsKsssK

sG++++

+=

(a) The stability of a third-order system is dependent upon the coefficients of the denominator of its transfer function. If the system is stable if all coefficients are positive and . Applying this criteria to Equation (a) leads to

012

23)( asasassD +++=

021 aaa >

(b) 1025)410(5

<>+

i

ii

KKK

Thus the system is stable when 10<iK and unstable if . 10>iK

(b) For , Equation (a) becomes 5=iK

(c) 125305

12520)( 23 ++++

=sss

ssG

The MATLAB generated impulsive response is

204

Chapter 6

The system is clearly stable.

For , Equation (a) becomes 10=iK

(d) 250505

25040)( 23 ++++

=sss

ssG

The impulsive response for the transfer function of Equation (d) is

This system is neutrally stable.

For Equation (a) becomes 15=iK

(e) 375705

37560)( 23 ++++

=sss

ssG

The impulsive response for the system whose transfer function is Equation (e) is

205

Chapter 6

The impulsive response shown above is clearly unstable.

(c) The denominator of Equation (a) is written in the form of Equation (6.32) as

(f) )254(105

)()()(23 ++++=

+=

sKsss

sRKsQsD

i

i

The root-locus diagram is generated by the MATLAB worksheet commands >> clear >> Q=[1 5 10 0] Q = 1 5 10 0 >> R=[4 25] R = 4 25 >> rlocus(R,Q) >>

The MATLAB generated root locus diagram is

206

Chapter 6

6.9 The transfer function for a third-order system is

)a( 354

42)( 23 ++++

=sss

ssG

(a) The stability of a third-order system is dependent upon the coefficients of the denominator of its transfer function. If the system is stable if all coefficients are positive and . Applying this criteria to Equation (a),

012

23)( asasassD +++=

021 aaa >)b( 3)4)(5( >

Thus the system is stable. (b) The stability of the system with respect to 1−=s is tested by examining the stability of a system with D(s-1). To this end

c)( 2 355484133

3)1(5)1(4)1()1(

23

223

23

++=

+−++−+−+−=

+−+−+−=−

ssssssss

ssssD

Since the coefficient of the s term is zero in Equation (c) the system is not stable with respect to s=-1. 6.10 The appropriate transfer function is

(a) 25)410(5

)254()( 23

ii

i

KsKsssK

sG++++

+=

(a) The denominator of Equation (a) is written in the form of Equation (6.32) as

(f) )254(105

)()()(23 ++++=

+=

sKsss

sRKsQsD

i

i

207

Chapter 6

The root-locus diagram is generated by the MATLAB worksheet commands >> clear >> Q=[1 5 10 0] Q = 1 5 10 0 >> R=[4 25] R = 4 25 >> rlocus(R,Q) >>

The MATLAB generated root locus diagram is

The above root locus curve shows that the system is stable for 2.10<iK and unstable for larger values.

6.11 The transfer function for the fourth-order plant is

(a) )56)(22(

2)( 2 ++−+

=sss

ssG

The closed loop transfer function is

208

Chapter 6

(b) )(1

)()(sKG

sKGsH+

=

Substitution of Equation (a) into Equation (b) leads to

(c) )2()56)(2(

)2(

)56)(2(21

)56)(2(2

)(

2

2

2

++++−+

=

++−+

+

++−+

=

sKsssssK

sssssK

sssssK

sH

The denominator of Equation (c) is of the form of Equation (6.32) with

(e) 2)((d) )56)(2(2)( 2

+=++−=

ssRssssQ

The MATLAB commands used to generate the root locus diagram for the system are given below

>> R=[1 2] R = 1 2 >> Q=[1 4 -7 -10 0] Q = 1 4 -7 -10 0 >> rlocus(R,Q) >>

The resulting root locus diagram is

209

Chapter 6

Since one of the poles of Q(s) is positive, s=2, one branch of the root locus diagram begins at s=2. A second branch originates at s=0. These branches coincide at s=1.16, which is a breakaway point. This corresponds to K=4.1. Two branches of the root locus for larger values of K are in the right-half plane. Thus the system is unstable for all values of K.

6.12 The transfer functions for the system of Example 5.37 are obtained from

(a) )()(

)()(

93000055518000

2

1

2

1⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡+−

−+sFsF

sHsH

ss

One transfer function is

(b) 201012.3104.5

930000

930000555180009300005

51

)(

528

1,1

+++

=

+−−++−

−

=

sxsxs

ss

ssG

The closed loop transfer function is

(c) )(1

)()(sKG

sKGsH+

=

Substitution of Equation (b) in Equation (c) leads to

210

Chapter 6

(d) )930000(201012.3104.5

)930000(

201012.3104.59300001

201012.3104.5930000

)(

528

528

528

+++++

=

+++

+

+++

=

sKsxsxsK

sxsxsK

sxsxsK

sH

Equation (d) is of the form of Equation (6.32) with

(f) 930000)((e) )930000(201012.3104.5)( 528

+=++++=

ssRsKsxsxsQ

The MATLAB commands to generate the root locus diagram for this system are >> R=[30000 9] R = 30000 9 >> Q=[5.4e8 3.12e5 20] Q = 540000000 312000 20 >> rlocus(R,Q) >>

The resulting root locus diagram is

211

Chapter 6

The root locus diagram shows that the system is stable for all values of K. Root locus diagrams for the other transfer functions for this system can be developed. The forms of H(s) resulting from using other transfer functions all have the same Q(s) as Equation (e), but with different forms of R(s). For , R(s)=5 while for , R(s)=18000s+5. However root locus diagrams drawn for all cases show that all transfer functions are stable for all values of K.

)( and )( 1,22,1 sGsG )(2,2 sG

6.13 Application of KVL to each loop of the circuit leads to

(b) 0)(102.01100

)(a 0)()(102.0150

02162

02161

=−−−

=+−−−

∫

∫

−

−

t

t

dtiix

i

tvdtiix

i

Define )()(,)()( 2211 tisItisI L L == and )()( tvsV L= . Taking the Laplace transforms of Equations (a) and (b), using the properties of linearity and transform of integrals leads to

( ) (d) 0)()(105)(100

(c) )())()((10550

21

6

2

21

6

1

=−−

=−+

sIsIs

xsI

sVsIsIs

xI

Equations (c) and (d) are summarized in matrix form as

(e) 0

)()()(

105100105

10510550

2

166

66

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

+−

−+ sVsIsI

sx

sx

sx

sx

Cramer’s rule is used on Equation (e) leading to

( ) (f) 105.7105

)(105100

105.2105.2105105.2105

)(105100

105100105

10510550

1051000

105)(

)(

83

6

2

13

2

13883

6

66

66

6

6

1

xsxsVxs

sx

sx

sx

sxx

sVs

x

sx

sx

sx

sx

sx

sxsV

sI

++

=

−+++

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

+−

−+

+

−

=

The transfer function is determined from Equation (f) as ( ) (g)

105.7105105100

)()()( 83

61

xsxxs

sVsIsG

++

==

Equation (g) is put in the standard form for the transfer function of a first-order system as

212

Chapter 6

(h) 105.1

10502.0)( 5

3

xsxssG

++

=

(b) The circuit’s time constant is determined from Equation (h) as

(i) 1067.6 105.1

1

6

5

s−=

=

xx

T

(c) If then )(10)( tutF =

( ) (j) 105.11052.0

)(10)(

5

4

1

xssxs

sGs

sI

++

=

=

A partial fraction decomposition of Equation (j) leads to

(k) 105.1

13333.033333.0

105.1

)(

5

51

xss

xsB

sAsI

+−=

++=

Inversion of Equation (k) leads to (l) 1333.0)(3333.0)(

5105.11

txetuti −−=

6.14 The and Ω50 Ω 60 resistors are in parallel and have an equivalent resistance of

(a) 3.27

601

501

1Ω=

Ω+

Ω

=eqR

The electric potential in the capacitor before the switch is closed is

(b) 3.433103

103.1)0( 6

3

VFxCx

CQV === −

−

Applying KVL around the resulting circuit, after the switch is closed leads to

(c) 03.43310313.2720

06 =−−−− ∫−

t

idtx

ii

Equation (c) is rearranged to

(d) 3.4331033.33.470

5 −=+ ∫t

idtxi

Define )()( tisI L= . Taking the Laplace transform of Equation (d) leads to

(e) 1004.7

16.9

1033.33.473.433)(

3.433)(1033.3)(3.47

3

5

5

xs

xssI

ssI

sxsI

+−

=

+−

=

−=+

213

Chapter 6

Inversion of the transform in Equation (e) leads to (f) 16.9)(

31004.7 txeti −−= The voltage change across the capacitor is

(g) 3.433)(1033.3)(0

5 += ∫t

c dttixtv

Using Equation (f) in Equation (g) gives

( ))(h 3.433

3.43313.433

3.4331004.7

)1033.3)(16.9(

3.433)16.9(1033.3

3

3

3

3

1004.7

1004.7

0

1004.73

5

0

1004.75

Ve

e

ex

x

dtexv

tx

tx

ttx

ttx

c

−

−

−

−

=

+−=

+=

+−= ∫

The voltage change across the capacitor when the charge is is CxQ 6105 −=

VFxCx

CQvc

67.1

103105

6

6

=== −

−

. The time when this occurs is determined using Equation (h)

(i) 1090.7 3.433

67.1ln1004.7

1 3.433 67.1

4

3

1004.7 3

sxx

t

VeV tx

−

−

=

⎟⎠⎞

⎜⎝⎛−=

=

6.15 Currents are as defined below. Define as the voltage drop across the resistor in the direction of the current and as the voltage change across the inductor in the direction of the current.

Rv

Lv

Application of KCL at the node leads to

(a) iii LR =+ Application of KVL around the loop leads to

(b) 0=− RL vv Equation (b) implies that

(c) LR vvv ==

214

Chapter 6

The currents are related to the voltage changes across the circuit components by

(e) 1

(d)

0∫=

=

t

L

R

vdtL

i

Rvi

Substitution of Equations (d) and (e) in Equation (a) leads to

(f) 1

0

ivdtLR

v t

=+ ∫

Taking the Laplace transform of Equation (f) leads to

(g) )()(

)()(1)(1

LRs

sRsIsV

sIsVLs

sVR

+=

=+

The transfer function is determined as

(h)

)()()(

LRs

RssIsVsG

+=

=

The impulsive response is the inverse of the transfer function which is rewritten as

(i) 1)(⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+−=

LRs

LR

RsG

(a) Equation (i) is inverted leading to

(j) )()( ⎥⎦

⎤⎢⎣

⎡−=

− tLR

i eLRtRtv δ

(b) The current through the inductor is obtained using Equation (e) as

(k) 1)(

)(

1

0

0

⎟⎟⎠

⎞⎜⎜⎝

⎛−+=

⎥⎦

⎤⎢⎣

⎡−=

=

−

−

∫

∫

tLR

t tLR

t

L

eLRtu

LR

dteLRt

LR

vdtL

i

δ

Noting that for t>0, u(t)=1, Equation (k) can be rewritten as

(l) )(tueLRi

tLR

L

−=

215

Chapter 6

The presence of the unit step function in Equation (l) is necessary to show that the current through the inductor is discontinuous at t=0. 6.16 It is given that

(a) 75.3

5.0

154

2)(

+=

+=

s

ssG

(a) The system’s impulsive response is

(b) 0.5

)()(75.3

1

ti

e

sGtx−

−

=

= -L

(b) The system’s step response is

(c) )(1)( 1

⎭⎬⎫

⎩⎨⎧= − sG

stxs L

A partial fraction decomposition leads to

(d) 75.3

1333.01333.0

)75.3(5.0)(1

+−=

+=

ss

sssG

s

Equations (c) and (d) lead to (e) 1333.0)(1333.0)( 75.3 t

s etutx −−= (c) The system’s ramp response is

(f) )(1)( 21

⎭⎬⎫

⎩⎨⎧= − sG

stxs L

A partial fraction decomposition leads to

(g) 75.3

0356.01333.00356.0

)75.3(5.0)(1

2

22

+++−=

+=

sss

sssG

s

Inversion of Equation (g) leads to (h) 0356.01333.0)(0356.0)( 75.3 t

r ettutx −++−= (d) The time constant for the system is

(i) s 2667.0 75.31

=

=T

The 2% settling time for a first-order system is

(j) s 05.1 92.3

== Tts

(e) The 10% rise time is the time required for the step response to reach 10% of its final value. Thus from Equation (e)

216

Chapter 6

(k) s 0281.0

)9.0ln(75.31

)1333.0()1333.0()1333.0)(1.0(

10

75.3 10

=

=

−= −

t

e t

The 90% rise time is similarly determined by

(l) s 614.0

)1.0ln(75.31

90

=

=t

The 10%-90% rise time is

(m) s 586.0 1090

=−= tttr

6.17 If the drug is again infused 2 hr after being expended the rate of infusion as a function of time is given by

(a) ]260()14([302.0)12()([302.0)( −−−+−−= tututututI (a) Substitution of Equation (a) into Equation (a) of Example 6.35 leads to

(b) )]26()14()12()([0719.052.9 −−−+−−=+ tutututuCdtdC

Taking the Laplace transform of Equation (b), assuming the initial condition is zero leads to

( ) (c) )105.0(

11055.7)(2614123

+−+−

=−−−−

sseeexsC

sss

Inversion of Equation (c) leads to [

(d) mg/L )]26()1( )14()1()12()1(11019.7)(

)26(105.0

)14(105.0)12(105.0105.02

−−−

−−+−−−−=−−

−−−−−−

tuetuetueextC

t

ttt

(b) The general form for the rate of infusion is (e) ))]1(1412())1(14([302.0)( ∑ −−−−−−=

k

ktuktutI

Using the solution procedure of Example 6.35 the time dependent concentration of the drug in the plasma is

[ ] (f) ))1(1412(]1[

))1(14(11019.7)(

))1(1412(105.0

))1(14(105.02

−−−−−

−−−=

−−−−

−−−− ∑ktue

ktuextC

ktk

kt

6.18 The differential equation for the perturbation in liquid level is

(a) 1gRph

RdtdhA

ρ=+

Assuming the flow is turbulent through the pipe

217

Chapter 6

(b) 78.17 9.0

82

2

23 ms

smm

qh

Rs

s

==

=

The area of the tank is ( ) 222

44.34

1.24

mmDA

=== ππ . Substitution of given and

calculated data into Equation (a) leads to

(c) )( 0143.00562.044.3

)( 78.17 81.9 1000

2500

78.17144.3

223

2

tuhdtdh

tu

ms

sm

mkg

mN

hdtdh

=+

⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

=+

Taking the Laplace transform of Equation (c) leads to

(d) )1063.1(

1016.4

)0562.044.3(0143.0)(

0143.0)(0562.0)(44.3

2

3

−

−

+=

+=

=+

xssx

sssH

ssHsHs

(a) The final value of the liquid level perturbation is

(e) 255.01063.1

1016.4lim

)(lim

2

3

0

0

=+

=

=

−

−

→

→

xsx

sHsh

s

sf

Thus the liquid level when the tank reaches steady state is 8.255 m. (b) The time dependent response of the system is obtained by inverting

(f) 1063.1

11255.0)1063.1(

1016.4)( 22

3

⎟⎠⎞

⎜⎝⎛

+−=

+= −−

−

xssxssxsH

Inversion of Equation (f) leads to ( ) (g) 1255.0)(

21063.1 txeth−−−=

(c) The 4% settling time is the time required for the response to be permanently within 4% of its final value. It is obtained from Equation (g) by

( ) (h) 1255.0)255.0(96.021063.1 stxe−−−=

Equation (h) is solved leading to

(i) 5.1971063.1

)04.0ln(04.

2

1063.1 2

sx

t

e

s

tx s

=−=

=

−

= −

218

Chapter 6

6.19 The properties of the pipe are assumed to be lumped. The mathematical model is

(a) 110T

RT

RdtdTmc p =+

where R is the total resistance of the pipe. It is assumed that the air in the interior of the pipe is at a uniform temperature. The mass of air in the pipe is

( )( ) ( )

(b) F

Btu 158.1Fslugs

Btu 7.72slugs) 150.0(

slugs 150.0

ft 20ft 1ft

slugs 1038.2

r

23

3

2i

=⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

=

=

⎟⎠⎞

⎜⎝⎛=

=

=

−

pmc

x

LVm

π

πρ

ρ

The total resistance is obtained using Equation 4.109. The R value is a measure of thermal conductivity. From Equation (4.102) the resistance of a wall of area A, thickness w made of a material of thermal conductivity k is

(c) kAwR =

In Example 4.14 it is noted that the R value, call it R , is related to resistance through

(d) ˆ

ARR =

Comparison of Equations (c) and (d) shows that the thermal conductivity is related to the R value by

(e) Rwk =

Given that the insulation is 0.5 in thick its thermal conductivity is

( )

(f) Fhrft

Btu 1078.2

Btufthr 15

in 12ft 1in 5.0

3

2

⋅⋅

⋅⋅

⎟⎠⎞

⎜⎝⎛

=

−x

Fk

Taking the thermal conductivity of the steel pipe as Fhrft

Btu 1.35⋅⋅

=sk the total resistance

of the pipe, its insulation, and the convective resistance is obtained using Equation (4.109)

219

Chapter 6

(g) Btu

Fhr 1672.0

1214ln

ft) 20(Fhrft

Btu 35.12

114

5.14lnft) 20(

FhrftBtu 1078.22

1

ft) (20 ft 1214

ftFhrBtu 3.22

1

ft) ft)(20 1(ftFhr

Btu 1.22

1

3

22

⋅=

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

⋅⋅

+⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

⋅⋅

+⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛

⋅⋅

+⎟⎠⎞

⎜⎝⎛

⋅⋅

=

− ππ

ππ

x

R

Substituting calculated values into Equation (a) leads to

(h) 98.598.5158.1 0TTdtdT

=+

(a) The time constant for the system is

(i) hr 194.0 )1672.0)(158.1(

==

= RmcT p

(b) Setting and using the initial condition T(0)=0, taking the Laplace transform of Equation (i) leads to

20 −=T

( )

(j) 16.5

112

)16.5(33.10

98.5158.1

96.11)(

)2(98.5)(98.5)(158.1

⎟⎠⎞

⎜⎝⎛

+−−=

+−=

+−

=Θ

−=Θ+Θ

ss

ss

sss

ssss

Inversion of Equation (j) leads to ( ) (k) 12)( 16.5 tetT −−−=

(c) The temperature in the interior of the pipe reaches 68.3 F when T=-1.7. To this end

(l) hr 367.0)1(27.1 16.5

=−−=− −

te t

6.20 The mathematical model for the temperature distribution in the brake of Example 4.26 is

( ) (a) 141 0

0∞2

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −−−+=+

TtuTtThAThAT

dtdTcDg p

ωωπδρ

The angular velocity of the shaft due to a constant braking torque T is

(b) )( 0ωω +−=eqI

TtT

In order to bring the shaft to rest in 1 sec,

220

Chapter 6

(c) lbft 8.45s 60

min 1rev

r 2minrev350

ftslugs 25.1s) 1(0 2

⋅=

⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛+

⋅−=

T

T π

It is noted that

( ) ( )

( )

0284.0 in 12ft 1in) 5(2

s 60hr 1

fthrBtu 9.4

(d) 1022.2 s 60

hr 1Ffthr

Btu 12.0in 10in 12ft 1in 5.0

ftin 12

inlb 283.0

41

41

22

2

2

233

32

=

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛

⋅⋅=

=

⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛

⋅⋅⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛=

−

π

ππδρ

FhA

x

cDg p

Thus Equation (a) becomes

( )

[ ] (e) )]1(1[1010.21068.168.10284.0022.0

)]1(1[8.4560

)2()350(8.4532)8.1(250284.00284.0022.0

33 −−−+=+

−−⎥⎦⎤

⎢⎣⎡ −++=+

tutxxTdtdT

tutTdtdT π

Taking the Laplace transform of Equation (e), noting that the initial temperature is the same as the ambient,

(f) 101.2102.41010.21068.168.10248.0)77(022.0 2

32

2

33ss e

sxe

sx

sx

sx

ss −− ++−+=Θ+−Θ

(g) )0248.0022.0(

)1(101.2)0248.0022.0(

102.41068.10248.0022.0

69.1)( 2

323

+−

−+

++

+=Θ

−−

ssex

ssexx

ss

ss

Inversion of Equation (g) leads to ( ) ( ) ( )

[ ][ ] (h) )1()1(8871.07769.07769.01055.9

8871.07769.07769.01055.9 )1(1105.111077.6177)(

)1(13.14

13.14

)1(13.1413.1413.1

−−+−+

+−−

−−+−+−=

−−

−

−−−−

tutextex

tuexexet

t

t

tttθ

6.21 The appropriate form of the differential equation is

( ) ( )[ ](a) ...)]7()6()[()]4()3()[(

1141

00

0∞2

+−−−−+−−−−+

−−−+=+

tutuTtTtutuTtT

tuTtThAThATdtdTcDg p

ωω

ωπδρ

6.22 The mathematical model for the change in pressure in a tank is derived in Example 4.10 as

(a) Rp

Rp

dtdp

RV s

a

=+Θ

where is the difference in supply pressure from the initial pressure in the vessel and is the constant absolute temperature in the vessel. Taking the Laplace transform of

both sides of Equation (a) leads to

spΘ

221

Chapter 6

(b) )(

1

)(1

)(

)(1)(1)(

⎟⎠⎞

⎜⎝⎛ Θ

+

⎟⎠⎞

⎜⎝⎛ Θ

=

+⎟⎟⎠

⎞⎜⎜⎝

⎛Θ

=

=+⎟⎟⎠

⎞⎜⎜⎝

⎛Θ

RVR

s

sPRVR

Rs

RV

sPRsP

sPR

sPR

ssPRV

a

sa

a

s

sa

The transfer function is obtained from Equation (b) as

(c)

)()()(

⎟⎠⎞

⎜⎝⎛ Θ

+

⎟⎠⎞

⎜⎝⎛ Θ

=

=

RVR

s

RVRsPsPsG

a

a

s

(a) The time constant is obtained from Equation (c) as

(d) Θ

=aR

RVT

Substitution of given values in Equation (d) leads to

( )

( )

(e) s 1056.3

K 313Kkg

J 0.287

m 20sm

1 0.16

3

3

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

⎟⎠⎞

⎜⎝⎛

⋅=

x

T

(b) Substitution of numerical values into Equation (c) leads to

(f) 7.280

7.280)(+

=s

sG

The supply pressure to the vessel has a sudden increase of 8000 Pa. Thus

(g) 8000)(

)(8000)(

ssP

tutp

s

s

=

=

222

Chapter 6

The system response is obtained from

( )

(h) 7.280

118000

7.280)7.280)(8000(

)()()(

⎥⎦⎤

⎢⎣⎡

+−

+=

=

ss

ss

sPsGsP s

Inversion of Equation (h) leads to [ ] (i) Pa )(8000)( 7.280 tetutp −−=

(c) The pressure in the tank reaches 18 kPa when the change in pressure is 6 kPa. THe time required is determined using Equation (i)

[ ]

(j) s 1094.4

)25.0ln(7.280

125.0

175.0180006000

3

7.280

7.280

7.280

−

−

−

−

=

−=

=

−=

−=

x

t

ee

e

t

t

t

6.23 The denominator of the transfer function of a second-order system has the general form

(a) 2)( 22nn sssD ωζω ++=

The transfer function may be written as

(b) 25.115.2

75.05.0)( 2 +++

=ss

ssG

Comparison of Equation (b) with Equation (a) leads to

(c) sr 35.3

25.11

=

=nω

The damping ratio is determined from

(d) 373.0 )35.3(2

5.25.22

=

=

=

ζ

ζωn

6.24 The transfer function

(a) 50020

53)( 2 +++

=ss

ssG

is that of a second-order system with natural frequency 36.22500 ==nω and damping

ratio 447.0)36.22(2

20==ζ . Thus the system is underdamped. The system’s damped

223

Chapter 6

natural frequency is 0.20)447.0(1)36.22( 2 =−=dω . It is also noted that 10)36.22)(447.0( ==nζω

(a)The impulsive response is obtained using linearity and Table 6.4 as

[ ] (b) )20sin(75.1)20cos(3

)20sin(2015)20sin(

2010)20cos(3)(

10

1010

tte

tettetx

t

tti

+=

⎥⎦⎤

⎢⎣⎡+⎥⎦

⎤⎢⎣⎡ +=

−

−−

(b)The system’s step response is obtained using linearity and Table 6.5 as

[ ][ ] (c) )20cos(01.)20sin(995.201.

)20sin(2010)20cos(1

50015)20sin(3)(

10

101010

tte

tetetetx

t

ttts

−+=⎭⎬⎫

⎩⎨⎧

⎥⎦⎤

⎢⎣⎡ +−+=

−

−−−

6.25 The differential equation for the system is )(a )(101200010 5 tFxxxx =++ &&&

The system’s transfer function is obtained from Equation (a) as

(b) 10000200

1.0

1012000101)(

2

52

++=

++=

ss

xsssG

The system’s natural frequency is sr

n 10010000 ==ω and its damping ratio is

1)100(2

200==ζ . Thus the system is critically damped.

(a)The given initial conditions are 002.0)0( =x and 0)0( =x& . With F(t)=0, Equation (a) can be rewritten as

(d) 010000200(c) 02 2

=++=++

xxxxx nn

&&&&&& ωζω

Taking the Laplace transform of Equation (d), using linearity, the property of transform of derivatives and applying the initial conditions leads to

(e) )100(

2100

002.

)100(4)100100(002.0

)100(4002.0

100002004002.0)(

0)(10000)002(.200)(200)002(.)(

2

2

2

2

2

++

+=

++−+

=

++

=

+++

=

=+−+−

ss

ss

ss

ssssx

sxsxsssxs

Inversion of Equation (e) gives

224

Chapter 6

(f) 2002.0)( 100100 tt teetx −− += (b) The system’s impulsive response is obtained from Table 6.4 as

(g) )( 100ti tetx −=

(c) The system’s step response is obtained using Table 6.5 as

[ ] (h) 100110000

1)( 100100 tts teetx −− −−=

6.26 The differential equation governing the angular displacement of the bar is derived in Example 2.14 resulting in Equation (d) repeated below

(a) )(94

91

91 222 tMkLcLmL =++ θθθ &&&

The free response is obtained by setting M(t)=0, leading to

(b) 04=++ θθθ

mk

mc &&&

The natural frequency is obtained from Equation (b) as

(c) sr 9.276

kg 2.1mN 103.24

4

4

=

⎟⎠⎞

⎜⎝⎛

=

=

x

mk

nω

The damping ratio for the system is obtained from

(d) 2mc

n =ζω

The system is overdamped when the damping ratio is greater than 1. The values of c for which the system is overdamped are obtained from Equation (d) as

( )

(e) m

sN 664.5

sr 9.276kg 2.12

2

⋅

⎟⎠⎞

⎜⎝⎛=

> nmc ω

6.27 The differential equation governing the motion of the block of mass in Example 2.17 is given in Equation (c) repeated below.

1m

( ) ( ) (a) 0999 212121 =++++⎟⎠⎞

⎜⎝⎛ ++ xrkrkxrcrcx

rIrmrm &&&

Substitution of given values leads to

225

Chapter 6

( ) ( )

(b) 01012250.780

0m 1.0mN 1019m 1.0

mN 101

m) 1.0(m

sN 2259m) 1.0(m

sN 225m 0.1

mkg 0.05m) kg)(0.1 9(0.2m) kg)(0.1 1(

5

55

2

=++

=⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ ⋅

+⎟⎠⎞

⎜⎝⎛ ⋅

+⎥⎦

⎤⎢⎣

⎡ ⋅++

xxxx

xxx

xx

&&&

&&&

The system’s natural frequency and damping ratio are obtained from Equation (b) as

(c) sr 1.358

mkg 0.780N 101 5

=

⋅=

xnω

(d) 403.0 )1.358(2

5.288mkgsN 5.288

mkg 0.780sN 2252

=

=

⋅⋅

=⋅⋅

=

ζ

ζωn

Since the damping ratio is less than one the system is underdamped. It is desired to find the free response of the system when the disk is rotated clockwise from equilibrium and then released. The dependent variable is x, the downward displacement of the block of mass 1 kg, measured positive downward from the system’s equilibrium position. The given rotation of the disk leads to the block moving upward such that

°1

( )( )

(e) mm 75.1 360

r 21m 1.0)0(

−=

⎟⎠⎞

⎜⎝⎛

°°−=

πx

Since the system is released from rest (f) 0)0( =x&

The free-response of an underdamped second-order system is given by Equations (6.68)-(6.71). Direct substitution into these equations leads to

(g) sr 7.327

)403.0(1sr 1.358

1

2

2

=

−⎟⎠⎞

⎜⎝⎛=

−= ζωω nd

226

Chapter 6

( )( )

(h) mm 91.1

sr 7.327

1075.1sr 1.358)403.0(0

m 1075.1

23

23

2

0020

=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛−⎟

⎠⎞

⎜⎝⎛+

+−=

⎟⎟⎠

⎞⎜⎜⎝

⎛ ++=

−

−

xx

xxxA

d

nd ω

ζω&

( )

( )(i) r 985.1

m 1075.1sr 1.358)403.0(0

sr 7.327m 1075.1

tan

tan

3

3

1

00

01

−=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−⎟⎠⎞

⎜⎝⎛+

⎟⎠⎞

⎜⎝⎛−

=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

−

−

−

−

x

x

xxx

n

dd ζω

ωφ

&

( )( )

(j) m )985.17.327sin()1091.1( )985.17.327sin(1091.1

sin)(

2,1443

)1.358)(403.0(3

−=

−=

+=

−−

−−

−

textex

teAtx

t

tdd

td

n φωζω

6.28 A free-body diagram of the door is illustrated below

The door rotates about a fixed axis through its hinges. Summing moments about this axis leads to

227

Chapter 6

(a) 02

2

=++

=−−

=∑

θθθ

θθθ

α

ttO

Ott

OO

kcI

Ick

IM

&&&

&&&

Substitution of given values into Equation (a) leads to

(b) 019.12.25

0302.25

=++

=++

θθθ

θθθ

&&&

&&&

t

t

cc

The natural frequency of the door is determined from Equation (b) as

(c) sr 09.1

19.1

=

=nω

The system is critically damped, thus 1=ζ and 18.2)09.1)(1)(2(2 ==nζω . Thus Equation (b) is written as

(d) 019.118.2 =++ θθθ &&& The free response of a critically damped system is obtained using Equation (6.76)

( ) (e) )( 000

tn

t nn etet ωω ωθθθθ −− ++= & (a) The initial conditions are

( )

(g) 0)0(

(f) r 873.0 360

r 250)0(

0

0

==

=

⎟⎠⎞

⎜⎝⎛

°°==

θθ

πθθ

&

Substituting Equations (c), (f) and (g) into Equation (e) leads to

( ) (h) 952.0873.0 )09.1)(873.0(873.0)(

09.1

09.109.1

t

tt

etteet

−

−−

+=

+=θ

The time required to for the door to close to is obtained by °5

( ) ( )

( ) (i) 952.0873.00873.0

952.0873.0360

r 25

09.1

09.1

t

t

et

et

−

−

+=

+=⎟⎠⎞

⎜⎝⎛

°°

π

Equation (i) is solved using a trial and error procedure leading to (j) s 57.3=t

(b) Since the door is closed at t=0,

(k) 0)0( =θ and Equation (e) becomes

(l) 09.1)( 09.10

ttet −= θθ & The door will open to if the velocity of the door is zero when the door reaches . To this end

r 22.170 =°°70

( ) (m) 09.1109.1)( 09.10

tett −−= θθ &&

228

Chapter 6

Equation (m) leads to the conclusion that the velocity is zero when

(n) s 917.009.11

==t

Substituting Equation (n) into Equation (l) and requiring r 22.1)917.0( =θ leads to

(o) sr 448.0

)917.0(09.122.1

0

)917.0)(09.1(0

=

= −

θ

θ

&

& e

6.29 A free-body diagram of the door is illustrated below

The door rotates about a fixed axis through its hinges. Summing moments about this axis leads to

(a) 02

2

=++

=−−

=∑

θθθ

θθθ

α

ttO

Ott

OO

kcI

Ick

IM

&&&

&&&

Substitution of given values into Equation (a) leads to

(b) 019.12.25

0302.25

=++

=++

θθθ

θθθ

&&&

&&&

t

t

cc

The natural frequency of the door is determined from Equation (b) as

(c) sr 09.1

19.1

=

=nω

The system has a damping ratio of 15.1=ζ and 51.2)09.1)(15.1)(2(2 ==nζω . Thus Equation (b) is written as

(d) 019.151.2 =++ θθθ &&&

229

Chapter 6

The free response of an overdamped system is obtained using Equation (6.80)

( ) ( ) (e) 1112

1)( 21 020

0202 ⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎦

⎤⎢⎣

⎡+−++⎥

⎦

⎤⎢⎣

⎡−−+−

−= ts

n

ts

n

eetωθ

ζζθωθ

ζζθζ

θ&&

where ( )

( )

( )(g) 9029.0

1

(f) 874.1 1)15.1(15.1)09.1(

1

22

2

21

−=

−−−=

−=

−+−=

−+−=

ζζω

ζζω

n

n

s

s

Substitution of Equations (f) and (g) in Equation (e) leads to

[ ] [ ] (h) 917.0473.1917.08275.055.1)( 9029.000

874.100

tt eet −− ++−−= θθθθθ && (a) The initial conditions are

( )

(j) 0)0(

(i) r 873.0 360

r 250)0(

0

0

==

=

⎟⎠⎞

⎜⎝⎛

°°==

θθ

πθθ

&

Substituting Equations (i) and (j) into Equation (h) leads to (k) 99.112.1)( 9029.0874.1 tt eet −− +−=θ

The time required to for the door to close to is obtained by °5

( )

(l) 99.112.10873.0

99.112.1360

r 25

9029.0874.1

9029.0874.1

tt

tt

ee

ee

−−

−−

+−=

+−=⎟⎠⎞

⎜⎝⎛

°°

π

Equation (l) is solved using a trial and error procedure leading to (m) s 44.3=t

(b) Since the door is closed at t=0,

(n) 0)0( =θ and Equation (e) becomes

(o) 42.1)( 9029.0874.10

tt eet −− +−= θθ &

The door will open to if the velocity of the door is zero when the door reaches . To this end

r 22.170 =°°70

( ) (p) 9029.01.874e42.1)( 9029.0-1874t0

tet −−= θθ && Equation (p) leads to the conclusion that the velocity is zero when

230

Chapter 6

( )

(q) s 752.0

482.0ln9711.01

482.0874.19029.0

09029.01.874e

9711.0

9029.0

874.1

9029.0-1874t

=

−=

=

=

=−

−

−

−

−

t

t

eee

e

t

t

t

t

Substituting Equation (q) into Equation (p) and requiring r 22.1)917.0( =θ leads to ( )

(s) sr 1060.4

9029.01.874e42.122.1

30

)752.0(9029.02)-1874(0.750

x

e

=

−= −

θ

θ

&

&

6.30 A free-body diagram of the vehicle drawn at an arbitrary instant is shown below

(a) Application of Newton’s second law to the free-body diagram leads to

(a) )()(kyyckxxcxm

xmxycxyk

xmF

+=++=−−−−

=∑

&&&&&&&&

&&

(b) The system’s transfer function is obtained as

(b)

)()()(

2 kcsmskcs

sYsXsG

+++

=

=

Substitution of given values into Equation (b) leads to

(c) 64020

64020

102.3101500

102.3101)(

2

542

54

+++

=

+++

=

sss

xsxsxsxsG

Equation (c) is the transfer function of a second-order system with a natural frequency of

d)( sr 3.25

640

=

=nω

231

Chapter 6

and a damping ratio determined as

(e) 395.0 220

202

=

=

=

n

n

ωζ

ζω

Since the damping ratio is less than one the system is underdamped. The damped natural frequency is calculated as

(f) sr 2.23

1 2

=

−= ζωω nd

The system response occurs due to an input of (g) )(0128.0)( tuty =

Thus the response is (h) )(0128.0)( txtx s=

where is the step response of the system which is obtained using Table 6.6 and superposition as

)(txs

[ ](i) )2.23sin(8.31)2.23cos(5.275.27

)2.23sin(430.0)2.23cos(15.27)2.23sin(20

)sin()cos(1640)sin(20)(

1010

101010

tetetetete

tetetetx

tt

ttt

dt

d

nd

t

dd

ts

nnn

−−

−−−

−−−

+−=

+−+=

⎥⎦

⎤⎢⎣

⎡+−+= ωωζω

ωω

ω ζωζωζω

The system response is obtained using Equations (h) and (i) as (j) )2.23sin(407.0)2.23cos(352.0352.0)( 1010 tetetx tt −− +−=

6.31 The mathematical model for the system is

(a) )(12

)(

2 tFm

xxx

tFkxxcxm

nn =++

=++

ωζω &&&

&&&

Where m=500 kg, the damping ratio is 0.28 and

(b) r/s 0.49kg 500

mN 102.1 6

==x

nω

The input force is (c) )]2()([500)( −−= tututtF

The convolution integral solution for the response is

(d) )](sin[)]2()([5001)(0

))(0.49)(28.0(∫ −−−= −−t

dt

d

teuum

tx τωτττω

τ

where the damped natural frequency is (e) r/s 0.47)28.0(1)0.49( 2 =−=dω

Equation (d) becomes

232

Chapter 6

(f) )](0.47sin[)]2()([0213.0)(0

)(7.13∫ −−−= −−t

t teuutx ττττ τ

The symbolic capabilities of MATLAB are used to evaluate the integral. For t<2 the commands and response are f = 213/10000*tau*exp(-137/10*t+137/10*tau)*sin(47*t-47*tau) >> y=int(f,tau,0,t) y = -1371507/287206147805+10011/23966900*t+1371507/287206147805*cos(47*t)*exp(-137/10*t)-43053903/5744122956100*sin(47*t)*exp(-137/10*t) >> y1=vpa(y,3) y1 = -.478e-5+.418e-3*t+.478e-5*cos(47.*t)*exp(-13.7*t)-.750e-5*sin(47.*t)*exp(-13.7*t) For t>2 sec the appropriate commands for integration are >> z=int(f,tau,0,2) z = 2385611289/2872061478050*cos(47*t-94)*exp(-137/10*t+137/5)+1802262651/7180153695125*sin(47*t-94)*exp(-137/10*t+137/5)+1371507/287206147805*cos(47*t)*exp(-137/10*t)-43053903/5744122956100*sin(47*t)*exp(-137/10*t) >> z1=vpa(z.3) >> z1=vpa(z,3) z1 = .831e-3*cos(47.*t-94.)*exp(-13.7*t+27.4)+.251e-3*sin(47.*t-94.)*exp(-13.7*t+27.4)+.478e-5*cos(47.*t)*exp(-13.7*t)-.750e-5*sin(47.*t)*exp(-13.7*t) To summarize for t<2 sec

(g) )47sin(105.7)47cos(1078.41018.41078.4)( 7.1367.136406 textextxxtx tt −−−−−− −++−= While for t>2 sec

(h) )47sin(1050.7)47cos(1078.4 )9447sin(1051.2)9447cos(1031.8)(

7.1367.136

247,134247,134

textextextextx

tt

tt

−−−−

+−−+−−

−+

−+−=

233

Chapter 6

The transmitted force is xckx &+ . MATLAB is used to differentiate the response and plot. The MATLAB program to plot the response is % Problem 6.31 t=1.3 dt=4/100; t=0 k=1.2e6 m=500 wn=(k/m)^0.5; c=2*.28*m*wn; for i=1:101 tt(i)=t; if t<2 y1(i)=-.478e-5+.418e-3*t+.478e-5*cos(47.*t)*exp(-13.7*t)-.750e-5*sin(47.*t)*exp(-13.7*t); dx=.418e-3-.121910e-3*sin(47.*t)*exp(-13.7*t)-.417986e-3*cos(47.*t)*exp(-13.7*t); F(i)=k*y1(i)+c*dx; t=t+dt; else y1(i)=.831e-3*cos(47.*t-94.)*exp(-13.7*t+27.4)+.251e-3*sin(47.*t-94.)*exp(-13.7*t+27.4)... +.478e-5*cos(47.*t)*exp(-13.7*t)-.750e-5*sin(47.*t)*exp(-13.7*t); dx=-.424957e-1*sin(47.*t-94.)*exp(-13.7*t+27.4)+.4123e-3*cos(47.*t-94.)*exp(-13.7*t+27.4)... -.121910e-3*sin(47.*t)*exp(-13.7*t)-.417986e-3*cos(47.*t)*exp(-13.7*t); F(i)=k*y1(i)+c*dx; t=t+dt; end end figure plot(tt,y1) xlabel('t (s)') ylabel('x (m)') title('Displacement vs. time for Problem 6.31') figure plot(tt,F) xlabel('t (s)') ylabel('F (N)') title('Transmitted force vs. time for Problem 6.31')

234

Chapter 6

The resulting plots are

The displacement is not discontinuous at t=2. When the force is removed the system response is that of a system with a free response. The large natural frequency leads to large derivatives.

235

Chapter 6

6.32 The currents are as defined below.

Application of KCL at node B leads to

(a) 213

321

iiiiii

−=−−

Application of KVL around loop ABEF in the clockwise direction and using Equation (a) gives

( ) (b) 01 0

0211 =+−+ ∫ C

qdtii

CRi

t

Application of KVL around loop BCDE in the clockwise direction and using Equation (a) leads to

(c) 0 )(1 0

0212

2 =−−−+ ∫ Cqdtii

CRi

dtdiL

t

Taking the Laplace transform of Equations (b) and (c) noting that leads to 0)0(2 =i

( )

( ) (e) 0)()(1)()(

(d) 0)()(1)(

02122

0211

=−−−+

=+−+

Csq

sIsICs

sRIsLsI

Csq

sIsICs

sRI

Equations (d) and (e) are summarized in matrix form as

(f) )()(

11

11

0

0

2

1

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−=⎥

⎦

⎤⎢⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++−

−+

CsqCsq

sIsI

CsRLs

Cs

CsCsR

Cramer’s rule is used to solve for leading to )(2 sI

236

Chapter 6

(g) 2

2

111

11

11

11

1

1

)(

2

0

2

0

0

0

2

⎥⎦

⎤⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛ ++

⎟⎠⎞

⎜⎝⎛ ++

−=

⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛ ++⎟⎠⎞

⎜⎝⎛ +

⎟⎠⎞

⎜⎝⎛ +++−

=

++−

−+

++−

−−

=

CsR

CLRLRsCs

CsRLsq

CsCsRLs

CsR

CsCsRLs

Csq

CsRLs

Cs

CsCsR

CsRLs

Csq

CsCsq

sI

Equation (g) is simplified to

(h) 21

2

2

2

)(

2

20

22

20

2

⎥⎦

⎤⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛ ++

⎟⎠⎞

⎜⎝⎛ ++

−=

⎥⎦

⎤⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛ ++

⎟⎠⎞

⎜⎝⎛ ++

−=

LCs

RCLRsLRCs

CRsLsq

CRs

CLRLRsCs

CRsLsq

sI

(a) It is given that L=0.2 H, C=0.4 µF, C 1.30 µ=q , and R=1000 Ω. Substitution into Equation (h) leads to

( )

( ) (i) 105.27500105.210002.00388.0

)104.0)(2.0(2

)104.0)(1000(1

2.01000)104.0)(1000)(2.0(

104.0210002.0101.3

)(

72

62

6626

626

2

xsssxss

xs

xssx

xssx

sI

++++

−=

⎥⎦

⎤⎢⎣

⎡+⎟⎟

⎠

⎞⎜⎜⎝

⎛++

⎟⎠⎞

⎜⎝⎛ ++

−=

−−−

−−

The quadratic polynomial in the denominator of Equation (i) has complex roots. A partial fraction decomposition of Equation (i) leads to

237

Chapter 6

(j) 10094.1)3750(

625)3750(1.01.00388.0

105.27500

2501.01.00388.0)(

72

722

⎟⎟⎠

⎞⎜⎜⎝

⎛++

−++−=

⎟⎠⎞

⎜⎝⎛

++−

+−=

xss

s

xsss

ssI

Inversion of Equation (j) leads to [ ] (k) )3307sin(189.0)3307cos(1.0)(1.00388.0)( 37503750

2 tetetuti tt −− −+−=

(b) It is given that L=0.2 H, C=0.4 µF, C 1.30 µ=q , and R=2000 Ω. Substitution into Equation (h) leads to

( )

( ) (l) 105.211250105.220002.00194.0

)104.0)(2.0(2

)104.0)(2000(1

2.02000)104.0)(2000)(2.0(

104.0220002.0101.3

)(

72

62

6626

626

2

xsssxss

xs

xssx

xssx

sI

++++

−=

⎥⎦

⎤⎢⎣

⎡+⎟⎟

⎠

⎞⎜⎜⎝

⎛++

⎟⎠⎞

⎜⎝⎛ ++

−=

−−−

−−

The quadratic polynomial in the denominator of Equation (l) has real roots of

and . A partial fraction decomposition leads to 31020.8 x− 31005.3 x−

(m) 1005.3

1106.01020.8

0106.01.00194.0)( 332 ⎟⎠⎞

⎜⎝⎛

++

+−−=

xsxsssI

Inversion of Equation (m) leads to ( ) (n) 1106.00106.0)(1.00194.0)(

33 1005.31020.82

txtx eetuti −− +−−= (c) It is given that L=0.2 H, C=0.4 µF, C 1.30 µ=q , and R=3000 Ω. Substitution into Equation (h) leads to

( )

( ) (o) 105.21058.1

105.230002.00129.0

)104.0)(2.0(2

)104.0)(3000(1

2.03000)104.0)(3000)(2.0(

104.0230002.0101.3

)(

742

62

6626

626

2

xsxssxss

xs

xssx

xssx

sI

++++

−=

⎥⎦

⎤⎢⎣

⎡+⎟⎟

⎠

⎞⎜⎜⎝

⎛++

⎟⎠⎞

⎜⎝⎛ ++

−=

−−−

−−

The quadratic polynomial in the denominator of Equation (l) has real roots of

and . A partial fraction decomposition leads to 41040.1 x− 31078.1 x−

(p) 1078.1

0105.01058.1

0015.01.00129.0)( 342 ⎟⎠⎞

⎜⎝⎛

++

+−−=

xsxsssI

Inversion of Equation (m) leads to ( ) (q) 0105.00015.0)(1.00129.0)(

34 1078.11058.12

txtx eetuti −− +−−=

238

Chapter 6

6.33 The circuit diagram with the switch closed is illustrated below.

Application to KVL around each of the loops leads to

(b) 0)(C1

(a) )(1

120

2

210

111

=−+

=−+++

∫

∫

iiRdti

viiRdtiC

Ridtdi

L

t

t

Taking the Laplace transforms of Equations (a) and (b), assuming all initial conditions are zero leads to

(d) 0)(1)(

(c) )()()(12

21

21

=⎟⎠⎞

⎜⎝⎛ ++−

=−⎟⎠⎞

⎜⎝⎛ ++

sICs

RsRI

sVsRIsICs

RLs

Equations (c) and (d) are summarized in matrix form as

(e) 0

)()()(

1

12

2

1⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+−

−++ sVsIsI

CsRR

RCs

RLs

Cramer’s rule is used with Equation (e) giving

239

Chapter 6

(f) 12

)(1

12

)(1

1

12

10

)(

)(

2223

2

222

2

Cs

CRs

CLRLRs

sVC

Rs

sCCsR

CLRLRs

sVCs

R

CsRR

RCs

RLs

CsR

RsV

sI

++⎟⎠⎞

⎜⎝⎛ ++

⎟⎠⎞

⎜⎝⎛ +

=

++⎟⎠⎞

⎜⎝⎛ ++

⎟⎠⎞

⎜⎝⎛ +

=

+−

−++

+

−

=

The system output is the voltage change across the capacitor,

(g) 1

022 ∫=

t

dtiC

v

such that

(h) )(1)( 22 sICs

sV =

Use of Equation (f) in Equation (h) leads to

(j) 121

1

(i) 12

1

)()()(

223

2

2223

2

2

⎥⎦

⎤⎢⎣

⎡++⎟

⎠⎞

⎜⎝⎛ ++

+=

⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟

⎠⎞

⎜⎝⎛ ++

+=

=

LRCs

LCs

RCLRsLRCs

CRs

Cs

CRs

CLRLRsCs

CRs

sVsVsG

(a) The system input is a step input of the form

(k) 12)(

)(12

1

1

ssV

tuv

=

=

Substituting given numerical values of L=0.25 H, R=3.6 kΩ and C=0.15 µF as well as Equation (k) in Equations (h) and (j) leads to

240

Chapter 6

( )

( ) ( )( )

( )(m) 1094.41033.51086.1

1015.0)3600)(25.0(1

1015.025.02

1015.0)3600(1

25.03600)(ˆ

(l) )(ˆ1015.0)3600)(25.0(

1015.01360012

)(

107233

6

62

63

26

62

2

xsxsxsx

sx

sx

ssD

sDsxx

ssV

+++=

+

+⎟⎟⎠

⎞⎜⎜⎝

⎛++=

⎟⎠⎞

⎜⎝⎛ +

=

−

−−

−

−

Use of Equation (m) in Equation (l) leads to

( ) (n) 1094.41033.51086.1

1093.51020.3)( 1072332

1128

2 xsxsxssxsxsV

++++

=

Use of MATLAB’s residue command leads to

(o) 1024.55.914

0.161095.61041.9

108.6120130.0)( 72222 xsss

xssssV

+++−

++

++−

=

Inversion of Equation (o) leads to

(p) )1022.7cos(0.4492e

)1022.7cos(095.6108.612)(0130.0)(3557t-

355794122

tx

txeettusV tt

+

−++−= −−

6.34 The currents in the op-amp circuit are shown below.

Application of KCL at node A gives

(a) 0321 =−− iii The amplifier is assumed to be ideal, such that 0=Av and 03 =i . The current through the 300 kΩ resistor is

(b) 103 5

11 x

vi =

The current through the parallel combination of the resistor and capacitor is

241

Chapter 6

(c) 1020106

265

22 dt

dvx

xv

i −−−=

Substitution of Equations (b) and (c) into Equation (a) leads to

(e) 212

010201067.11033.3

122

262

61

6

vvdt

dvdt

dvxvxvx

=+

=++ −−−

Taking the Laplace transform of Equation (e), assuming initial conditions are zero leads to

(f) 112)(2

)( 12 +

=s

sVsV

If then )(201 tuv =

(g) 20)(1 ssV =

Which when substituted into Equation (f) leads to

(h)

121

1140

121

310

)112(40)(2

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+−=

⎟⎠⎞

⎜⎝⎛ +

=

+=

ss

ss

sssV

Inversion of Equation (h) leads to

(i) V )(40)( 122 ⎥

⎦

⎤⎢⎣

⎡−=

−t

etutv

6.35 The mathematical model for the two-tank system is

(b) )(111

(a) )(11

2221

11

22

121

11

11

tqhRR

hRdt

dhA

tqhR

hRdt

dhA

p

p

=⎟⎟⎠

⎞⎜⎜⎝

⎛++−

=−+

The resistances are calculated using the steady-state values

242

Chapter 6

(c) ms 45.1

sm 4.0

m 8.6m 7.92

2

2

3

1

211

=

−=

−=

s

ss

qhh

R

(d) ms 4.3

sm 0

sm 4.0

m 8.62

2

2

33

21

22

=

+=

+=

ss

s

qqh

R

The perturbation in flow rates are

(f) m )(2.0)(

(e) 0)(3

2

1

stutq

tq

p

p

=

=

Substitution of Equations (c)-(f) in Equations (a) and (b) leads to

(g) 045.11

45.11450 21

1 =−+ hhdtdh

(h) )(2.040.31

45.11

45.11790 21

2 tuhhdt

dh=⎟

⎠⎞

⎜⎝⎛ ++−

Equations (g) and (h) are rearranged as

(j) )(2.0984.0690.0790

(i) 0690.0690.0450

212

211

tuhhdt

dh

hhdtdh

=+−

=−+

Taking the Laplace transforms of Equations (i) and (j), taking all initial conditions to be zero leads to

(k) 2.00

)()(

984.0790690.0690.0690.0450

2

1

⎥⎥⎦

⎤

⎢⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡+−

−+

ssHsH

ss

Cramer’s rule is applied to Equation (k) leading to

243

Chapter 6

( )

( )

(l) 2029.093.4

9.9871002.193.4138.0

)2.05x10987.9)(ss(s0.138

2029.09.987138.0

)690.0()984.0790)(690.0450(

2.0690.0

984.0790690.0690.0690.0450

984.07902.0690.00

)(

6

4-

2

2

1

⎟⎟⎠

⎞⎜⎜⎝

⎛+

−+

+=

++=

++=

−−++

⎟⎠⎞

⎜⎝⎛

=

+−−+

+

−

=

−

ssx

s

sss

sss

ss

sssH

Inversion of Equation (l) gives (m) 680.01041.1)(680.0)( 2029.09.9877

1tt eextuth −−− −+=

Application of Cramer’s rule to Equation (k) also leads to

( )

( )

(n) 2029.0

589.09.987

0911.680.0

)2.05x10987.9)(ss(s0.13890s

2029.09.987138.090

)690.0()984.0790)(690.0450(

2.0690.0450

984.0790690.0690.0690.0450

2.0690.00690.0450

)(

4-

2

2

2

⎟⎠⎞

⎜⎝⎛

+−

+−=

+++

=

+++

=

−−++

⎟⎠⎞

⎜⎝⎛+

=

+−−+

−

+

=

sss

ssss

sss

s

ss

s

s

sH

Inversion of Equation (n) gives

(o) 589.00911.0)(680.0)( 2029.09.9871

tt eetuth −− −−=

244

Chapter 6

6.37 Free-body diagrams of the blocks at an arbitrary instant are shown below.

Application of Newton’s second law applied to free-body diagrams of each of the blocks at an arbitrary instant leads to

(a) 01031052

2)(103102

25

15

1

1125

15

11

=−+

=−+−

=∑

xxxxx

xxxxxx

xmF

&&

&&

&&

(b) 01041033

3101)(103

25

15

2

225

125

2

=+−

=−−−

=∑

xxxxx

xxxxxx

xmF

&&

&&

&&

Equations (a) and (b) are summarized in matrix form as

(c) 00

4335

103002

2

15

2

1⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡xx

xx&&&&

Application of the Laplace transform to Equation (c) leads to

(d) 00

)()(

433352

2

12

2

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

+−−+

sXsX

ss

The natural frequencies are obtained by setting the determinant of the coefficient matrix of Equation (d) to zero leading to

(e) 01011102360)103)(103()1043)(1052(

01043103

1031052

10254

555252

525

552

=++

=−−−++

=+−

−+

xsxsxxxsxs

xsxxxs

The solutions of Equation (e) are . Thus the natural frequencies are

jxjxs 22 1072.5 ,1036.2 ±±=

(g) sr 1072.5

(f) sr 1036.2

22

21

x

x

=

=

ω

ω

245

Chapter 6

6.38 Free-body diagrams of the block and the junction between the spring and viscous damper are shown below

Application of Newton’s law to the free-body diagram of the block leads to

(a) )()()()(

221

21

tFzkxkkxmxmzxkxktF

maF

=−++=−−−

=∑

&&&&

Summing forces at the junction between the spring and viscous damper leads to

(b) 00)(

22

2

=−−=−+−

zkzcxkzxkzc

&&

Taking the Laplace transform of Equations (a) and (b) leads to

(d) 0)()()((c) )()()()()(

22

2212

=−−=−++

sZkscsZsXksFsZksXkkxXms

Equations (c) and (d) are rearranged and written in matrix form as

(e) 0

)()()(

22

2212

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

+−−++ sF

sZsX

kcskkkkms

(a) Cramer’s rule is used to solve Equation (e) for X(s) leading to

( )(f)

)()()(

)()()(

0)(

)(

21212

23

2

22221

22

22

2212

2

2

kkskkcsmkmcssFkcs

kkcskkmssFkcs

kcskkkkms

kcsksF

sX

+++++

=

−++++

=

+−−++

+−

=

The desired transfer function is obtained from Equation (f) as

246

Chapter 6

(g) )(

)()()(

21212

23

2

1

kkskkcsmkmcskcs

sFsXsG

+++++

=

=

(b) Cramer’s rule is again applied to determine Z(s) as

( )(h)

)()(

)()(

0)(

)(

21212

23

2

22221

22

22

2212

2

212

kkskkcsmkmcssFk

kkcskkmssFk

kcskkkkms

ksFkkms

sZ

++++=

−+++=

+−−++

−++

=

The transmitted force is calculated as (i) )( 1 zcxktFT &+=

Taking the Laplace transform of Equation (i) leads to (j) )()()( 1 scsZsXksFT +=

Substitution of Equations (f) and (h) into Equation (j) gives

[ ] (k) )(

)()(

)()(

)()()()(

21212

23

221

21212

23

2

21212

23

21

kkskkcsmkmcssFcskkcsk

kkskkcsmkmcssFkcs

kkskkcsmkmcssFkcsksFT

++++++

=

+++++

+++++

=

The desired transfer function is determined from Equation (k) as

(l) )(

)(

)()()(

21212

23

2121

2

kkskkcsmkmcskkskkc

sFsFsG T

++++++

=

=

(c) The impulsive responses are the inverses of the transfer functions. These cannot be determined in closed form without specific values for all parameters. 6.39 A free body diagram of the system at an arbitrary instant is shown below assuming small θ,

247

Chapter 6

Summing moments about the pin support

( ) (a) 22

)( θθθ

α

&&&O

OO

ILLcLkLLtF

IM

=⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛−−

=∑

The moment of inertia of the bar about O is obtained using the parallel axis theorem as

(b) 31

2121

2

22

mL

LmmLIO

=

⎟⎠⎞

⎜⎝⎛+=

Substitution of Equation (b) in Equation (a) leads to

(c) )(3343

)(43

1 22

2

tFmLm

kmc

tLFkLLcmL

=++

=++

θθθ

θθθ

&&&

&&&

Taking the Laplce transform of Equation (c) leads to the transfer function

(d) 3

43

3

)()()(

2

mks

mcs

mL

sFsXsG

++=

=

Substitution of given values leads to

(e) 1011.16.173

139.0

8.10)104(3

)8.10(4)2500(3

)2)(8.10(3

)(

52

52

xss

xsssG

++=

++=

The transfer function of Equation (e) is that of a second-order system with

(f) sr 3.333

1011.1 5

=

= xnω

248

Chapter 6

and a damping ratio determined by

(g) 2604.06.1732

==

ζζωn

Since the damping ratio is less than one, the system is underdamped. (a) If )(10)( ttF δ= ,the response is obtained from Table 6.5 as

(h) mm )sin(1)139.0)(10()( tetx dt

d

n ωω

ζω−=

where the damped natural frequency is

(i) sr 8.321

1 2

=

−= ζωω nd

Substituting known values into Equation (h) leads to

(j) )8.321sin(32.4

)8.321sin(8.321

39.1)(

8.86

)3.333)(2604.0(

te

tetx

t

t

−

−

=

=

The MATLAB worksheet used to obtain a plot of Equation (j) is N=[1.39] N = 1.3900 >> D=[1 173.6 1.11e5] D = 1.0e+005 * 0.0000 0.0017 1.1100 >> impulse(N,D)

249

Chapter 6

(b) The step response is obtained using Table 6.6

[ ] (k) mm )8.321sin(270.0)8.321cos(1125.0

)8.321sin(8.321

)3.333)(2604.0()8.321cos(1)3.333(

9.13

)sin()cos(1)139.0)(100()(

8.86

)3.333)(2604.0(2

2

tte

tte

tetetx

t

t

dt

d

nd

t

n

nn

+−=⎭⎬⎫

⎩⎨⎧

⎥⎦⎤

⎢⎣⎡ +−=

⎥⎦

⎤⎢⎣

⎡+−=

−

−

−− ωωζω

ωω

ζωζω

The step response is obtained from MATLAB

250

Chapter 6

6.40 The transfer function for the series LRC circuit is

(a) 10513

2.0

1016.22.0)(

72

72

xsss

xssssG

++=

++=

The transfer function of Equation (a) is that of a second-order system of natural

frequency 37 1007.7105 xxn ==ω and damping ratio 43 1019.9)1007.7(2

13 −== xx

ζ .

Thus the system is underdamped with a damped natural frequency of ( ) 3243 1007.7)1019.9(11007.7 xxxd =−= −ω . It is noted that 5.6=nζω .

(a)The response for )(10)( ttv δ= is obtained from Table 6.4 as

)(b )1007.7sin(1007.75.6)1007.7cos(10)( 3

335.6

⎥⎦⎤

⎢⎣⎡ += − tx

xtxeti t

The response is plotted using MATLAB by the commands N=[2 0] D=[1 13 5E7] impulse(N,D)

251

Chapter 6

(b) The response when is obtained using Table 6.5 as )(10)( tutv =

(c) )1007.7sin()( 35.6 txeti t−= The command step(N,D) is used along with the previously defined N and D in MATLAB to generate the step response shown below

(c) If then ttv 5.0)( = 2

5.0)(s

sv = and since )()()( sGsvsI = , Equation (a) gives

(d) )10513(

1.0)( 72 xssssI

++=

252

Chapter 6

Partial fraction decomposition of Equation (a) leads to

)(e )1007.7()5.6(

103.1)1007.7()5.6(

)5.6(102102

)1007.7()5.6(106.2)5.65.6(102102

10513106.2102102)(

232

8

232

99

232

899

72

899

xsx

xssx

sx

xsxsx

sx

xssxsx

sxsI

++−

+++

−=

+++−+

−=

+++

−=

−−−

−−−

−−−

Inversion of Equation (e) leads to (f) )1007.7sin(103.1)1007.7cos(102102)( 35.6835.699 txextxexxti tt −−−−− −−=

The response is plotted from MATLAB using the following commands N=[0.1 0] D=[1 13 5E7 0] step(N,D)

253

Chapter 6

6.41 The currents in the branches of the circuit are as defined below

Application of KVL to each of the loops leads to

[ ]

[ ]

(c) 0106.1)(106.22.0

(b) 0)(106.2)()(1010.1

(a) )()()(1013.0

33

3233

323

021

72

021

71

=+−−

=−+−−

=−+

∫

∫

ixiixdtdi

iixdttitixdtdi

tvdttitixdtdi

t

t

Taking the Laplace transforms of Equations (a)-(c), assuming all initial conditions are zero leads to

[ ]

[ ] [ ]

[ ] (f) 0)(106.1)()(106.2)(2.0

(e) 0)()(106.2)()(101)(1.0

(d) )()()(101)(3.0

33

323

3

323

21

7

2

21

7

1

=+−−

=−+−−

=−+

sIxsIsIxssI

sIsIxsIsIs

xssI

sVsIsIs

xssI

Equations (d)-(f) are summarized in matrix form as

(g) 00

)(

)()()(

106.32.0106.20

106.2101106.21.0101

01011013.0

3

2

1

33

37

37

77

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

+−

−++−

−+sV

sIsIsI

xsx

xs

xxss

xs

xs

xs

Cramer’s rule is applied to Equation (g) in order to solve for leading to )(3 sI

254

Chapter 6

( )(h)

)(

106.2101)(

106.32.0106.20

106.2101106.21.0101

01011013.0

0106.20

0101106.21.0101

)(1011013.0

)(

37

33

37

37

77

3

73

7

77

3

sD

xs

xsV

xsx

xs

xxss

xs

xs

xs

xs

xxss

x

sVs

xs

xs

sI

−⎟⎟⎠

⎞⎜⎜⎝

⎛−

=

+−

−++−

−+

−

++−

−+

=

where

( ) ( )

( )

(i) 1060.21096.11058.1264006.0

101106.32.0

1013.0106.2106.32.0101106.21.01013.0)(

1310623

273

7233

73

7

sxxsxss

sxxs

sxsxxs

sxxs

sxssD

++++=

⎟⎟⎠

⎞⎜⎜⎝

⎛−+−

⎟⎟⎠

⎞⎜⎜⎝

⎛+−−+⎟⎟

⎠

⎞⎜⎜⎝

⎛++⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

Use of Equation (i) in Equation (h) and simplifying leads to

(j) 1033.41027.31063.21040.4

)(1033.4)( 151228344

12

3 xsxsxsxssVxsI

++++=

The system output is the voltage change across the 1600 Ω resistor

(k) 1033.41027.31063.21040.4

1093.6

)()(1600

)(

151228344

15

3

xsxsxsxsx

sVsI

sG

++++=

=

The step response is obtained by inverting )(1 sGs

. Thus

( ) (l) 1033.41027.31063.21040.4

1093.6)( 151228344

15

2 xsxsxsxssxsV s ++++

=

The following MATLAB worksheet is used to help determine a partial fraction decomposition of the right-hand side of Equation (l) >> D=[1 4.40e4 2.63e8 3.27e12 4.33e15 0]

255

Chapter 6

D = 1.0e+015 * 0.0000 0.0000 0.0000 0.0033 4.3300 0 >> [r,p,k]=residue(N,D) r = 0.0031 0.0578 + 0.1283i 0.0578 - 0.1283i -1.7192 1.6005 p = 1.0e+004 * -3.9358 -0.1594 + 0.8551i -0.1594 - 0.8551i -0.1454 0 k = [] >> syms s >> f1=(s-p(2))*(s-p(3)) f1 = (s+7010686265853559/4398046511104-2350353352750307/274877906944*i)*(s+7010686265853559/4398046511104+2350353352750307/274877906944*i) >> f2=vpa(simplify(f1),3) f2 = .517e-25*(.440e13*s+.701e16-.376e17*i)*(.440e13*s+.701e16+.376e17*i)

256

Chapter 6

>> expand(f2) ans = 1.00091200*s^2+3189.26960*s+75631935.17+0.*i >> g1=r(2)*(s-p(3))+r(3)*(s-p(2)) g1 = (8330162633398495/144115188075855872+4622007775436625/36028797018963968*i)*(s+7010686265853559/4398046511104+2350353352750307/274877906944*i)+(8330162633398495/144115188075855872-4622007775436625/36028797018963968*i)*(s+7010686265853559/4398046511104-2350353352750307/274877906944*i) >> g2=vpa(simplify(g1),3) g2 = .116*s-.201e4 The above shows that the partial fraction decomposition is

(m) )1055.8()1060.1(

1001.2116.01045.1

72.11094.3

0031.0

1056.71019.31001.2116.0

1045.172.1

1094.30031.0)(

2323

4

34

732

4

342

xxsxs

xsxs

xsxsxs

xsxssV s

++−

++

−+

=

++−

++

−+

=

Inversion of Equation (m) leads to

(n) )1055.8sin(2.37-

)1055.8cos(116.072.10031.0)(31060.1

31060.11045.11094.32

3

334

txe

txeeetvtx

txtxtxs

−

−−− +−=

6.42 Application of KCL at node A leads to

(a) 0321 =−− iii where

257

Chapter 6

(d)

(c) )(

(b)

13

2

1

11

Rvvi

vvdtdCi

Rvvvi

EA

BA

AD

−=

−=

−+=

Substitution of Equations (b)-(d) in Equation (a) leads to

(e) 0)(11

1 =−

−−−−+

Rvvvv

dtdC

Rvvv EA

BAAD

Since the amplifier is assumed to be ideal, there is no current in the amplifier. Thus application of KCL at node B leads to

(f) 0)(2

=−

−−R

vvvv

dtdC CB

BA

Also at node C

(g) 022

=−

−−

Rvv

Rvv DCCB

Application of KCL at node E, noting the amplifier is ideal leads to

(h) 0)(1

=−−−

DEEA vv

dtdC

Rvv

Since the operator is ideal, (i) CE vv =

The desired system output is (j) 2 BD vvv −=

Taking the Laplace transforms of Equations (e), (f), (g), and (h), using Equation (i) leads to

( )

(n) 0)()()1()((m) 0)()(2)((l) 0)()()(1(k) 0)()()2()()(

11

22

111

=++−=+−

=+++−=+++−+

sCsVRsVCsRsVsVsVsV

sVsCsVRsVCsRsVCsVRsVCsRsVsV

DCA

DCB

CAB

CBAD

Equation (m) leads to

( ) (o) )()(21)( sVsVsV DBC +=

Using Equation (o) in Equations (k) and (n) and rewriting the resulting system in a matrix form leads to

(p) 00

)(

)()()(

)1(5.0)1(5.015.05.05.15.02 1

11

22

11

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−+−−+−−−−+ sV

sVsVsV

CsRCsRCsRCsRCsRCsR

D

B

A

Cramer’s rule is applied to Equation (p) resulting in

258

Chapter 6

[ ](q)

)()()1(5.05.0

)1(5.0)1(5.015.05.05.15.02)1(5.001

5.005.1)(2

)(

112

11

22

11

1

2

11

sDsVCsRCsR

CsRCsRCsRCsRCsRCsR

CsRCsR

sVCsR

sVB

−+=

−+−−+−−−−+

−−−−−+

=

[ ](r)

)()()1)(5.0(5.0

)1(5.0)1(5.015.05.05.15.02

0)1(5.0105.0

)(5.02

)(

1122

11

22

11

1

22

111

sDsVCsRCsRCsR

CsRCsRCsRCsRCsRCsR

CsRCsRCsR

sVCsRCsR

sVD

+−+=

−+−−+−−−−+

+−+−−−+

=

Hence

[ ]

[ ]

(s) )(

)(

)()()1(5.05.0

)()()1)(5.0(5.0

)()()(

12

112

1122

2

sDsCsVR

sDsVCsRCsR

sDsVCsRCsRCsR

sVsVsV BD

−=

−+

−+−+

=

−=

Evaluation of the determinant and substitution of numerical values leads to

(t) 25.21098.61019.2

)(106.1)( 4271

4

2 ++= −−

−

sxsxsVsxsV

Equation (t) is rewritten as

(u) )(1004.11023.3

7.740)( 17322 sVxsxs

sV++

=

The system is of second-order with natural frequency

(v) sr 1023.31004.1 37 xxn ==ω

and damping ratio

(w) 5.0)1023.3(2

1023.33

3

==xxζ

Thus the system is underdamped and has a damped natural frequency of

259

Chapter 6

(x) sr 1080.2)5.0(11023.3 323 xxd =−=ω

The step response for this second-order system is determined using Table 6.6 as (y) )1080.2sin(7.740)( 31062.1 3

txetx txs

−=

6.43 The mathematical model for the two-tank system is

(d) 0)()(

(c) 0)()(

(b) 0)(

(a) )(

222112

222112

12111

12111

=++−−−

=−++−−

=++−

=−++

BBBBBB

B

BBABAA

B

BAAAB

A

AiBAAA

A

CVkqCVktqCdt

dCV

CVkCVkqtqCdt

dCV

CVkqCVkdt

dCV

qCCVkCVkqdt

dCV

τ

τ

Substitution of given values into Equations (a)-(d) leads to

(h) 01048.3108.2)5.2(108.1104.1

(g) 01068.1106.4)5.2(108.1104.1

(f) 01048.3108.2104.1

(e) )(ˆ108.11068.1106.4104.1

26

26

1623

26

26

1623

16

1613

61

61

613

=+−−−

=−+−−

=+−

=−+

−−−−

−−−−

−−−

−−−−

BABB

BAAA

BAB

BAA

CxCxtCxdt

dCx

CxCxtCxdt

dCx

CxCxdt

dCx

tuCxCxCxdt

dCx

Dividing each equation by the reactor volume and taking their Laplace transforms, assuming all initial conditions are zero leads to

(j) 0)(105.2)(102)(

(i) ˆ1029.1)(102.1)(1033.3)(

13

13

1

3

13

13

1

=+−

=−+

−−

−−−

sCxsCxsCss

CxsCxsCxsCs

BAB

BAA

(l) 0)(105.2)(102)(1029.1)(

(k) 0)(102.1)(1033.3)(1029.1)(

23

23

15.23

2

23

23

15.23

2

=+−−

=−+−−−−−

−−−−

sCxsCxsCexsCs

sCxsCxsCexsCs

BABs

B

BAAs

A

Equations (i) and (j) are summarized in matrix form as

(m) 0

ˆ1029.1

)()(

105.2102102.11033.3

3

1

133

33

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

+−−+

−

−−

−−

sCx

sCsC

xsxxxs

B

A

Cramer’s rule is used to solve Equation (m) resulting in

260

Chapter 6

[ ]

( ) (n) 1092.51083.5

ˆ)1023.31029.1(

)102)(102.1()105.2)(1033.3()105.2(ˆ1029.1

105.2102102.11033.3105.2102

102.1ˆ1029.1

)(

632

63

3333

33

33

33

33

33

1

−−

−−

−−−−

−−

−−

−−

−−

−−

+++

=

−−−+++

=

+−−+

+−

−

=

xsxssCxsx

xxxsxssxsCx

xsxxxsxsx

xs

Cx

sCA

[ ]

( ) (o) 1092.51083.5

ˆ1058.2

)102)(102.1()105.2)(1033.3()102(ˆ1029.1

105.2102102.11033.3

0102

ˆ1029.11033.3

)(

632

6

3333

33

33

33

3

33

1

−−

−

−−−−

−−

−−

−−

−

−−

++=

−−−++=

+−−+

−

+

=

xsxssCx

xxxsxssxCx

xsxxxs

xs

Cxxs

sCB

Equations (n) and (o) are inverted leading to [ ][ ] (q) 6136.01778.04358.0ˆ)(

(p) 3664.01792.05456.0ˆ)(33

33

103.1105.41

103.1105.41

txtxB

txtxA

eeCtC

eeCtC−−

−−

−−

−−

−+=

−−=

Substituting Equations (n) and (o) in Equations (k) and (l)< rearranging and writing in a matrix form leads to

( )( )

( )(r)

1092.51083.5

ˆ1058.21092.51083.5

ˆ1023.31029.1

1029.1

)()(

105.2102102.11033.3

632

6

632

63

5.2

2

233

33

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

++

+++

=

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

+−−+

−−

−

−−

−−

−−

−−

−−

xsxssCx

xsxssCxsx

ex

sCsC

xsxxxs

s

B

A

Application of Cramer’s rule to Equation (r) leads to

261

Chapter 6

( )( )

( )

( )(s)

1092.51083.5

ˆ1044.11033.81066.1

)1058.2)(102.1(

)105.2(1023.31029.11092.51083.5

ˆ1029.1)(

2632

5.211926

63

3632632

5.23

2

−−

−−−−

−−

−−−

−−

−−

++

++=

+

++++

=

xsxsseCxsxsx

xx

xsxsxxsxss

eCxsC

s

s

A

( )( )

( )( )

(t) 1092.51083.5

ˆ1094.11067.6

)1058.2)(1033.3(

)102(1023.31029.11092.51083.5

ˆ1029.1)(

2632

5.2119

63

3632632

5.23

2

−−

−−−

−−

−−−

−−

−−

++

+=

++

+++

=

xsxssCexsx

xxs

xxsxxsxss

eCxsC

s

s

B

Inversion of Equations (s) and (t) leads to

(v) )5.2()5.2(ˆ)(

(u) )5.2()5.2(ˆ)(

2

2

−−=

−−=

tutgCtC

tutqCtC

B

A

where q(t) and g(t) are obtained using MATLAB as q = .411+.205e-31*exp(-.292e-2*t)*((-.201e32-.341e29*t)*cosh(.160e-2*t)-1.*(.152e32+.117e29*t)*sinh(.160e-2*t)) g = .554+.205e-31*exp(-.292e-2*t)*((-.273e29*t-.270e32)*cosh(.160e-2*t)-1.*(.320e32+.497e29*t)*sinh(.160e-2*t))

6.44 The system transfer function determined in the solution of Problem 6.33 is

(a) 121

1

)(

223

2

⎥⎦

⎤⎢⎣

⎡++⎟

⎠⎞

⎜⎝⎛ ++

+=

LRCs

LCs

RCLRsLRCs

CRs

sG

The integro-differential equations governing the free response of the system are

(c) 0)(C1

(b) 0)(1

120

2

210

111

=−+

=−+++

∫

∫

iiRdti

iiRdtiC

Ridtdi

L

t

t

Assuming an initial condition of the form 0)0( ii = , application of the Laplace transform to Equations (a) and (b) leads to

262

Chapter 6

(d) 0)(

)(1

120

2

1⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+−

−++ LisIsI

CsRR

RCs

RLs

Cramer’s rule is used to determine

(e) 121

12

120

12

)(

223

0

2223

0

2223

0

2

LRCs

LCs

RCLRs

iC

sCRs

CLRLRs

RLiC

sCRs

CLRLRs

R

LiCs

RLs

sI

++⎟⎠⎞

⎜⎝⎛ ++

=

++⎟⎠⎞

⎜⎝⎛ ++

=

++⎟⎠⎞

⎜⎝⎛ ++

−

++

=

The transfer function has either three real poles or one real pole and two poles which are complex conjugates. Consider the former case. Then Equation (e) can be written as

)(f ))()((

))()(()(

3

12

2

13

1

32

133221

0

321

02

⎥⎦

⎤⎢⎣

⎡−−

+−−

+−−

−−−−=

−−−=

ssss

ssss

ssss

ssssssi

ssssssi

sI

The free response obtained from Equation (f) is

[ ] (g) )()()())()((

)( 321121332

133221

02

tststs essessessssssss

iti −+−+−

−−−−=

In the case of only one real pole, Equation (e) can be written as

(g) ))((

)( 21

02 bassss

isI

++−=

where if the complex poles are written as jr jss ±

(i) sb

(h) 222

r j

r

s

sa

+=

−=

A partial fraction decomposition of Equation (g) leads to

(j) 1)( 21

1121

02 ⎥

⎦

⎤⎢⎣

⎡++

++−

−++=

basssas

ssbassi

sI

Inversion of Equation (j) leads to

(k) )sin(2)cos()( 21

2

121

02

1

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡ ++−

++=

−−tbe

b

sa

tbeebass

iti

tatats

263

Chapter 6

A MATLAB program which satisfies all requirements is % Program for Problem 6.44 % Input data disp('Please eneter resistance in ohms ') R=input('>>'); disp('Please enter capacitance in farads ') C=input('>>'); disp('Please enter inductance in henrys ') L=input('>>'); disp('Plese enter initial current in amperes ') i0=input('>>'); % Define transfer function N=1/(L*R*C)*[R 0 1/C] D=[1 R/L+1/(R*C) 2/(L*C) 1/(L*R*C^2) 0]; % Determines poles and resiudes of transfer fucntion [r,p,k]=residue(N,D) % Determines free response p=[1 R/L+1/(R*C) 2/(L*C) 1/(L*R*C^2)]; s=roots(p); if imag(s(1))==0 & imag(s(2))==0 % Poles are all real tt=min(abs(s)); dt=tt/500; c=1/(s(1)-s(2))/(s(2)-s(3))/(s(3)-s(1)); for j=1:201 t(j)=(j-1)*dt/200; i2(j)=-i0*c*((s(2)-s(3))*exp(s(1)*t(j))+(s(1)-s(2))*exp(s(2)*t(j))... +(s(3)-s(1))*exp(s(3)*t(j))); end else tt=min(abs(real(s))); dt=tt/500; if real(s(1))<=1.e-5 s1=s(1); a=-2*real(s(2)); b=abs(s(2))^2; else s1=s(3); a=-2*real(s(2)); b=abs(s(2))^2; end for j=1:201 t(j)=(j-1)*dt/200; i2(j)=i0/(s1^2+a*s1+b)*(exp(s1*t(j))-exp(-a/2*t(j))*(cos(b^0.5*t(j))...

264

Chapter 6

+(a/2+s1)/b^0.5*sin(b^0.5*t(j)))); end end plot(t,i2) xlabel('t (s)') ylabel('i_2 (A)') title('Free Response for System of Problem 6.44') Two examples of the program’s execution follow. The first corresponds to the former case with three real poles >> Please eneter resistance in ohms >>10000 Please enter capacitance in farads >>1e-6 Please enter inductance in henrys >>1 Plese enter initial current in amperes >>10 N = 1000000 0 100000000 r = 1.0e+004 * -0.0103 1.0202 -1.0100 0.0001 p = 1.0e+003 * -9.8990 -0.1010 -0.1000 0

265

Chapter 6

k = [] >>

The next example corresponds to a case with one real pole and two complex conjugate poles >> Please eneter resistance in ohms >>100 Please enter capacitance in farads >>1e-6 Please enter inductance in henrys >>1 Plese enter initial current in amperes >>10 N = 1.0e+010 * 0.0001 0 1.0000 r = 1.0e+002 *

266

Chapter 6

-1.0001 0.4950 - 0.0243i 0.4950 + 0.0243i 0.0100 p = 1.0e+004 * -1.0000 -0.0050 + 0.0999i -0.0050 - 0.0999i 0 k = [] >>

267

Chapter 6

6.45 The differential equations governing the motion of the system of Figure P6.45 are

(a) )()()(

3032

023

00

00

30002000

3

2

1

3

2

1

3

2

1

3

2

1

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

−+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

tFtFtF

xxx

kkkkk

kk

xxx

cccc

c

xxx

mm

m

&&&

&&&&&&

Application of the Laplace transform to Equation (a) assuming all initial conditions are zero leads to

(b) )()()(

)()()(

330322

023

3

2

1

3

2

1

2

2

2

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++−−−−++−

−++

sFsFsF

sXsXsX

kcsmskcskcskcsmsk

kkcsms

The matrix of transfer functions is obtained by

(c) 330

322023

)(

1

2

2

2 −

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++−−−−++−

−++=

kcsmskcskcskcsmsk

kkcsmssG

A MATLAB program which specifies the values of m,c, and k and determines the matrix of transfer functions symbolically follows % Program for Problem 6.45 m=1 c=2 k=10 syms s t A(1,1)=m*s^2+c*s+3*k; A(1,2)=-2*k; A(1,3)=0; A(2,1)=A(1,2); A(2,2)=2*m*s^2+c*s+3*k; A(2,3)=-c*s-k; A(3,1)=A(1,3); A(3,2)=A(2,3); A(3,3)=3*m*s^2+c*s+3*k; A G=A^-1 G1=vpa(G,3) 6.46 If the door of Problem 6.29 is subject to a moment the appropriate differential equation is

(a) )(302.25 tMct =++ θθθ &&& The system has a damping ratio of 1.15. Thus

(b) 2.63 )30)(2.25()15.1(2

=

=tc

Equation (a) becomes

268

Chapter 6

(c) )(302.632.25 tM=++ θθθ &&& The system’s transfer function is determined from Equation (c) as

(d) 302.632.25

1)( 2 ++=Θ

sss

The MATLAB commands to determine the impulsive response of the door >> clear >> N=[1]; >> D=[25.2 63.2 30]; >> impulse(N,D) >> The impulsive response is

6.47 The step response for the perturbations in temperature of the oil and water in the heat exchanger of Example 6.41 are

(b) 12.80104.014.8)(

(a) 87.1222.062.1)(314.063.2

314.063.2

ttw

to

eet

teet−−

−−

−−=

−+=

θ

θ

A MATLAB program to compute the temperature distributions and plot versus time is % Problem 6.47 % Compute and plot temperature perturbations for heat exahanger t=0; dt=4/100; for i=1:101 t(i)=(i-1)*dt; tho(i)=1.64+0.222*exp(-2.63*t(i))-1.87*exp(-0.314*t(i));

269

Chapter 6

thw(i)=8.14-0.0104*exp(-2.63*t(i))-8.12*exp(-0.314*t(i)); end plot(t,tho,'-',t,thw,'.') xlabel('t (s)') ylabel('T (C)') title('Temperature perturbations in heat exachanger of Problem 6.47') legend('\theta_o','\theta_w') The resulting plot is

270

Chapter 7

7. Frequency Response 7.1 The transfer function for the LR circuit is

)(a 1003.0

1

1)(

+=

+=

s

RLssG

(a) The sinusoidal transfer function at the given frequency is

(b) 1055.31048.1 )240()100(

240100

2401001

100)800(3.01)800(

33

22

jxx

jj

jjG

−− −=

+−

=

+=

+=

(b)The steady-state amplitude is

( ) (c) 462.0)1055.3(1048.1120 2323 =−+= −− xxI The steady-state phase is

)(d r 11.51048.11055.3tan 3

31 =⎟⎟

⎠

⎞⎜⎜⎝

⎛ −= −

−−

xxφ

Thus the steady-state current is (e) )11.5800sin(462.0)( Atti +=

7.2 The loop equations for the circuit are

( )

)(b 0)(1.0)(810

(a) 0)(1.0)(85

21212

21211

=−+−+−

=+−−−−−

iidtdiii

tviidtdiii

Taking the Lapalce transforms of Equations (a) and (b) assuming all initial conditions are zero leads to

)c( 0

)()()(

181.01.01.0131.0

2

1⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡+−

−+ sVsIsI

ssss

Using Cramer’s rule to solve for )(2 sI gives

271

Chapter 7

( )( )

)(d 2341.3

)(1.0

)1.0(181.0131.0)(1.0

181.01.01.0131.0

01.0)(131.0

)( 22

+=

−−++=

+−−+

−+

=

ssVs

ssssVs

ssss

ssVs

sI

The transfer function is determined as

)(e 2341.3

1.0)()()( 2

+==

ss

sVsIsG

(a) A frequency of 2.1 kHz is converted into r/s as

srx

cyclerad

scycle 1032.1 2 2100 4=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛= πω . Thus the appropriate sinusoidal transfer

function is

(f) 1085.10323. 1067.1

1009.31040.5

1009.42341009.4234

2341009.41032.1

234)1032.1(1.3)1032.1(1.0)1032.1(

4

9

57

4

4

4

3

4

44

jxx

jxx

jxjx

jxjx

jxjxjxG

−+=

+=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

+=

(b) The steady-state amplitude is

( ) ( ) (g) 58.21085.10323.80 242 AxI =+= − The steady-state phase is

(h) 0057.0323.

1085.1tan4

1 radx=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−−φ

Thus the steady-state current through the 10 Ohm resistor is (i) )0057.1032.1sin(58.2)( 4 += txti

7.3 A free-body diagram of the junction between the spring and viscous damper at an arbitrary instant is illustrated below

Since the system is assumed to be massless application of Newton’s second law to the free-body diagram leads to

272

Chapter 7

(a) )(0)(

0

tFkxxctFxckx

F

=+=++−−

=∑

&&

Taking the Laplace transform of Equation (a), assuming the initial condition is zero leads to

(b) )()(

)()()(

kcssFsX

sFskXscsX

+=

=+

The system’s transfer function is determined using Equation (b)

(c) 1

)()()(

kcs

sFsXsG

+=

=

Substitution of given values leads to

(d) 5.0

105

100020001)(

4

+=

+=

−

sx

ssG

The input is sinusoidal with a frequency of 40 r/s and amplitude of 50 N. The sinusoidal transfer function at this frequency is

(e) 1025.11056.1 106003.1

02.0105.2

405.0405.0

5.040105

4.040105)40(

57

3

4

4

4

jxxx

jx

jj

jx

jxjG

−−

−

−

−

−=

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

+=

The steady-state amplitude is

( ) ( )(f) mm 625.0

1025.11056.150

)40(502527

=

+=

=

−− xx

jGX

The phase of the steady-state response is

(g) r 558.1 1056.11025.1tan

)40(Re()40(Im(tan

7

51

1

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

−−

−

xx

jGjGφ

273

Chapter 7

The system’s steady-state response is

(h) mm )558.140sin(625.0 )sin()(−=

+=t

tXtx φω

7.4 The transfer function for the system is

(a) 2000030020

1)( 2 ++=

sssG

The system input has an amplitude of 20 N at a frequency of 80 r/s. The sinusoidal transfer function for this frequency is

(b) 1096.11082.8 1022.1

104.21008.1

104.21008.1104.21008.1

104.21008.11

20000)80(300)80(201)80(

66

10

45

45

45

45

2

jxxx

jxx

jxxjxx

jxx

jjjG

−− −−=

−−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−−−

⎟⎟⎠

⎞⎜⎜⎝

⎛+−

=

++=

The steady-state amplitude of the block is

( ) (c) 1081.1)1096.1(1082.8)20( 42626 mxxxX −−− =−+−= The steady-state phase is

(d) 36.31082.81096.1tan 6

61 r

xx

=⎟⎟⎠

⎞⎜⎜⎝

⎛−−

= −

−−φ

The steady-state displacement is (e) )36.380sin(1081.1)( 4 += − txtx

7.5 The differential equation governing the motion of the system of Figure P7.5 is (a) kyyckxxcxm +=++ &&&&

The transfer function for the system is

(b) 2

2

)(

22

2

2

nn

nn

ss

kcsmskcssG

ωζωωζω++

+=

+++

=

The given values are used to calculate the natural frequency and damping ratio as

(c) sr 6.31

kg 20mN 102

4

=

=

=

x

mk

nω

274

Chapter 7

(d) 237.0 sr 31.6kg) 20(2

msN 300

2

=

⎟⎠⎞

⎜⎝⎛

⋅

=

=nm

cω

ζ

Substituting Equations (c) and (d) into Equation (b) leads to

(e) 100015

100015)( 2 +++

=ss

ssG

The input is sinusoidal with a frequency of 80 r/s and amplitude of 0.05 m. The sinusoidal transfer function at this frequency is

[ ]

(f) 251.0129.0

1006.31068.71096.3

)1200()5400()5400)(1200()1200)(1000()1200)(1200()5400)(1000(

1200540012005400

1200540012001000

1200540012001000

10001200640012001000

1000)80(15)80(1000)80(15)80(

7

66

22

2

jx

jxx

jjj

jj

jjj

jjj

jjG

−−=

−−=

+−+−++−

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−−−

⎟⎟⎠

⎞⎜⎜⎝

⎛+−+

=

+−+

=

++−+

=

+++

=

The amplitude of the steady-state response is

(g) cm 41.1 )251.0((-0.129)0.05

)80(22

=

−+=

= jGYX

The phase of the steady-state response is

(g) r 05.2 129.0251.0tan

)80(Re()80(Im(tan

1

1

−=

⎟⎠⎞

⎜⎝⎛−−

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

−

jGjGφ

The steady-state response for the absolute displacement is

275

Chapter 7

(h) cm )05.280sin(41.1 )sin()(−=

+=t

tXtx φω

7.6 The differential equation governing the motion of the system of Example 2.17 is given by Equation (c) in the solution of the example as

( ) ( ) (a) 0999 212121 =++++⎟⎠⎞

⎜⎝⎛ ++ xrkrkxrcrcx

rIrmrm &&&

where θrx = . If the disk is subject to a moment Equation (a) can be rewritten using θ as the dependent variable as

( ) ( ) ( ) (b) )(99 221

22

21

22

21 tMkrrkrcrcIrmrm =++++++ θθθ &&&

Substitution of given values into Equation (b) leads to (c) )(4000100090.5 tM=++ θθθ &&&

The transfer function is obtained from Equation (c) as

(d) 400010009.5

1

)()()(

2 ++=

Θ=

ss

sMssG

The sinusoidal input is given as (e) )200sin(5.20)( ttM =

The sinusoidal transfer function evaluated at the input frequency is

(f) 1038.9

1021032.2

1021032.21

4000)200(1000)200(9.51)200(

10

55

55

2

xjxxjxx

jjjG

−−=

+−=

++=

The magnitude of the sinusoidal transfer function is

(g) 1026.3

)102()1032.2(1)]200(

6

2525

−=

+=

x

xxjG

The steady-sate amplitude of angular oscillation is

(h) rad 1069.6 )1026.3(5.20

)200(

5

6

0

−

−

=

=

=Θ

xx

jGM

276

Chapter 7

7.7 Application of KVL to each loop in the circuit leads to

(b) 0)(102.120

(a) 0)()(102.125.0

02162

0216

1

=−+−

=+−−−

∫

∫

−

−

t

t

dtiix

i

tvdtiixdt

di

Application of the Laplace transform to Equations (a) and (b) leads to

(c) 0

)()()(

10520105

10510525.0

2

166

66

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

+−

−+ sVsIsI

sx

sx

sx

sxs

Cramer’s rule is used to solve Equation (c) leading to

( ) (d) 1011025.14)(10520

1011025.14

)(10520

10520105

10510525.0

105200

105)(

)(

862

6

86

6

66

66

6

6

1

xsxssVxs

sxxs

sVs

x

sx

sx

sx

sxs

sxs

xsV

sI

+++

=

++

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

+−

−+

+

−

=

The transfer function is ( ) (e)

1011025.14)(10520

)()()( 862

61

xsxssVxs

sVsIsG

+++

==

The appropriate sinusoidal transfer function is [ ]

(f) 0125.01032.3 1051.1

1087.11000.5

1075.31096.91075.31096.9

1075.31096.96000105

101)300(1025.1)300(4105)300(20)300(

3

17

1514

87

87

87

6

862

6

jxx

jxx

jxxjxx

jxxjx

xjxjxjjG

−=

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

⎟⎟⎠

⎞⎜⎜⎝

⎛++

=

+++

=

−

The steady-state amplitude is

( ) (g) 58.2)0125.0(1032.3200 223 =−+= −xI The steady-state phase is

(h) 83.11032.3

0125.0tan 3 1 r

x=⎟

⎠⎞

⎜⎝⎛ −

= −−φ

The steady-state response is

277

Chapter 7

(i) )83.1300sin(58.2)( Atti +=

7.8 The mathematical model for the perturbations in liquid levels in the tanks is

(b) )(111

(a) 011

22

211

1

22

21

11

11

gRtp

hRR

hRdt

dhA

hR

hRdt

dhA

p

ρ=⎟⎟

⎠

⎞⎜⎜⎝

⎛++−

=−+

The resistances are obtained by considering the steady state

( )

(c) ms 5

m 0.8

m 6-m 82

2

2

3

1

1

=

=

=

s

qh

R λ

(d) ms 36.14

m 8.0

sm 81.9

mkg 1000

mN 105.2

m 62

2

2

3

23

23

2

22

=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛

−

=

=

s

x

qh

R λ

Substitution of all values into Equations (a) and (b) leads to

(f) )50sin(1026.4270.02.01.14

(e) 02.02.02.10

321

2

211

txhhdt

dh

hhdtdh

−=+−

=−+

Dividing Equations (e) and (f) by the areas leads to

(h) )50sin(1002.30191.00142.0

(g) 00196.00196.0

421

2

211

txhhdt

dh

hhdtdh

−=+−

=−+

Defining , taking the Laplace transforms of Equations (g) and (h) assuming all initial conditions are zero leads to

)50sin(1020.3)( 4 txtF −=

(i) )(

0)()(

0191.00124.00196.00196.0

2

1⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡+−−+

sFsHsH

ss

278

Chapter 7

The solutions to Equation (i) are obtained using Cramer’s rule as

(j) 1017.60387.0

)(0196.0

)0124.0)(0196.0()0191.0)(0196.0()(0196.0

0191.00124.00196.00196.0

0191.0)(0196.00

)(

42

1

−++=

−−−++=

+−−+

+−

=

xsssF

sssF

ss

ssFsH

(k) 1017.600387)()0196.0(

0191.00124.00196.00196.0

)(0124.000196.0

)(

42

2

−+++

=

+−−+

−+

=

xsssFs

ss

sFs

sH

The system’s transfer functions are determined from Equations (j) and (k) as

(m) 1017.60387.0

0196.0

)()()(

(l) 1017.60387.0

0196.0

)()()(

42

22

42

11

−

−

+++

=

=

++=

=

xsss

sFsHsG

xss

sFsHsG

The sinusoidal transfer functions corresponding to an input frequency of 50 r/s are

(n) 1025.6

)935.12500(0195.0

935.12500935.12500

935.125000196.0

1017.6)50(0387.0)50(0196.0)50(

6

421

xj

jj

j

xjjjG

−−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−−−

⎟⎟⎠

⎞⎜⎜⎝

⎛+−

=

++= −

(o) 1025.6

1025.175.47

935.12500935.12500

935.125000196.050)50(

6

5

2

xjx

jj

jjjG

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−−−

⎟⎟⎠

⎞⎜⎜⎝

⎛+−

+=

The steady-state amplitudes of the perturbations are obtained using the sinusoidal transfer functions. The amplitude of the perturbation in the first tank is

279

Chapter 7

( )( )

( )(q) m 1004.6

1025.6)1025.1()75.47(

1002.3

)50(1002.3(p) m 1092.5

1025.6)935.1()2500(

0196.01002.3

)50(1002.3

6

6

2524

24

2

6

6

224

14

1

−

−

−

−

−

−

=

+=

=

=

+=

=

xx

xx

jGxHx

xx

jGxH

7.9 The differential equation governing the motion of the system of Figure P7.9 is

(a) kyyckxxcxm +=++ &&&& The transfer function for the system is

(b) 2

2

)(

22

2

2

nn

nn

ss

kcsmskcssG

ωζωωζω++

+=

+++

=

The given values are used to calculate the natural frequency and damping ratio as

(c) sr 4.35

kg 80mN 101

5

=

=

=

x

mk

nω

(d) 0530.0 sr 35.4kg) 80(2

msN 300

2

=

⎟⎠⎞

⎜⎝⎛

⋅

=

=nm

cω

ζ

Substituting Equations (c) and (d) into Equation (b) leads to

(e) 125075.3

125075.3)( 2 +++

=ss

ssG

Equation (e) is the transfer function for the displacement of the block. The acceleration of the block is

(f) xA &&= Taking the transform of Equation (f) leads to

280

Chapter 7

(g) )()( )()(

2

2

sYsGssXssA

=

=

Equation (g) is rewritten to define the transfer function for the acceleration of the block as

(h) 125075.3

)125075.3(

)( )()()(

2

2

2

+++=

=

=

=

ssss

sGssYsAsGa

The input is sinusoidal with a frequency of 45 r/s and amplitude of 0.022 m. The sinusoidal transfer function at this frequency is

[ ]

)i( 1010.11021.3

1029.61092.61009.2

75.16877575.168775

75.1687751042.310534.2

75.1687751042.310534.2

125075.16820251042.310534.2

1250)45(75.3)45(1250)45(75.3)45()45(

33

5

89

56

56

56

2

2

jxxx

jxx

jj

jjxx

jjxx

jjxx

jjjjjGa

−=

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−−−

⎟⎟⎠

⎞⎜⎜⎝

⎛+−−−

=

+−−−

=

++−−−

=

+++

=

The amplitude of the steady-state response is

( ) ( )(j)

sm 78.6

1010.11021.30.002

)80(

2

2323

=

−+=

=

jxx

jGYX

The phase of the steady-state response is

(k) r 330.0 1021.31010.1tan

)80(Re()80(Im(tan

3

31

1

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

−

xx

jGjGφ

The steady-state response for the acceleration is

281

Chapter 7

(l) sm )330.080sin(78.6

)sin()(

2−=

+=

t

tAtx φω

7.10 Free-body diagrams of the blocks at an arbitrary instant are illustrated below

Application of Newton’s Second Law to each of the free-body diagrams leads to

(a) 0108104.14

4)(108106

25

16

1

1125

15

11

=−+

=−+−

=∑

xxxxx

xxxxxx

xmF

&&

&&

&&

(b) )(1081086

6)()(108

25

15

2

2125

22

tFxxxxx

xtFxxx

xmF

=+−

=+−−

=∑

&&

&&

&&

Taking the Laplace transforms of both sides of Equations (a) and (b) leads to

(d) )()(108)(108)(6

(c) 0)(108)(104.1)(4

25

15

22

25

16

12

sFsXxsXxsXs

sXxsXxsXs

=+−

=−+

Equations (c) and (d) are summarized in matrix form as

(e) )(

0)()(

1086108108104.14

2

1525

562

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

+−−+

sFsXsX

xsxxxs

Application of Cramer’s rule to Equation (e) leads to

( )

(f) 1080.41016.1

)(8x10

1080.41016.110)(8x10

)108()1086(104.14)(8x10

1086108108104.14

1086)(1080

)(

10264

4

11274

5

255262

5

525

562

52

5

1

xsxssF

xsxssF

xxsxssF

xsxxxs

xssFx

sX

++=

++=

−−++=

+−−+

+−

=

The transfer function is obtained from Equation (f) as

282

Chapter 7

(g) 1080.41016.1

8x10

)()(

)(

10264

4

11

xsxs

sFsX

sG

++=

=

The input is sinusoidal at a frequency of 150 r/s and an amplitude of 200 N. The sinusoidal transfer function at this frequency is

(h) 1076.3 1012.2

8x10

1080.4)150(1016.1)150(8x10)150(

7

11

4

10264

4

−−=−

=

++=

xx

xjxjjG

The sinusoidal transfer function is real, but is negative, thus

(k) (j) m 1042.7)1076.3(200

(i) 1076.3)150((57

7

πφ ===

=−−

−

xxX

xjG

The block’s steady-state response is (l) )150sin(1042.7)( 5 π+= − txtx

7.11 The transfer function for a series LRC circuit with 2.1=ζ and sr 500=nω is of the

form

(m) 105.21200

)( 42 xssssG++

=

The Bode diagram is obtained using the following MATLAB work session >> N=[1 0] N = 1 0 >> D=[1 1200 25000] D = 1 1200 25000 >> bode(N,D) >>

The resulting Bode diagram is

283

Chapter 7

The function )(ωL the series LRC circuit is given in Equation (7.39). Substitution of given parameters leads to

( ) [ ]

( ) [ ](b)

1200105.2log20

)500)(2.1(2105.2log20)(

2224

2224

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

+−=

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

+−=

ωω

ω

ωω

ωω

x

xL

Properties of logarithms are used to rewrite Equation (b) as ( ) [ ][ ] (c) 1200105.2log10)log(20)( 2224 ωωωω +−−= xL

The low-frequency asymptote is obtained by taking the limit of Equation (c) as 0→ω , leading to

(d) 96.87)log(20

)105.2log(20)log(20)(lim 4

0

−=

−=→

ω

ωωω

xL

The high-frequency asymptote is obtained by taking the limit of Equation (c) as ∞→ω leading to

(e) )log(20

)log(40)log(20)(lim

ω

ωωωω

−=

−=∞→

L

284

Chapter 7

7.12 The transfer function for this system is

(a) 100015

05.0

102300201)(

2

42

++=

++=

ss

xsssG

The Bode diagram is obtained using the following MATLAB work session >> clear >> N=[0.05] N = 0.0500 >> D=[1 15 1000] D = 1 15 1000 >> bode(N,D) >>

The resulting Bode diagram is

285

Chapter 7

The sinusoidal transfer function is obtained from Equation (a) as

(b) )15()1000(

75.02015

151000151000

15100005.0

15100005.0

1000)(15)(05.0)(

222

2

2

2

2

2

2

ωω

ωω

ωωωω

ωω

ωω

ωωω

+−

−−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−−−

⎟⎟⎠

⎞⎜⎜⎝

⎛+−

=

+−=

++=

j

jj

j

j

jjjG

The real and imaginary parts of the sinusoidal transfer function are obtained from Equation (b) as

[ ]

(d) )15()1000(

75.0)]Im[G(j

(c) )15()1000(

2015

)(Re

222

222

2

ωωωω

ωω

ωω

+−−

=

+−

−=jG

The magnitude of the sinusoidal transfer function is obtained as

(e) )15()1000(

105.0

)15()1000(75.0

)15()1000(2015

)](Im[)](Re[)(

222

2

222

2

222

2

22

ωω

ωωω

ωω

ω

ωωω

+−=

⎭⎬⎫

⎩⎨⎧

+−−

+

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

+−

−=

+= jGjGjG

The Bode diagram function is

(f) )15()1000(

105.0log20

)(log20)(

222 ⎥⎥⎦

⎤

⎢⎢⎣

⎡

+−=

=

ωω

ωω jGL

Properties of logarithms are used to rewrite Equation (f) as [ ]

[ ] (g) )15()1000(log1002.26 )15()1000(log10)05.0log(20)(

222

222

ωω

ωωω

+−−−=

+−−=L

The low-frequency asymptote is obtained by taking the limit of Equation (g) as 0→ω , leading to

(h) dB 02.86

)1000log(2002.26)(lim0

−=

−−=→

ωω

L

286

Chapter 7

The high-frequency asymptote is obtained by taking the limit of Equation (g) as ∞→ω leading to

(i) )log(4002.26)(lim ωωω

−−=∞→

L

7.13 The transfer function for an undamped mass-spring system of natural frequency

sr 100 is

(a) 101

1)( 42 xssG

+=

The MATLAB work session to draw the Bode diagram is >> clear >> N=[1] N = 1 >> D=[1 0 10000] D = 1 0 10000 >> bode(N,D) >>

The resulting Bode diagram is

287

Chapter 7

7.14 The differential equation of Problem 6.30 is (a) kyyckxxcxm +=++ &&&&

Dividing Equation (a) by m leads to (b) 22 22 yyxxx nnnn ωζωωζω +=++ &&&&

where the natural frequency and damping ratio are determined using the parameter values of Problem 6.30

(c) sr 3.25

kg 500mN 10.3

5

=

=

=

x

mk

nω

(d) 395.0 sr 25.3kg) 500(2

msN 10000

2

=

⎟⎠⎞

⎜⎝⎛

⋅

=

=nm

cω

ζ

Substitution into Equation (b) leads to (e) 6402064020 yyxxx +=++ &&&&

The system transfer function is obtained from Equation (e) as

(f) 64020

64020

)()()(

2 +++

=

=

sss

sYsXsG

As the vehicle traverses the road it is subject to a sinusoidal input of the form (g) )sin()( tYty ω=

where Y=0.002 m and

(h) 24.5 2.1

2

v

v

=

=πω

where v is the horizontal speed of the vehicle. The Bode diagram is determined for the transfer function of Equation (f) using MATLAB. The MATLAB work session to generate the Bode diagram is

>> N=[20 640] >> D=[1 20 640] >> bode(N,D)

288

Chapter 7

>> The resulting Bode diagram is illustrated below

7.15 Consider the transfer function

(a) 5010

3)( 2 +++

=ss

ssG

The sinusoidal transfer function is

[ ]( )( )

( )(b)

10050207150

10050)50(3010)50(3

10501050

10503

10503

50)(10)(3)()(

222

22

222

222

2

2

2

2

2

ωω

ωωωωω

ωωωωω

ωωωω

ωωω

ωωω

ωωωω

+−

−++=

+−

−+−++−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−−−

⎟⎟⎠

⎞⎜⎜⎝

⎛+−

+=

+−+

=

+++

=

j

j

jj

jj

jj

jjjjG

The real and imaginary parts of the sinusoidal transfer function are

289

Chapter 7

( )( )

( ) (d)

1005020)]Im[G(j

(c) 10050

7150)](Re[

222

2

222

2

ωω

ωωω

ωω

ωω

+−

−=

+−

+=jG

The magnitude of the transfer function is

( )( )

( )( )

( )[ ] (e) 10050

)20(7150

10050

2010050

7150

)](Im[)](Re[)(

2222

22222

2

222

22

222

2

22

ωω

ωωω

ωω

ωω

ωω

ω

ωωω

+−

−++=

⎥⎥⎦

⎤

⎢⎢⎣

⎡

+−

−+

⎥⎥⎦

⎤

⎢⎢⎣

⎡

+−

+=

+= jGjGjG

(a) The Bode diagram is constructed using

( )( )[ ]

( )( )

( )[ ] ( )[ ] (f) 10050log20)20(7-15010log

10050)20(7-15010log

10050

)20(7150log20

)(log20)(

22222222

222

22222

2222

22222

ωωωωω

ωω

ωωω

ωω

ωωω

ωω

+−−−+=

⎥⎥⎦

⎤

⎢⎢⎣

⎡

+−

−+=

+−

−++=

= GL

The low-frequency asymptote is obtained as

(g) dB 4.24

)50log(40)150log(20)(lim0

−=

−=→

ωω

L

The high-frequency asymptote is obtained as ( ) ( )

(h) log20 log80log60

log80log10)(lim 46

ωωω

ωωωω

−=−=

−=∞→

L

(b) The Nyquist diagram is a plot of Re[G(jω0] on the horizontal scale with Im[G(jω)] on the vertical scale. Repeating Equation (c) and (d)

( )( )

( ) (d)

1005020)]Im[G(j

(c) 10050

7150)](Re[

222

2

222

2

ωω

ωωω

ωω

ωω

+−

−=

+−

+=jG

(i) For 0=ω , 06.050150)]0(Re[ 2 ==G and 0)]0(Im[ =G .

As 0)](Re[, →±∞→ ωω jG and 0)](Im[ →ωjG

290

Chapter 7

(ii) The Nyquist diagram intercepts the horizontal axis when Im[G(jω)]=0. From Equation (d) the intercepts occur for 20,0 ±=ω . The value of the intercept for ω=0 is 0.06, The intercept values for 20±=ω are obtained from Equation (c) as

[ ](i) 01.0

)20(100)2050()20(7150)20(Re 2

=+−

+=± jG

The Nyquist diagram intercepts the vertical axis when Re[G(jω)]=0. From Equation (c) the only intercept occurs for ω=0 for which 0)]0(Im[ =G (iii) Since )](Re[ ωjG is always positive the Nyquist diagram exists only in the first and fourth quadrants. (a) The MATLAB work session to develop the Bode and Nyquist diagrams is

presented below >> clear >> N=[1 3] N = 1 3 >> D=[1 10 50] D = 1 10 50 >> bode(N,D) >> figure >> Nyquist(N,D) >> The resulting Bode diagram is

291

Chapter 7

The resulting Nyquist diagram is

292

Chapter 7

7.16 Consider the transfer function

(a) 1065

42)( 23

2

+++++

=sss

sssG

The sinusoidal transfer function corresponding to the transfer function of Equation (a) is

( )

( )( )( )

[ ]( )

[ ]( )

(b) 6)510(

4181840

6)510()510(2)6()4()6(2)510)(4(

65106510

651024

651024

10)(6)(5)(4)(2)()(

22222

442

22222

2222222

32

32

32

2

32

2

23

2

ωωω

ωωωωωωω

ωωωωωωωωω

ωωωωωω

ωωωωω

ωωωωω

ωωωωωω

−+−

+−+−=

−+−

−+−−−+−+−−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−−−−−

⎟⎟⎠

⎞⎜⎜⎝

⎛−+−

+−=

−+−+−

=

+++++

=

j

j

jj

jj

jj

jjjjjjG

The real and imaginary parts of the sinusoidal transfer function are obtained from Equation (b) as

( )[ ]

( )(d)

6)510(4)](Im[

(c) 6)510(

181840)](Re[

22222

4

22222

42

ωωω

ωωω

ωωω

ωωω

−+−

+−=

−+−

+−=

jG

jG

The magnitude of the sinusoidal transfer function is

( ) [ ]( )( ) (e) 6)510(

4181840

)](Im[](Re[)(

22222

24242

22

ωωω

ωωωω

ωωω

−+−

+−++−=

+= jGjGjG

(a) The Bode diagram is drawn using

( ) [ ]( )( )

( ) [ ]( )[ ] ( )[ ] (f) 6)510(log204181840log10

6)510(

4181840log20

)(log20)(

2222224242

22222

24242

ωωωωωωω

ωωω

ωωωω

ωω

−+−−+−++−=

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

−+−

+−++−=

= jGL

The low-frequency asymptote is obtained as

(g) dB 96.7

)10log(40)40log(20)(lim0

−=

−=→

ωω

L

The high-frequency asymptote is obtained as

293

Chapter 7

( ) ( )

(h) log30 log120log90

log20log10)(lim 69

ωωω

ωωωω

−=−=

−=∞→

L

(b) The Nyquist diagram is drawn from the real and imaginary parts of the sinusoidal transfer function. Equations (c) and (d) are repeated below

( )[ ]

( )(d)

6)510(4)](Im[

(c) 6)510(

181840)](Re[

22222

4

22222

42

ωωω

ωωω

ωωω

ωωω

−+−

+−=

−+−

+−=

jG

jG

(i) For 0=ω , Equation (c) shows that =0.4 and Equation (d) shows that )](Im[ ωjG =0. As ±∞→ω , Equations (c) and (d) show that )](Re[ ωjG 0→ and )](Im[ ωjG 0→ .

(ii) Equation (d) shows that the only horizontal intercept corresponds to ω=0 while Equation shows that a vertical intercept may occur for any real ω such that =0. The quadratic equation is applied to show that the values of ω which satisfy this equation are complex. Thus there are no Im-intercepts.

42 181840 ωω +−

(iii) Since there are no y-intercepts the Nyquist diagram does not exist in the second and third quadrants.

(c) The MATLAB work session necessary to draw the Bode and Nyquist diagrams is shown below

>> clear >> N=[1 2 4] N = 1 2 4 >> D=[1 5 6 10] D = 1 5 6 10 >> bode(N,D) >> figure >> nyquist(N,D)

The Bode diagram is

294

Chapter 7

The Nyquist diagram for this transfer function is

295

Chapter 7

7.17 The transfer function for the system is ( ) (a)

)2)(3)(6(32)(

3

++++

=−

ssssessG

s

The amplitude part of the Bode diagram is unaffected by the time delay. The following MATLAB commands leads to the correct amplitude part of the Bode diagram

>> N=[2 3] N = 2 3 >> D=[1 11 36 36 0] D = 1 11 36 36 0 >> bode(N,D) >>

The resulting Bode diagram is

296

Chapter 7

The correct phase part of the Bode diagram can be determined by a superposition

(b) 6

13

12

1232)( 3 ⎟

⎠⎞

⎜⎝⎛

+⎟⎠⎞

⎜⎝⎛

+⎟⎠⎞

⎜⎝⎛

+⎟⎠⎞

⎜⎝⎛ += −

sssessG s

The phase of a constant is 0, the phase of a first-order lead is given by Equation (7.51); the phase of a first-order lag is given by Equation (7.52); the phase of a time delay is given by Equation (7.55). Thus

(c) 361tan

31tan

21tan

32tan)( 1111 ωωωωωωφ −⎟

⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛= −−−−

The MATLAB M file used to generate the frequency part of the Bode diagram is given below

% Program to generate phase portion of Bode iagram for Problem 7.17 % Defining the range of frequencies clear omega=logspace(-2,2,200); phi=atan(2/3.*omega)-atan(1/2.*omega)-atan(1/3.*omega)-atan(1/6.*omega)-3.*omega semilogx(omega,phi) xlabel(‘\omega’) ylabel(‘\phi’) title(‘Phase part of Bode diagram for Problem 7.17’)

The resulting phase portion of the Bode diagram is

297

Chapter 7

7.18 The transfer function for the system is

(a) )125(

43)( 2

2

++++

=ssssssG

The MATLAB commands to draw the Bode diagram are >> N=[1 3 4]; >> D=[1 5 12 0]; >> bode(N,D) >>

The resulting Bode diagram is

7.19 The transfer function for the system is ( ) (a)

)2)(3)(6(32)(

3

++++

=−

ssssessG

s

The sinusoidal transfer function is

298

Chapter 7

[ ]

[ ][ ]

[ ][ ]

[ ] [ ] ( )

[ ]( ) [ ][ ]( )

[ ][ ] [ ][ ]( ) (b)

)1136(361136)3sin(2)3cos(336)3sin(3)3cos(2

)1136(361136)3sin(3)3cos(236)3sin(2)3cos(3

)1136(36113636)3sin(3)3cos(2)3sin(2)3cos(3

113636)3sin(3)3cos(2)3sin(2)3cos(3

113636)3sin()3cos()23(

)(36)(36)(11)(3)(2)(

23224

324

23224

324

23224

324

324

324

234

3

ωωωωωωωωωωωωωω

ωωωωωωωωωωωωωω

ωωωωωωωωωωωωωω

ωωωωωωωωωω

ωωωωωωω

ωωωωωω

ω

−+−

−+−−++

−+−

−++−+=

−+−

−−−+++=

−+−+++

=

−+−−+

=

++++

=−

j

jjjj

jjj

jjjjejjG

j

The real part of the sinusoidal transfer function is obtained from Equation (b) as [ ]( ) [ ][ ]

( )

(c) )1136(36

1136)3sin(3)3cos(236)3sin(2)3cos(3)](Re[23224

324

ωωωω

ωωωωωωωωωωω−+−

−++−+=jG

The imaginary part of the sinusoidal transfer is obtained from Equation (b) as

[ ][ ] [ ][ ]

( )(d)

)1136(361136)3sin(2)3cos(336)3sin(3)3cos(2)](Im[

23224

324

ωωωω

ωωωωωωωωωωω−+−

−+−−+=jG

The following are obtained using Equations (c) and (d)

[ ]( ) [ ][ ]( )

( )[ ]

(e) 361

)36()36(2)36(3

)1136(36)1136()3cos(2)36()3cos(3lim

)1136(361136)3sin(3)3cos(236)3sin(2)3cos(3lim]0Re[

2

222222

2222

0

23224

324

0

−=

+−=

−+−

−+−=

−+−

−++−+=

→

→

ωωωω

ωωωωωωωωωω

ωωωωωωωωωω

ω

ω

299

Chapter 7

[ ][ ] [ ][ ]( )

( )( )[ ]

(f) 0 )1136(36)1136(3362lim

)1136(361136)3sin(2)3cos(336)3sin(3)3cos(2lim]0Im[

222222

223

0

23224

324

0

=−+−

−−−=

−+−

−+−−+=

→

→

ωωωω

ωωωωωωωω

ωωωωωωωωωω

ω

ω

Thus the point corresponding to 0=ω is(-1/36,0). Also

( )[ ](g) 0

)1136(36)3cos(22)3sin(2lim)](Re[lim

222222

45

=−+−

−=

∞→∞→ ωωωωωωωωω

ωωjG

( )[ ](h) 0

)1136(36)3sin(22)3cos(2lim)](Im[lim

222222

45

=−+−

+=

∞→∞→ ωωωωωωωωω

ωωjG

The real axis intercepts correspond to values of ω such that 0)](Im[ =ωjG which from Equation (d) occurs for values of ω such that

[ ][ ] [ ][ ][ ] [ ] [ ] [ ]

(i) 18019

108392)3tan(

0)3sin(11362363)3cos(1136336201136)3sin(2)3cos(336)3sin(3)3cos(2

24

35

324324

324

ωωωωωω

ωωωωωωωωωωωω

ωωωωωωωωωω

+−−

=

=−−−+−−−

=−+−−+

The imaginary axis intercepts correspond to the to values of ω such that 0)](Re[ =ωjG which from Equation (c) occurs for values of ω such that

[ ]( ) [ ][ ] (j) 01136)3sin(3)3cos(236)3sin(2)3cos(3 324 =−++−+ ωωωωωωωωωω

7.20 The transfer function for the system is

(a) )125(

43)( 2

2

++++

=ssssssG

The sinusoidal transfer function is determined as

300

Chapter 7

[ ]

( ) ( ) (b) 144

481615144

162

)12(25)12)(4()5)(3()12)(3()5)(4(

)12(5)12(5

)12(5)3(4

)12(5)3(4

)]5(12)[()3(4

]12)(5))[((4)(3)()(

242

234

24

2

2224

2222222

22

22

22

2

22

2

2

2

2

2

3

++−+−−

+++

+=

−+−−−−+−+−−

=

⎥⎦

⎤⎢⎣

⎡−−−−−−

⎥⎦

⎤⎢⎣

⎡−+−

+−=

−+−+−

=

+−+−

=

++++

=

ωωωωωω

ωωω

ωωωωωωωωωωω

ωωωωωω

ωωωωω

ωωωωω

ωωωωωωωω

ωωω

j

j

jj

jj

jj

jjj

jjjjjjG

It is determined from Equation (b) that

( ) (c) 144

162)](Re[ 24

2

+++

=ωω

ωωjG

( ) (d) 144

481615)](Im[ 242

2343

++−+−−

=ωωω

ωωωωjG

The Nyquist diagram is a polar plot of Equations (c) and (d) with ω as a parameter. The following are determined from Equations (c) and (d)

(i) For ω=0, Re(0)=1/9, Im(0)=-∞ . Thus the Nyquist diagram has an asymptote at Re=1/9. (ii) As ∞→ω , 0)](Re[ →ωjG and 0)](Im[ →ωjG (iii) As −∞→ω , 0)](Re[ →ωjG and 0)](Im[ →ωjG (iv) There are no real values of ω such that )](Re[ ωjG =0. Thus the only imaginary axis intercept is the origin. (v) 0)](Im[ =ωjG when which has no real roots. Thus the only real axis intercept is at the origin.

0481615 2343

=−+−− ωωω

(vi) Since the only intercepts are the origin and Re(0)=1/9, the Nyquist diagram exists only in the first and fourth quadrants.

The Nyquist diagram generated from MATLAB is

301

Chapter 7

7.21 The transfer function of a proposed filter is

(a) 1002010

62)( 23

3

+++++

=sss

sssG

(a) This circuit may act as a third-order band-reject filter as 03 ≠a and where is the coefficient of in the numerator of G(s).

00 ≠a kaks

(b) The MATLAB workspace used to create the Bode diagram for the transfer function is >> clear >> N=[1 0 2 6] N = 1 0 2 6 >> D=[1 10 20 100] D = 1 10 20 100 >> bode(N,D) >>

The Bode diagram for this transfer function is

302

Chapter 7

While its the transfer function satisfies necessary conditions, the circuit is not an effective band-reject filter. The numerator must be modified such that the low frequency asymptote is approximately the same as the high frequency asymptote.

7.22 (a) The natural frequency of the undamped system is

(a) sr 100

kg 500mN 105

6

=

=

=

x

mk

nω

The frequency ratio is

(b) 1.1 sr 100

sr 110

=

=

=n

rωω

303

Chapter 7

The amplitude of the steady-state response of an undamped second-order mechanical system is

( )

(c) m 1043.1

)1.1(11

sr 100kg 500

N 150

1

1

4

22

220

−=

−⎟⎠⎞

⎜⎝⎛

=

−=

x

rmF

Xnω

Since the input frequency is greater than the natural frequency the phase angle is r πφ = . Thus the steady-state response is

(d) m )100sin(1043.1)( 4 π+= − txtx (a) When the absorber is tuned to the operating frequency the steady-state amplitude of the absorber is

(e) 2

02 k

FX =

In order to limit the steady-state amplitude of the absorber to 1 cm

(f) mN 105.1

m 0.01N 150

N 150m 01.0

4

2

2

x

k

k

=

>

>

When the absorber is tuned to the operating frequency

(g) kg 5.1 sr 100

mN 105.1

2

4

22

2

2

2

=

⎟⎠⎞

⎜⎝⎛

=

=

=

x

km

mk

ω

ω

(b) If a 10-kg machine is removed from the machine its mass is 490 kg. The resulting steady-state amplitude of the machine is calculated using Equation (7.92)

( )[ ] (f)

)( 212

122214

21

2220

1 kkmkmkkmmmkF

X+++−

−=

ωωω

However, when mass is removed from the machine the absorber is still tuned to the input frequency, . Thus the steady-state amplitude of the machine is still zero. 2

22 ωmk =

304

Chapter 7

7.23 The differential equation governing the motion of the machine is (a) )sin(2

0 temkxxcxm ωω=++ &&& Equation (a) is of the form of a second-order system with

(b) )sin()( 20 temtF ωω=

The transfer function for the system of Equation (a) is

(c) 2

/1

1

)()()(

22

2

nn ssm

kcsms

sFsXsG

ωζω ++=

++=

=

The sinusoidal transfer function is

( ) ( )(d)

2

21

2/1)(

2222

22

22

nn

nn

nn

jm

jmjG

ζωωωω

ζωωωω

ζωωωωω

+−

−−=

+−=

The steady-state response of the system is (e) )sin()( φω += tXtx

where

( ) ( )( )

( )(f)

)2(

1

)2(

)2(1

)(

2222

20

2222

22222

0

0

nn

nn

nn

mem

mem

jGFX

ζωωωω

ω

ζωωωω

ζωωωωω

ω

+−=

+−

+−=

=

Substituting nrωω = in Equation (f) leads to

( )( ) ( )

(g) )2()1(

2

1

222

20

22222

20

rrr

mem

rrmrem

Xnnn

n

ζ

ωζωω

ω

+−=

+−=

The phase angle is

305

Chapter 7

(h) 12tan

2tan

2tan

)](Re[)](Im[tan

21

222

21

221

1

⎟⎠⎞

⎜⎝⎛−

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

−

−

−

rr

rr

jGjG

nn

n

n

n

ζ

ωωωζ

ωωζωω

ωωφ

306

Chapter 8

8. Feedback Control Systems 8.1 The differential equations governing the perturbations in concentrations of reactants in a CSTR are

(b) 0

(a) )(

=+−

=++

BAB

AiAA

qCkCdt

dCV

qCCqVkdt

dCV

Application of the Laplace transform to Equations (a) and (b) leads to

(d) 0)()()((c) )()()()(

=+−

=++

sCqsCksCVssCqSCqVksCVs

BAB

AiAA

Equations (c) and (d) are rearranged as

(f) )()(

(e) )(

)(

qVssCksC

qVkVssCq

sC

AB

AiA

+=

++=

The block diagram model for the system is developed from Equations (a) and (b)

8.2 The differential equations for the perturbations of the concentrations of the reactants in a CSTR due to the perturbation in inlet flow rate are

(b) 1.09.04.02.0

(a) 2.04.02.12.0

qCCdt

dC

qCCdt

dC

BAB

BAA

=+−

=−+

Application of the Laplace transform to Equations (a) and (b) leads to

(d) )(1.0)(9.0)(4.0)(2.0(c) )(2.0)(4.0)(2.1)(2.0

sQsCsCsCssQsCsCsCs

BAB

BAA

=+−

=−+

Equations (c) and (d) are rearranged as

(f) 9.02.0

)(4.0)(1.0)(

(e) 2.12.0

)(4.0)(2.0)(

++

=

++

=

ssCsQsC

ssCsQ

sC

AB

BA

The block diagram model for the system is developed from Equations (e) and (f)

307

Chapter 8

8.3 The differential Equations for the temperature of the oil and water in the heat exchanger of Example 6.41 are given in Equations (e) and (f) of the solution of the Example as

(b) )(1035.11099.24.04x10

(a) 01099.21048.11073.5

434

343

tFxxdt

d

xxdt

dx

wow

woo

=+−

=−+

θθθ

θθθ

where the right-hand side of Equation (b) has been replaced by an arbitrary F(t). Application of the Laplace transform to Equations (a) and (b) and rearrangement leads to

(d) 1035.11004.4

)(1099.2)()(

(c) 1048.11073.5)(1099.2

)(

44

3

43

3

xsxsxsF

s

xsxsx

s

ow

wo

+Θ+

=Θ

+Θ

=Θ

The block diagram model for the system is developed from Equations (c) and (d)

8.4 The differential equations which model the concentrations of the drug using the two-compartment model are given by Equations (g) and (h) of the solution of Example 4.30 as

( )

(b) 0

(a) )(

21

21

=+−

=−++

ttppt

t

ttppeP

p

CVkCVkdt

dCV

tICVkCVkkdt

dCV

Application of the Laplace transform to Equations (a) and (b) and rearrangement leads to

308

Chapter 8

(d) )(

)(

(c) )(

)()()(

2

1

1

2

tt

ppt

pep

ttP

VksVsCVk

sC

VkksVsCVksI

sC

+=

+++

=

The block diagram model for this system is developed using Equations (c) and (d)

8.5 The feedback loop involving is in series with leading to the reduction of Figure (a).

)(2 sG 1G

309

Chapter 8

Elimination of the unity feedback loop leads to the reduction of Figure (b) with

(a) 1

11

1)(

212

21

2

21

2

21

1

GGGGGGGG

GGG

sH

++=

++

+=

Noting that is in series with and elimination of the remaining feedback loop leads to Figure (c) with

1H 3G

(b) 1

11

1

1)(

4321212

321

212

2134

212

213

134

31

GGGGGGGGGG

GGGGG

GG

GGGGG

G

HGGGH

sH

+++=

+++

++=

+=

8.6 The transfer functions are in parallel, but both have negative signs. Their equivalent is in series with as illustrated in Figure (a). The feedback loop is eliminated leading to a transfer function of

32 G and G

1G

(a) )(1

)()(

3251

3211 GGGG

GGGsH

+−+−

=

The resulting reduction has a second feedback loop which is eliminated resulting in a system transfer function of

(b) )()(1

)(

)(1)(

1

)(1)(

1)(

32143251

321

3251

3214

3251

321

14

1

GGGGGGGGGGGGGGG

GGGG

GGGGGGG

HGH

sH

+++−+−

=

+−+−

−

+−+−

=

−=

310

Chapter 8

8.7 From the annotated block diagram

it is clear that

[ ] (b) )(2)()()16)(4(

2)(

(a) )(2)()(

2 sBsCsAss

ssB

sCs

sBsC

−−++

+=

+=

311

Chapter 8

Equation (b) is solved for B(s) as [ ] [ ]

(c) 68184)]()()[2()(

)()()2()()2(2)16)(4(

23

2

+++−+

=

−+=++++

ssssCsAssB

sCsAssBsss

Substitution of Equation (c) into Equation (a) leads to

( ) ( )( )[ ]

(d) )(13634112

)2()(

)()2()()2(68184)2()(681842)]()()[2()(68184

)(268184)]()()[2()(

234

23

2323

23

sAssss

sssC

sAsssCsssssssCssssCsAsssCssss

sCssss

sCsAssC

−++++

=

+=+++++−

++++−+=+++

++++

−+=

The closed-loop transfer function is obtained from Equation (d) as

(e) 13634112

)2(

)()()(

234 −++++

=

=

ssssss

sAsCsH

8.8 It is clear from the annotated block diagram

that

(a) 1165405

)()2()(

)()(2

1)16)(2(5

)()(2

1)()16(5

2)()(2.0)()16(

162)()(2.0

)(

23

2

2

2

2

++++

=

=+

++++

=+

+++

⎥⎦⎤

⎢⎣⎡

+−=++

++

⎥⎦⎤

⎢⎣⎡

+−

=

ssssAssC

sAsCs

sss

sAsCs

sCss

ssCsAsCss

sss

sCsAsC

Thus the closed-loop transfer function is

312

Chapter 8

(b) 1165405

2

)()()(

23 ++++

=

=

ssss

sAsCsH

8.9 From the annotated block diagram

it is clear that

[ ]

[ ]

[ ][ ]

(a) )(6.086

2.04)(

)(]2.0)4([)(6.0)4)(2()(6.0)(2.0)4()()4)(2(

)4()(3.0)(2.0)()4()()2(

2)4(

)(3.0)(2.0)()(

23

2

sAsss

sssC

sAsssCssssCsAsssCsss

sssCsAsAsssCs

sss

sCsAsAsC

+++++

=

++=+++−++=++

+−++

=+

++−

+=

The closed-loop transfer function is obtained from Equation (a) as

(b) 6.086

2.04

)()()(

23

2

+++++

=

=

sssss

sAsCsH

8.10 From the annotated block diagram

it is clear that

313

Chapter 8

(b) 5)(3

2)()(

32)(

(a) 1)()(

⎥⎦⎤

⎢⎣⎡

+−

+−

⎟⎠⎞

⎜⎝⎛

+=

+=

ssC

ssBsA

ssB

ssBsC

Equation (b) is solved for B(s) as [ ]

[ ]

(c) 174

)()5(3)()2()(

)()5(3)()2()()2()5(3)5)(2(

)()5(3)()()2()(2

3

++−+

=

+−+=+++++

+−−+=

+

ssCssAssB

sCssAssBssss

sCssBsAssBs

Substitution of Equation (c) in Equation (a) leads to

( )

[ ]

(d) )(32244

2)(

)()2()()5(3)1)(174()()5(3)()2()()1)(174(

)1(174)()5(3)()2()(

2 sAss

ssC

sAssCssssCssAssCss

sssCssAssC

+++

=

+=+++++−+=++

+++−+

=

The closed-loop transfer function is obtained from Equation (d) as

(e) 32244

2

)()()(

2 +++

=

=

sss

sAsCsH

8.11 The differential equation for the system of Example 4.15 is

(a) )(50010095.21068.1 35 tuxdtdx =+ θθ

Equation (a) is rewritten in the standard form of the differential equation for a first-order system as

(b) )(2387.02.80 tudtd

=+θθ

The time constant for the system is T=80.2 s. The settling time for a first-order system is 2.93T. Thus in order to reduce the settling time by 30% the time constant must be reduced by 30%. The required time constant when placed in a feedback control system is

(c) s 1.56 )2.80)(7.0(7.0ˆ

=== TT

The transfer function for the system is obtained by taking the Laplace transform of Equation (a)

314

Chapter 8

(d) 125.0

003.0)(

)(0125.0

003.0

12.802387.0)(

)(2387.02.80

+=

+=

+=Θ

=Θ+Θ

ssG

sFs

ss

sFs

The time constant of the system when placed in a feedback loop with a proportional controller is given by Equation (8.38)

(e) 1

ˆKTK

TTp+

=

Using Equation (c) in Equation (e) leads to

(f) 80.1

)2.80)(003.0(12.801.56

=

+=

p

p

K

K

The closed loop transfer function when this proportional controller is used is given by Equation (8.37) which when applied to this problem leads to

(g) 0179.0

0054.0

)003.0)(80.1(0125.0)003.0)(80.1()(

+=

++=

s

ssH

Application of the Final Value Theorem to Equation (g) leads to a final value of

(h) 302.0 0179.00054.0

=

=fT

8.12 The differential equation for the system of Example 4.15 is

(a) )(50010095.21068.1 35 tuxdtdx =+ θθ

Equation (a) is rewritten in the standard form of the differential equation for a first-order system as

(b) )(2387.02.80 tudtd

=+θθ

The transfer function for the system is obtained by taking the Laplace transform of Equation (a)

315

Chapter 8

(c) 0125.0

003.0)(

)(0125.0

003.0

12.802387.0)(

)(2387.02.80

+=

+=

+=Θ

=Θ+Θ

ssG

sFs

ss

sFs

The damping ratio of the second-order system that results when a first-order plant is used in a feedback control loop with an integral controller is given by Equation (8.46)

(d) 2

1KKT i

=ζ

Requiring the damping ratio of the closed-loop system to be 2.5 leads to

(e) 1007.2

)003.0()2.80(215.2

3−=

=

xK

K

i

i

8.13 The transfer function of a first-order plant is

(a) 5

2.0)(+

=s

sG

When used in a feedback loop with a PI controller the closed loop transfer function is

( ) (b) )(

)()()(

sGsKKssGsKK

sHpi

pi

++

+=

(a) Using and Equation (a) in Equation (b) leads to 10 ,2.5 == pi KK

(c) 04.17

204.1

204.1)5(204.1

52.0)102.5(

52.0)102.5(

)(

2 +++

=

++++

=

+++

++

=

sss

ssss

sss

ss

sH

(b) Equation (b) is the transfer function of a second-order system of natural frequency

(d) 020.1 04.1

==nω

and damping ratio

316

Chapter 8

(e) 43.3 )020.1(2

7

27

=

=

=nω

ζ

(c) Since the damping ratio is greater than one, the system is undamped. Its impulsive response is obtained using Table 6.5 where

( )( ) (g) 152.01)43.3(4.3020.1

(f) 85.61)43.3(43.3020.12

2

21

−=−+−−=

−=−+−=

s

s

Use of superposition leads to

( )

( )

(h) 354.088.3

85.6152.0)85.6(152.0

12

)85.6(152.0104.1)(

85.6152.0

152.085.6

85.6152.0

tt

tt

tti

ee

ee

eetx

−−

−−

−−

−=

+−⎟⎟⎠

⎞⎜⎜⎝

⎛−−−

+

−⎟⎟⎠

⎞⎜⎜⎝

⎛−−−

=

(d) The step response is obtained using Table 6.6. Use of superposition leads to

[ ]

( )

(i) 326.1504.01

)85.6(152.012

)85.6()152.0()152.0(85.6)85.6(152.0

1)85.6)(152.0(

04.1)(

85.6152.0

85.6152.0

152.085.6

tt

tt

tts

ee

ee

eetx

−−

−−

−−

++=

−⎟⎟⎠

⎞⎜⎜⎝

⎛−−−

+

−−−+−−−⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛−−

=

8.14 The transfer function for the first-order plant is

(a) 134.0

833.0

2.612.1

1

11)(

+=

+=

+=

s

s

RAs

sG

(a) When used in a feedback loop with a proportional controller with a gain of 0.2 the closed-loop transfer function becomes

317

Chapter 8

(b) 301.0

167.0

134.0833.02.01

134.0833.02.0

)(

+=

++

+=

s

s

ssH

The time constant for the system is

(c) s 321.3 301.01ˆ

=

=T

The system’s impulsive response is determined from Equation (6.41) as ( )

(d) 0.250

301.0)833.0()(301.0

301.0

t

ti

e

etx−

−=

The offset when subject to a unit step input is given by Equation (8.42) as

(e) 446.0 44.7321.3

ˆ

=

=

=TTη

(b) The closed-loop transfer function when the plant is in a feedback loop with an integral controller of gain 1.5 is

(f) 25.1134.0

25.1

134.0833.05.1

134.0833.05.1

)(

2 ++=

++

+=

ss

ss

ssH

Equation (f) is that of a second-order system with a natural frequency of

(g) r/s 118.1 25.1

==nω

and a damping ratio of

(h) 0601.0 )118.1(2

134.0

=

=ζ

Since the system is underdamped its damped natural frequency is

(i) r/s 116.1 )0601.(01118.1 2

=

−=dω

impulsive response is determined using Table 6.5 as

318

Chapter 8

(j) )116.1sin(120.1

)116.1sin(116.125.1)(

0674.0

)118.1)(0601.0(

te

tetx

t

ti

−

−

=

=

Since an integral controller is used the offset is zero. (c) The transfer function when the plant is placed in a feedback loop with a PI controller with and is 2.0=pK 5.1=iK

(k) 25.1301.0

25.1167.0

134.0833.0)5.12.0(

134.0833.0)5.12.0(

)(

2 +++

=

+++

++

=

sss

sss

ss

sH

Equation (k) is that of a second-order system with a natural frequency of

(l) r/s 118.1 25.1

==nω

and a damping ratio of

(m) 135.0 )118.1(2

301.0

=

=ζ

Since the system is underdamped its damped natural frequency is

(n) r/s 116.1 )0601.(01118.1 2

=

−=dω

impulsive response is determined using Table 6.5 as

[ ] (o) )116.1sin(120.1)118.1sin(135.0)118.1cos(167.0

)116.1sin(116.125.1

)118.1sin(116.1

)118.1)(135.0()118.1cos()167.0()(

151.0)118.1)(135.0(

)118.1)(135.0(

)118.1)(135.0(

tette

te

ttetx

tt

t

ti

−−

−

−

+=

+⎥⎦⎤

⎢⎣⎡ +=

Since a PI controller is used the offset is zero. (d) The closed loop transfer function for the system when the plant is placed in a feedback loop with a PD controller with 2.0=pK and 5.1=dK is

319

Chapter 8

(p) 134.0

0742.0556.0

301.025.2167.025.1

134.0833.0)5.12.0(1

134.0833.0)5.12.0(

)(

++

=

++

=

+++

++

=

ssss

ss

ss

sH

Equation (p) is that of the transfer function of a first-order system of time constant

(q) s 7.46 134.01ˆ =T

Equation (p) can be rewritten as (r) 556.0)( =sH

Thus the system’s impulsive response is (s) )(556.0)( ttxi δ=

(e) The transfer function when the plant is placed in a feedback loop with a PID controller with , and 2.0=pK 5.1=dK 5.1=iK is

(t) 556.0134.0

556.00742.0556.0

25.1301.025.225.1167.025.1

134.0833.0)5.12.05.1(

134.0833.0)5.12.05.1(

)(

2

2

2

2

2

2

++++

=

++++

=

++++

+++

=

ssss

ssss

ssss

sss

sH

Equation (t) is that of a second order system with natural frequency

(u) r/s 745.0 556.0

==nω

and damping ratio

(v) 090.0 )745.0(2

134.0

=

=ζ

Equation (t) can be rewritten as

(w) 556.0134.0

247.0556.0)( 2 +++=

sssH

The impulsive response is obtained by inverting Equation (w) leading to (x) )742.0sin(333.0)(556.0)( 167.0 tettx t

i−+= δ

Since a PID controller is used the offset is zero.

320

Chapter 8

8.15 (a) The differential equations governing the perturbations in liquid level are

(b) 021

(a) 11

212

211

=+−

=−+

hR

hRdt

dhA

qhR

hRdt

dhA

Substitution of given values in Equations (a) and (b) leads to

(d) 0322.0161.02.1

(c) 161.0161.02.1

12

211

=+−

=−+

hhhdt

dh

qhhdtdh

Taking the Laplace transforms of Equations (c) and (d) assuming all initial conditions are zero leads to

(e) 0

)()()(

322.02.1161.0161.0161.02.1

2

1⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡+−

−+ sQsHsH

ss

The transfer functions are obtained from Equation (e) as

(g) 0259.0577.044.1

161.0)(

(f) 0259.0577.044.1

322.02.1)(

22

21

++=

+++

=

sssG

ssssG

The closed-loop transfer function for this arrangement is obtained using the annotated block diagram

( )[ ] (i) 161.02.1

1)(161.0)()()()(

(h) )(322.02.1

161.0)(

++−=

+=

ssCsGsBsQsB

sBs

sC

a

Equation (i) is used to solve for B(s) leading to

(j) )(161.02.1

)(161.0)()()(

sGssCsGsQ

sBa

a

+++

=

Substituting Equation (j) into Equation (h) leads to

321

Chapter 8

(k) )(161.02.1

)(161.0)()()(

161.0322.02.1

sGssCsGsQ

sCs

a

a

+++

=+

Equation (k) is solved for C(s) as

(l) )()322.02.1(0259.0577.044.1

)()(161.0)( 2 sGsss

sGsQsC

a

a

++++=

The closed-loop transfer function for this system is

(m) )()322.02.1(0259.0577.044.1

)(161.0

)()()(

2

2

sGssssG

sQsCsH

a

a

++++=

=

Substitution of Equation (l) into Equation (h) leads to

(n) )()322.02.1(0259.0577.044.1

)()245.7(161.0

)()()(

)()322.02.1(0259.0577.044.1)()(161.0

161.0322.02.1)(

2

1

2

sGssssGs

sQsBsH

sGssssGsQssB

a

a

a

a

+++++

=

=

⎟⎟⎠

⎞⎜⎜⎝

⎛++++

+=

Note that in the remainder of the solution of this problem the numerators for both

and should be multiplied by 0.161, as should the values on all figures. )(1 sH )(2 sH(b) If , Equation (m) becomes 2.0)( =sGa

(o) 0627.0567.0

139.0

0903.0817.044.12.0)(

2

22

++=

++=

ss

sssH

and Equation (n) becomes

(p) 0627.0567.0

278.003.1)( 21 +++

=ss

ssH

The poles of the transfer function are the roots of the denominator of the transfer functions which are

(q) 416.0 ,151.0 21 −=−= ss Since both poles are real and negative the system is stable. The impulsive and step responses for the first tank obtained using MATLAB are given below N=[1.03 0.278]; >> D=[1 0.567 0.0627]; >> impulse(N,D) >>

322

Chapter 8

323

Chapter 8

(c) If s

sGa5.1)( = Equation (m) becomes

( )(r)

483.0826.1577.044.15.1

5.1)322.02.1(0259.0577.044.15.1)(

23

22

+++=

++++=

sss

sssssH

While Equation (n) becomes

( )(s)

483.0826.1577.044.1318.11

5.1)322.02.1(0259.0577.044.15.1)245.7()(

23

21

++++

=

+++++

=

ssss

ssssssH

The poles, impulsive, and step response corresponding to are obtained using the MATLAB commands

)(1 sH

>> N=[11.18 3]; >> D=[1.44 0.577 1.826 0.483]; >> p=roots(D) p = -0.0643 + 1.1086i -0.0643 - 1.1086i -0.2720 >> impulse(N,D) >> figure >> step(N,D) >> Since the poles , -0.0643 + 1.1086i, -0.0643 - 1.1086i, and -0.2720 all have negative real parts the system is stable. The impulsive and step responses are

324

Chapter 8

325

Chapter 8

(d) If s

sGa5.12.0)( += the transfer functions of Equations (m) and (n) become

( )(t)

483.089.1817.044.15.12.0

)5.12.0)(322.02.1(0259.0577.044.15.12.0)(

23

22

++++

=

++++++

=

ssss

sssssssH

( )(u)

483.089.1817.044.1358.1149.1

)5.12.0)(322.02.1(0259.0577.044.1)5.12.0)(245.7()(

23

2

21

+++++

=

+++++++

=

sssss

ssssssssH

The poles, impulsive, and step response corresponding to are obtained using the MATLAB commands

)(1 sH

>> N=[1.49 11.58 3] N = 1.4900 11.5800 3.0000 >> D=[1.44 0.817 1.89 0.483]; >> p=roots(D) p = -0.1476 + 1.1002i -0.1476 - 1.1002i -0.2722 >> impulse(N,D) >> figure >> step(N,D) >> Since the poles -0.1476 + 1.1002i, -0.1476 - 1.1002i, and -0.2722 all have negative real parts the system is stable. The impulsive and step responses are given below

326

Chapter 8

327

Chapter 8

(e) If the transfer functions of Equations (m) and (n) become 2.05.1)( += ssGa

(v) 0644.030.124.3

2.05.1

)2.05.1)(322.02.1(0259.0577.044.1

2.05.1)(

2

22

+++

=

++++++

=

sss

ssssssH

(w) 0644.030.124.3

4.049.418.11

)2.05.1)(322.02.1(0259.0577.044.1

)2.05.1)(245.7( )(

2

2

21

++++

=

+++++++

=

ssss

sssssssH

The poles, impulsive, and step response corresponding to are obtained using the MATLAB commands

)(1 sH

>> N=[11.18 4.49 0.4]; >> D=[3.24 1.30 0.0644]; >> p=roots(D) p = -0.3433 -0.0579 >> impulse(N,D) >> figure >> step(N,D) >> Since the poles, -0.3433 and -0.0579 both have negative real parts the system is stable. The impulsive and step responses are given below.

328

Chapter 8

329

Chapter 8

(f) If s

ssGa5.12.05.1)( ++= the transfer functions of Equations (m) and (n) become

( ) ( )(x)

483.089.13.124.35.12.05.1

5.12.05.1)322.02.1(0259.0577.044.1

5.12.05.1)(

23

2

22

2

2

+++++

=

++++++++

=

sssss

sssssssssH

( ) ( )(y)

483.089.13.124.3358.1149.418.11

5.12.05.1)322.02.1(0259.0577.044.1)5.12.05.1)(245.7()(

23

23

22

2

1

++++++

=

+++++++++

=

ssssss

ssssssssssH

The poles, impulsive, and step response corresponding to are obtained using the MATLAB commands

)(1 sH

>> N=[11.18 4.49 11.58 3]; >> D=[3.24 1.3 1.89 0.483]; >> p=roots(D) p = -0.0646 + 0.7376i -0.0646 - 0.7376i -0.2719 >> impulse(N,D) >> figure >> step(N,D) >> Since the poles -0.0646 + 0.7376i, -0.0646 - 0.7376i, and -0.2719 all have negative real parts the system is stable. The impulsive and step responses follow

330

Chapter 8

331

Chapter 8

8.16 (a) The differential equations governing the perturbations in liquid level are

(b) 021

(a) 11

212

211

=+−

=−+

hR

hRdt

dhA

qhR

hRdt

dhA

Substitution of given values in Equations (a) and (b) leads to

(d) 0322.0161.02.1

(c) 161.0161.02.1

12

211

=+−

=−+

hhhdt

dh

qhhdtdh

Taking the Laplace transforms of Equations (c) and (d) assuming all initial conditions are zero leads to

(e) 0

)()()(

322.02.1161.0161.0161.02.1

2

1⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡+−

−+ sQsHsH

ss

The transfer functions are obtained from Equation (e) as

(g) 0259.0577.044.1

161.0)(

(f) 0259.0577.044.1

322.02.1)(

22

21

++=

+++

=

sssG

ssssG

The closed-loop transfer function for this arrangement is obtained using the annotated block diagram

[ ] (i) )(161.0)()()(161.02.1

1B(s)

(h) )(322.02.1

161.0)(

sCsGsCsQs

sBs

sC

a +−+

=

+=

Use of Equation (i) in Equation (h) leads to

[ ]

(j) )(161.00259.0577.044.1

)(161.0)(

)(161.00259.0577.044.1

)()(161.0)(

)(161.0)()()(161.02.1

1)(161.0

322.02.1

22

2

sGsssGsH

sGsssQsGsC

sCsGsCsQs

sCs

a

a

a

a

a

+++=

+++=

+−+

=+

Substituting Equation (j) in Equation (h) leads to

( )(k)

)(161.00259.0577.044.1)(322.02.1

)(

)(161.00259.0577.044.1)()(161.0

161.0322.02.1)(

21

2

sGsssGs

sH

sGsssQsGssB

a

a

a

a

++++

=

⎟⎟⎠

⎞⎜⎜⎝

⎛+++

+=

(b) If , Equations (j) and (k) become 2.0)( =sGa

332

Chapter 8

(l) 0581.0577.044.1

0322.0

)2.0(161.00259.0577.044.1)2.0(161.0)(

2

22

++=

+++=

ss

sssH

( )

(m) 0581.0577.044.1

0644.024.0

)2.0(161.00259.0577.044.1)2.0(322.02.1)(

2

21

+++

=

++++

=

sss

ssssH

The following MATLAB commands are used to determine the poles, impulsive response, and step response for the second tank >> N=[0.0322]; >> D=[1.22 0.577 0.0581]; >> p=roots(D) p = -0.3276 -0.1454 >> impulse(N,D) >> figure >> step(N,D) >> Since the poles, -0,3276 and -0.1454 both have negative real parts, the system is stable. The impulsive and step responses follow.

333

Chapter 8

334

Chapter 8

(c) If s

sGa5.1)( = , Equations (j) and (k) become

( )(n)

24.00259.0577.044.1242.0

)5.1(161.00259.0577.044.1)5.1(161.0)(

23

22

+++=

+++=

sss

ssssH

( )( )

(o) 24.00259.0577.044.1

483.080.1

)5.1(161.00259.0577.044.1)5.1(322.02.1)(

23

21

++++

=

++++

=

ssss

sssssH

The following MATLAB commands are used to determine the poles, impulsive response, and step response for the second tank >> N=[0.242]; >> D=[1.44 0.577 0.0259 0.242]; >> p=roots(D) p = -0.7093 0.1543 + 0.4616i 0.1543 - 0.4616i >> impulse(N,D) >> figure >> step(N,D) >> Since the complex poles have positive real parts the system is unstable. The impulsive and step responses follow

335

Chapter 8

336

Chapter 8

(d) If s

sGa5.12.0)( += , Equations (j) and (k) become

( )(p)

242.00581.0577.044.1242.00322.0

)5.12.0(161.00259.0577.044.1)5.12.0(161.0)(

23

22

++++

=

+++++

=

ssss

ssssssH

( )( )

(q) 242.00581.0577.044.1

483.0864.124.0

)5.12.0(161.00259.0577.044.1)5.12.0(322.02.1)(

23

2

21

+++++

=

++++++

=

sssss

sssssssH

The following MATLAB commands are used to determine the poles, impulsive response, and step response for the second tank >> N=[0.0322 0.242]; >> D=[1.44 0.577 0.0581 0.242]; >> p=roots(D) p = -0.6927 0.1460 + 0.4704i 0.1460 - 0.4704i >> impulse(N,D) >> figure >> step(N,D) >> Since the complex poles have positive real parts the system is unstable. The impulsive and step responses follow

337

Chapter 8

338

Chapter 8

(e) If , Equations (j) and (k) become 2.05.1)( += ssGa

( )(s)

0581.08185.044.1242.00322.0

)5.12.0(161.00259.0577.044.1)5.12.0(161.0)(

2

22

+++

=

+++++

=

sss

sssssH

( )( )

(t) 0581.08185.044.1

0644.0723.080.1

)5.12.0(161.00259.0577.044.1)5.12.0(322.02.1)(

2

2

21

++++

=

++++++

=

ssss

ssssssH

The following MATLAB commands are used to determine the poles, impulsive response, and step response for the second tank >> N=[0.0322 0.242]; >> D=[1.44 0.8185 0.0581]; >> p=roots(D) p = -0.4853 -0.0831 >> impulse(N,D) >> figure >> step(N,D) >> Since the poles have negative real parts, the system is stable. The impulsive and step responses follow.

339

Chapter 8

340

Chapter 8

(f) If s

ssGa5.12.05.1)( ++= , Equations (j) and (k) become

( )(u)

242.00581.08185.044.1242.00322.0483.0

)5.12.05.1(161.00259.0577.044.1)5.12.05.1(161.0)(

23

2

22

2

2

+++++

=

+++++++

=

sssss

ssssssssH

( )( )

(v) 242.00581.08185.044.1

483.0864.1723.080.1

)5.12.05.1(161.00259.0577.044.1)5.12.05.1(322.02.1)(

23

23

22

2

1

++++++

=

++++++++

=

ssssss

sssssssssH

The following MATLAB commands are used to determine the poles, impulsive response, and step response for the second tank >> N=[0.483 0.0322 0.242]; >> D=[1.44 0.8185 0.0581 0.242]; >> p=roots(D) p = -0.7879 0.1098 + 0.4486i 0.1098 - 0.4486i >> impulse(N,D) >> figure >> step(N,D) >> Since the complex poles have positive real parts the system is unstable. The impulsive and step responses follow.

341

Chapter 8

342

Chapter 8

8.17 The transfer function of the plant is

(a) )4.0(

2.0)( 22 +=

ssCA

The closed-loop transfer function when the plant is placed in a feedback loop with a proportional controller is

(b) 2.016.02.0

2.0

2.0)4.0(2.0

)4.0(2.01

)4.0(2.0

)(

2

2

2

2

p

p

p

p

p

p

KssK

KsKs

K

sK

sH

+++=

++=

++

+=

The natural frequency of the second-order system is (c) 2.016.0 pn K+=ω

The damping ratio is calculated as

(d) 2.016.0

1.02.02

p

n

K+=

=

ζ

ζω

Requiring the damping ratio to be 0.8 leads to

(e) 2.016.0

1.08.0pK+

=

However, from Equation (d) it is clear that the largest possible value of the damping ration is 0.25. Thus it is not possible to design a proportional controller such that the system has a damping ratio of 0.8.

8.18 The transfer function of the plant is

(a) )4.0(

2.0)( 22 +=

ssCA

The closed-loop transfer function when this plant is used in a feedback loop with an integral controller is

343

Chapter 8

(b) 2.016.08.0

2.0

2.0)4.0(2.0

)4.0(2.0)4.0(

2.0

)(

23

2

2

2

i

i

i

i

i

i

KsssK

KssK

sKs

sK

sH

+++=

++=

++

+=

Routh’s criteria shows that a third-order system with is stable if . Application of this criterion to the transfer function of Equation (b) leads to

322

13)( asasassD +++=

321 aaa >

(c) 64.02.0)16.0)(8.0(

<>

i

i

KK

8.19 The transfer function of the plant is

(a) )4.0(

2.0)( 22 +=

ssCA

The closed-loop transfer function of the system when the plant is used in a feedback control loop with a PI controller is

( )(b)

)(2.0)4.0(2.0

)4.0(2.0)(

)4.0(2.0)(

)(

2

2

2

pi

pi

pi

pi

sKKsssKK

ssKKs

ssKK

sH

+++

+=

+++

++

=

Substituting in Equation (b) leads to 1=iK

(c) 2.0)2.016.0(8.0

2.02.0)( 23 ++++

+=

sKsssK

sHp

p

Routh’s criteria shows that a third-order system with is stable if . Application of this criterion to the transfer function of Equation (c) leads to

322

13)( asasassD +++=

321 aaa >

(d) 45.0

2.0)2.016.0)(8.0(

>

>+

p

p

K

K

8.20 The transfer function of the plant is

(a) )4.0(

2.0)( 22 +=

ssCA

344

Chapter 8

The closed-loop transfer function of the system when the plant is used in a feedback control loop with a PI controller is

( )(b)

)(2.0)4.0(2.0

)4.0(2.0)(

)4.0(2.0)(

)(

2

2

2

pi

pi

pi

pi

sKKsssKK

ssKKs

ssKK

sH

+++

+=

+++

++

=

Substituting in Equation (b) leads to 2=pK

(c) 2.056.08.0

4.02.0)( 23

i

i

KssssK

sH+++

+=

Routh’s criteria shows that a third-order system with is stable if . Application of this criterion to the transfer function of Equation (c) leads to

322

13)( asasassD +++=

321 aaa >

(d) 24.22.0)56.0)(8.0(

<>

i

i

KK

8.21 The transfer function of the plant is

(a) )4.0(

2.0)( 22 +=

ssCA

The closed-loop transfer function of the system when the plant is used in a feedback control loop with a PID controller is

(b) )(2.0)4.0(

)(2.0

)4.0(

2.0)(s

)4.0(

2.0)()(

22

2

22

22

ipd

ipd

ipd

ipd

KsKsKssKsKsKs

KsKsK

sKsKsK

sH

++++

++=

++++

+++

=

Substitution of and in Equation (b) leads to 1=iK 2=pK

(c) 2.056.0)2.08.0(

)12(2.0)( 23

2

++++++

=ssKs

ssKsH

d

d

Routh’s criteria shows that a third-order system with is stable if . Application of this criterion to the transfer function of Equation (c) leads to

322

13)( asasassD +++=

321 aaa >(d) 2.0)56.0)(2.08.0( >+ dK

Equation (d) is satisfied for all positive values of . If is allowed to be negative, Equation (d) leads to

dK dK

(e) 21.2−>dK

345

Chapter 8

8.22 The transfer function of the plant is

(a) )2.0(

1.0)( 33 +=

ssCA

(a) The closed-loop transfer function when the plant is placed in a feedback loop with a proportional controller is

(b) 1.0008.012.06.0

1.0

1.0)2.0(1.0

)2.0(1.01

)2.0(1.0

)(

23

3

3

3

p

p

p

p

p

p

KsssK

KsKs

K

sK

sH

++++=

++=

++

+=

(b) Routh’s criteria shows that a third-order system with is stable if . Application of this criterion to the transfer function of Equation (b) leads to

322

13)( asasassD +++=

321 aaa >

(c) 64.0

1.0008.0)12.0)(6.0(

<

+>

p

p

K

K

(c) Substitution of in Equation (b) leads to 1=pK

(d) 108.012.06.0

1.0)( 23 +++=

ssssH

The MATLAB commands to determine the impulsive response for the system whose transfer function is Equation (b) are >> N=[0.1]; >> D=[1 0.6 0.12 0.108]; >> impulse(N,D) >>

346

Chapter 8

The resulting plot of the impulsive response is

8.23 The transfer function of the plant is

(a) )2.0(

1.0)( 33 +=

ssCA

(a) The closed-loop transfer function when this plant is used in a feedback loop with an integral controller is

(b) 1.0008.012.06.0

1.0

1.0)2.0(1.0

)2.0(1.0)2.0(

1.0

)(

234

3

3

3

i

i

i

i

i

i

KssssK

KssKs

Ks

sK

sH

++++=

++=

++

+=

The root-locus method can be used to determine the values of for which the system is stable. The denominator of Equation (c) is written in the form of Equation (6.32) with

iK

(d) 1.0)((c) 008.012.06.0)( 234

=+++=

sRsssssQ

The MATLAB commands to draw the root-locus plot are >> Q=[1 0.6 0.12 0.008 0]; >> R=[0.1];

347

Chapter 8

>> rlocus(R,Q) >> The resulting root locus plot is

The root locus plot shows that the system is stable for 0138.0<iK (c) Substituting in Equation (b) leads to 5.0=iK

(e) 05.0008.012.06.0

05.0)( 234 ++++=

sssssH

The MATLAB commands to determine the step response of the system whose transfer function is that of Equation (e) are >> N=[0.05]; >> D=[1 0.6 0.12 0.008 0.05]; >> step(N,D) >>

348

Chapter 8

The resulting step response is

8.24 The transfer function of the plant is

(a) )2.0(

1.0)( 33 +=

ssCA

(a) The closed-loop transfer function of the system when the plant is used in a feedback control loop with a PID controller is

(b) )(1.0)2.0(

)(1.0

)2.0(

1.0)(s

)2.0(

1.0)()(

23

2

32

32

ipd

ipd

ipd

ipd

KsKsKssKsKsKs

KsKsK

sKsKsK

sH

++++

++=

++++

+++

=

(b) Substituting and 5.0=iK 5.1=pK in Equation (b) gives

( ) (c) 05.0108.01.012.06.0

05.01.01.0)( 234

2

+++++++

=ssKss

ssKsH

d

d

The root-locus method can be used to determine the values of for which the system is stable. The denominator of Equation (c) is written in the form of Equation (6.32) with

iK

(e) 1.0)((d) 0.05108.012.06.0)(

2

234

ssRsssssQ

=

++++=

The MATLAB commands to generate the root-locus plot are

349

Chapter 8

>> Q=[1 0.6 0.12 0.108 0.05]; >> R=[0.1 0 0]; >> rlocus(R,Q) >> The resulting root-locus plot is

The root locus plot shows that the system is stable for . 38.3>dK(c) Substituting in Equation (c) leads to 5.0=dK

(f) 05.0108.017.06.0

05.01.005.0)( 234

2

++++++

=ssss

sssH

The MATLAB commands to determine the step response of the system whose transfer function is Equation (f) are >> N=[0.05 0.1 0.05]; >> D=[1 0.6 0.17 0.108 0.05]; >> step(N,D) >> The resulting step response is

350

Chapter 8

8.25 The transfer function for the system is

(a) 2

2)( 22

2

nn

nn

sss

sGωζω

ωζω++

+=

When the golf cart is empty the natural frequency is 20 r/s and has a damping ratio of 0.7. Thus the transfer function for the empty gold cart is

(b) 40028

40028)( 2 +++

=ss

ssG

(a) When the suspension system is placed in a feedback loop with a proportional damper the closed loop transfer function is

(c) )1(400)1(28

)40028(

)40028(40028)40028(

)(

2

2

pp

p

p

p

KsKssK

sKsssK

sH

++++

+=

++++

+=

The natural frequency of the closed loop system is (d) 120ˆ pn K+=ω

The damping ratio of the closed loop system is

(e) 17.0

1)20(2

)1(28ˆ

p

p

p

K

K

K

+=

+

+=ζ

The system is critically damped when

351

Chapter 8

(f) 04.1

17.01

=

+=

p

p

K

K

(b) The total mass of the golf cart can be written as (g) 500 α+=m

where α is the additional mass of the golfers and clubs. When empty the natural frequency is 20 r/s. Thus with the added mass

(h)

5001

20

500)20(500

2

α

α

ω

+=

+=

=mk

n

The damping ratio of the golf cart with added mass is

(i)

5001

7.0

50050020)500(2

)20)(500)(7.0(2

2

α

αα

ωζ

+=

++

=

=nm

c

The closed loop transfer function of the system with proportional damping is

(j) )1()1(2

)2()( 22

2

pnpn

nnp

KsKssK

sH++++

+=

ωζωωζω

Defining 500αβ = ,substitution of Equations (h) and (i) in Equation (j) leads to

(k) )1(

1400)1(

128

1400

128

)(2

pp

p

KsKs

sKsH

++

+++

+

⎟⎟⎠

⎞⎜⎜⎝

⎛+

++

=

ββ

ββ

The natural frequency of the closed loop system is

(l) 1

120ˆ

βω

+

+= p

n

K

The damping ratio of the closed loop system is

352

Chapter 8

(m) 1

17.0

11

)20(2

11

28ˆ

β

β

βζ

+

+=

+

++

+

=

p

p

p

K

K

K

The system is critically damped when its damping ratio is one, thus

(n) 0408.01.041 500

041.2041.1

041.2041.11

17.01

α

α

ββ

+=

+=

+=

+

+=

p

p

K

K

(f) 04.1

17.01

=

+=

p

p

K

K

8.26 The total mass of the golf cart can be written as

(g) 500 α+=m where α is the additional mass of the golfers and clubs. When empty the natural frequency is 20 r/s. Thus with the added mass

(a)

5001

20

500)20(500

2

α

α

ω

+=

+=

=mk

n

The damping ratio of the golf cart with added mass is

353

Chapter 8

(b)

5001

7.0

50050020)500(2

)20)(500)(7.0(2

2

α

αα

ωζ

+=

++

=

=nm

c

Defining 500/αβ = m, the transfer function of the plant is

(c)

1400

128

1400

128

)(2

ββ

ββ

++

++

++

+=ss

ssG

(a) The closed-loop transfer function when the plant is placed in a feedback loop with a proportional controller is

(d) )1(

1400)1(

128

1400

128

)(2

pp

p

KsKs

sKsH

++

+++

+

⎟⎟⎠

⎞⎜⎜⎝

⎛+

++

=

ββ

ββ

The impulsive response is the inverse of the transfer function. The closed-loop system when a proportional controller is used is a second-order system. The increase in the mass leads to a decrease in the damping ratio. Thus as the mass increases the damping ratio of the system decreases, leading to larger overshoot for the impulsive response. (b) The closed-loop transfer function of the system when an integral controller is used is

(e)

1400

128400

128

1400

128

1400

128

1400

128

1400

128

)(

23

2

βββ

ββ

ββββ

ββ

++

++

++

+

⎟⎟⎠

⎞⎜⎜⎝

⎛+

++

=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

++

+⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛+

++

+

⎟⎟⎠

⎞⎜⎜⎝

⎛+

++

=

ii

i

i

i

Ks

Kss

sK

sKsss

sKsH

Application of Routh’s stability criterion for a third-order system requires

(f) 400)1( )28400)(28(1400

128400

128

ii

ii

KK

KK

ββββ

+>+

+>⎟⎟

⎠

⎞⎜⎜⎝

⎛++

⎟⎟⎠

⎞⎜⎜⎝

⎛+

354

Chapter 8

The range of β is 0<β<0.5. Equation (f) shows that the system is stable for all values of β in this range. The mass does not affect the stability, but does affect relative stability. The larger the value of β, the closer the real part of the dominant pole is to zero. (c) The closed-loop transfer function when the plant is placed in a feedback loop with a PD controller is

( )( ) (g)

400)2840028(281)40028(

1400

12840028

128

1

1400

128)(

1400

128)(

1400

128

1400

128)(

)(

2

2

2

ppdd

pd

ppdd

pd

pd

pd

KsKKsKsKsK

Ks

KKs

K

sKsK

sKsKss

sKsKsH

++++++

++=

++

+

+++⎟⎟

⎠

⎞⎜⎜⎝

⎛+

+

⎟⎟⎠

⎞⎜⎜⎝

⎛+

++

+=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

++

++⎟⎟⎠

⎞⎜⎜⎝

⎛+

++

+

⎟⎟⎠

⎞⎜⎜⎝

⎛+

++

+=

β

βββ

ββ

ββββ

ββ

The transfer function of Equation (g) is that of a second-order system. After simplification the increase in mass appears only in the denominator multiplying the

term. For a given controller, this has the effect of decreasing both the natural frequency and the damping ratio.

2s

8.27 The transfer function for the system is

(a) 2

2)( 22

2

nn

nn

sss

sGωζω

ωζω++

+=

When the golf cart is empty the natural frequency is 20 r/s and has a damping ratio of 0.7. Thus the transfer function for the empty gold cart is

(b) 40028

40028)( 2 +++

=ss

ssG

If the plant is placed in a feedback loop with an integral controller of gain 1.3 the resulting closed-loop transfer function is

(c) 5204.43628

5204.36

)3.1)(40028()40028()40028(3.1)(

23

2

++++

=

+++++

=

ssss

ssssssH

The step response is obtained by inverting

( ) (d) 5204.43628

5204.36)( 23 ++++

=ssss

ssX s

The MATLAB commands to obtain the poles and residues for Equation (d) are >> N=[36.4 520]; >> D=[1 28 436.4 520 0]; >> [r,p,k]=residue(N,D)

355

Chapter 8

r = -0.0048 + 0.0471i -0.0048 - 0.0471i -0.9905 1.0000 p = -13.3530 +14.9512i -13.3530 -14.9512i -1.2940 0 k = [] The above leads to a partial fraction decomposition of

(e) 84.4017.26

54.100953.0294.1

9905.01)( 2 +++

−+

−=sss

sssX s

Inversion of Equation (e) leads to ( ) (f) )05.20sin(0705.0)05.20cos(00953.09905.01)( 35.13294.1 tteetx tt

s +−−= −− (b) The 20% settling time is the time required for the step response to be permanently within 2% of its final value. To this end

( ) (g) )05.20sin(0705.0)05.20cos(00953.09905.0198.0 35.13294.1ss

tt ttee ss +−−= −− Trial and error leads to s 02.3=st(c) The 10%-90% rise time is determined as the time required for the response to grow from 10% of its final value to 90%of its final value. It is determined as 1.65 s. 8.28 The transfer function for the system is

(a) 2

2)( 22

2

nn

nn

sss

sGωζω

ωζω++

+=

When the golf cart is empty the natural frequency is 20 r/s and has a damping ratio of 0.7. Thus the transfer function for the empty gold cart is

(b) 40028

40028)( 2 +++

=ss

ssG

When the plant is placed in a feedback loop with a differential controller of gain 1.2 the closed loop transfer function is

356

Chapter 8

(c) 8806.51

4806.33

)40028(2.140028)40028(2.1)(

2

2

+++

=

+++++

=

sss

ssssssH

The step response of the system is obtained using MATLAB by >> N=[33.6 480]; >> D=[1 51.6 880]; >> step(N,D) >> The final value of the step response is obtained using the final value theorem as 480/880=0.5455. (b) The 2% settling time is the time it takes for the step response to be permanently within 2% of its final value: (0.02)(0.5455)=0.0109, or for the response to be permanently less than 0.5564 and greater than 0.5346. The plot of the step response shows that the settling time is 0.18 s

(c) The 10-90 percent rise time is the time it takes for the response to grow from (0.1)(0.5545) to (0.9)(0.5545) or from 0.05545 to 0.4990. The plot of the step response shows that the rise time is 0.0219 s- 0.00172 s=0.2173 s

357

Chapter 8

8.29 The SIMULINK model for the system of Example 8.2 follows the block diagram for the system presented in the solution of Example 8.2. The SIMULINK model is shown below Customized plots from the data sent to the workspace are developed using the MATLAB commands >> t1=CA(1).time; >> CA1=CA(1).signals.values; >> plot(t1,CA1) >> xlabel('t') >> ylabel('C_A') >> title('Simulation of C_A for Problem 8.29')

358

Chapter 8

>> t2=CB(1).time; >> CB1=CB(1).signals.values; >> plot(t2,CB1) >> xlabel(‘t’) >> ylabel(‘C_B’) >> title (‘Simulation of C_B for Problem 8.29’) >> The customized plots are shown below.

359

Chapter 8

8.30 The transfer function for the system of Example 8.25 is

(a) 25049.7

25049.7)( 2 +++

=ss

ssG

The SIMULINK model for the system when this plant is placed in a feedback loop with a PID controller is A customized plot is developed from the MATLAB worksheet through the commands >> plot(t1,x1) >> xlabel('t') >> ylabel('x') >> title('Simulation of Problem 8.30') >>

360

Chapter 8

8.31 The appropriate SIMULINK model is Customized plots for the concentrations in each reactor are developed from the time and concentration values sent to the worksheet. The commands to draw and annotate the plot for the concentration in the first reactor are >> t1=CA1(1).time; >> CA11=CA1(1).signals.values; >> plot(t1,CA11) >> xlabel('t') >> ylabel('C_A_1') >> title('Concentration of reactant A in first reactor')

361

Chapter 8

Note, that as predicted in Problem 8.24, the system is unstable. 8.32 The closed-loop transfer function is

(a) 8020

)( 3 KssKsH

+++=

The denominator of Equation (a) is of the form of Equation (6.32) with

362

Chapter 8

(c) 1)((b) 8020)( 3

=++=

sRsssQ

The poles of the diagram, the roots of Q(s) are -2.85, j107.5423.1 ± . There are no zeroes of the diagram and all branches end at infinity. The poles are located in Figure (a).

Since Q(s) can be factored into and it is shown in Appendix D that quadratic factors do not affect the portion of the real axis which may be on the a branch of the root locus diagram application of the angle criterion becomes

)1.28846.2)(85.2( 2 +++ sss

(d) )12(85.2 πφ +=+ ns If s is a real value less than -2.85 then πφ =+ 85.2s and the angle criterion is satisfied. Now if s is a real value greater than -2.85, then s+2.85>0 and 085.2 =+sφ . Thus the angle criterion cannot be satisfied. Thus the only part of the real axis which is on a branch of a root locus is s<-2.85, as illustrated in Figure (b). The breakaway points are obtained by solving

(e) 02030)0)(9020()203)(1(

0

2

32

=+

=++−+

=−

ssss

dsdRQ

dsdQR

363

Chapter 8

Since there are no real solutions of Equation (e), there are no breakaway points for this root-locus diagram. There are three poles and no zeroes for the diagram. Thus the angles at which the asymptotes to the branches approach infinity are determined by

( ) (f) 3

1230)12( ππφ +−=

−+

= nna

For n=0, Equation (f) leads to 3πφ −=a , For n=1, Equation (f) gives πφ −=a , and for

n=-1, Equation (a) gives 3πφ =a . The asymptote corresponding to πφ −=a corresponds

to a branch along the negative real axis. The values where the asymptotes meet on the real axis is determined using Equation (D.13) as

[ ](g) 0

107.5423.1107.5423.185.231

=

−+++−= jjs

The asymptotes are sketched in Figure (c). The crossover frequencies and the gain at cross-over are obtained from

(h) 080)20(080)(20)(

2

3

=++−

=+++

KjKjj

ωω

ωω

Setting the real part of Equation (h) to zero leads to a gain at cross-over of K=-80. Thus there are no positive values for which cross-over occurs. This implies that the branches originating in the right half plane remain in the right-half plane. The angle of departure at a complex pole is obtained using the angle criterion

(i) 9.39rad 6967.0 423.185.2

107.5tan2

1

°==

⎟⎠⎞

⎜⎝⎛

+−−= −ππφ p

The root-locus diagram is sketched in Figure (e). The MATLAB generated root-locus diagram, shown below, verifies the sketch of Figure (f).

364

Chapter 8

8.33 The dominant poles of the root-locus diagram of Problem 8.32 have positive real parts for all values of K. Thus this problem has no answer. 8.34 The transfer function for a second-order plant is

(a) 8020

1)( 2 ++=

sssG

When the plant is placed in a feedback loop with a PI controller with the closed-loop transfer function is of the form

30=pK

(b) 11020

30

)30()8020()30(

)(

23

2

i

i

i

i

KsssKs

KssssKs

sH

++++

=

+++++

=

The denominator of Equation (b) is of the form of Equation (6.32) with

(d) 1)((c) 10020)( 23

=++=

sRssssQ

The poles of the diagram are the roots of Q(s) which are js 16.310,0 ±−= . Since R(s)=1, there are no zeroes of the diagrams. Each branch of the root-locus diagram begins at a pole. Since there are no zeroes, each branch ends at infinity. It is noted that . Thus since a quadratic factor has an angle of zero when evaluated along the real axis, the points on the real axis which are on a branch of the diagram are determined by

)11020()( 2 ++= ssssQ

sφ . Since πφ =s for all negative values of s and 0=sφ for all positive values of s, the angle criterion implies that all points on the

365

Chapter 8

negative real axis are on a branch of the root locus while no points on the positive real axis lie on a root locus. The breakaway points are obtained by finding the real values of s, on the negative real axis, such that

( )(e) 88.3 ,46.9

0110403)1(

0

2

−−==++

=−

sss

dsdRQ

dsdQR

It is known that (a) the poles are at s=0 and js 16.310 ±−= (b) all points on the negative real axis are on a branch of the root locus and (c) the breakaway or break-in points are at s=-9.46 and s=-3.88. This knowledge, coupled with the knowledge that two branches may not intersect over a finite region, leads to the following conclusions: (a) the branch which begins at s=0 has a breakaway point at s=-3.88, (b) The branches that begin at the complex poles have a break-in point at s=-9.46. One branch then moves along the negative real axis toward as K increases while the other branch moves along the negative real axis toward s=0, but has a breakaway point at s=-3.88. These deductions are illustrated in Figure (a).

−∞=s

There are three poles and no zeroes for the diagram. Thus application of Equation (D.12) to determine the angles of the asymptotes leads to

(f) 3

)12(30)12( ππφ +−=

−+

= nna

Evaluation of Equation (f) for n=0,1, and -1 leads to the angles of the asymptotes as

(g) 3

,,3

πππφ −−=a

The branch with the asymptote of –π is the branch which travels along the negative real axis to . The points of intersection of the asymptotes with the real axis are obtained using Equation (D.13),

−∞=s

366

Chapter 8

[ ](h) 67.6

16.31016.310031

−=

+−−−= jjs

The asymptotes are illustrated in Figure (b) .

The crossover frequencies and the corresponding gains are obtained from

(i) 020)110(0)(110)(20)(

22

23

=−+−

=+++

ωωω

ωωω

KjKjjj

Setting the imaginary part of Equation (i) to zero leads to a crossover frequency of (j) r/s 49.10110 ==cω

Setting the real part of Equation (i) to zero leads to a crossover gain of

(k) 2200 20 2

== ccK ω

The above information is used to sketch the root-locus diagram of Figure (c).

The MATLAB generated root locus diagram confirms the sketch

367

Chapter 8

8.35 The transfer function for a plant is

(a) 8020

1)( 2 ++=

sssG

The closed-loop transfer function when the plant is placed in a feedback loop with a PI controller is

(b) )8020(

80201)(s

8020

1)()(

2

2

2

ip

ip

ip

ip

KsKsssKsK

ssKsK

ssKsK

sH

++++

+=

++++

+++

=

Substituting in Equation (b) leads to 30=pK

(c) 11020

30)( 23

i

i

KsssKs

sH+++

+=

The root-locus diagram for the transfer function of Equation (c) is obtained by defining

(e) 1)((d) 11020)( 23

=++=

sRssssQ

The following MATLAB commands are used to draw the root-locus diagram >> Q=[1 20 110 0]; >> R=[1]; >> rlocus(R,Q)

368

Chapter 8

>> The resulting root-locus diagram is

The root-locus diagram shows that the largest value of the integral gain such that the dominant pole is less than -2 is 714=iK . The corresponding damping ratio is 0.304 with a frequency of 6.69 r/s. 8.36 The transfer function for a plant is

(a) 8020

1)( 2 ++=

sssG

The closed-loop transfer function for the system when it is placed in a feedback loop with a PID controller with and 30=pK 50=iK is

369

Chapter 8

(b) 5011020

5030

5030)8020(5030

)(

223

2

22

2

sKsssssK

ssKsssssK

sH

d

d

d

d

++++++

=

+++++++

=

The denominator of Equation (b) is of the form of Equation (6.32) with

(d) )((c) 5011020)(

2

23

ssRssssQ

=

+++=

The poles of the diagram are the roots of Q(s) which are js 28.29.751- ,4986.0 ±−= , Each branch originates at a pole. The zeroes of the diagram are the roots of R(s) which has a double root at s=0. Thus the branches end either at infinity or at s=0. The angle criterion is used to determine the parts of the real axis which lie on a branch of the root locus. Since ( )3.1005.19)4896.0()( 2 +++= ssssQ and and a quadratic factor has an angle of zero along the real axis the angle criterion reduces to

2)( ssR =

(a) )12(4896.0 πφ +=+ ns If s<-0.4896, s+0.4896 is negative and πφ =+ 4896.0s . If s>-0.4896 then s+0.4896 is positive and 04896.0 =+sφ . Thus, the portion of the real axis such that s<-0.4896 is on some branch of a root locus. Note however that the zero s=0 is a special case in that

0φ can be any value. Thus s=0 could be on a branch of the root locus. The breakaway and break-in points are the real values of s which lie on a branch and satisfy

(b) 9161.10,9161.0,10,00)1001102(01001102

0)2)(5011020()110403)((

0

3

24

2322

−−==−−

=−−

=+++−++

=−

ssss

ssssssssss

dsdRQ

dsdQR

The following is known about the root locus diagram (a) The origins of the branches are . (b) The branches terminate either at s=0 or at infinity (c)

Every point of the portion of the real axis , s<-0.4896 is on a branch of the root locus (d) The breakaway points and break-in points are s=0,-10, -0.9161. The above knowledge along with the knowledge that two branches cannot intersect over a finite region leads to the following conclusions (a) The branch of the root locus which originates at s=-0.4896 travels along the negative real axis for increasing K until it reaches s=-0.9161 where its branch breaks away from the axis. (b) The branches originating at

have break-in points at s=10. One branch travels along the negative real axis for increasing K, terminating at infinity. The other branch, for increasing K, travels along the negative real axis toward s=0 until it reaches a breakaway point at

js 28.29.751- ,4986.0 ±−=

js 28.29.751- ±=

370

Chapter 8

S = -0.9161. The branches that have breakaway points have break-in points at s=0 where they terminate. A sketch of the root locus curves drawn form this knowledge is shown in Figure (a). Angles of departure and arrival may be calculated as well as asymptotes and the intersection of the asymptotes with the real axis. It can also be shown that the crossover frequency is zero which corresponds to a K of infinity. The root locus drawn using MATLAB confirms the above observations

8.37 The transfer function for a second-order plant is

(a) )8.1)(2.1(

2)(++

=ss

sG

The closed-loop transfer function of the system when the plant is used in a feedback control loop with a PI controller is

( )

(b) 2)216.2(3

22

)(2)8.1)(2.1(

2

)8.1)(2.1(2)(

)8.1)(2.1(2)(

)(

23ip

ip

pi

pi

pi

pi

KsKssKsK

sKKssssKK

sssKKs

sssKK

sH

++++

+=

++++

+=

++++

+++

=

Substituting in Equation (b) leads to 6=pK

(c) 216.143

212)( 23

i

ip

KsssKs

sH+++

+=

371

Chapter 8

Equation (c) is in the form of Equation (6.32) with

(e) 2)((d) 16.143)( 23

=++=

sRssssQ

(a)The MATLAB commands to draw the root-locus diagram for this system are >> Q=[1 3 14.16 0]; >> R=[2]; >> rlocus(R,Q) >>

(b) The minimum overshoot of 25.5 percent corresponds to a gain of zero.

(c) The minimum overshoot corresponds to a frequency of 3.76 r/s and a damping ratio of 0.399.

8.38 The transfer function for a second-order plant is

(a) )8.1)(2.1(

2)(++

=ss

sG

The closed-loop transfer function of the system when the plant is used in a feedback control loop with a PI controller is

372

Chapter 8

( )

(b) 2)216.2(3

22

)(2)8.1)(2.1(

2

)8.1)(2.1(2)(

)8.1)(2.1(2)(

)(

23ip

ip

pi

pi

pi

pi

KsKssKsK

sKKssssKK

sssKKs

sssKK

sH

++++

+=

++++

+=

++++

+++

=

It is desired to determine the minimum overshoot for several values of the proportional gain. Equation (b) can be written in the form of Equation (6.32) with

(d) 2)(

(c) )216.2(3)( 23

=

+++=

sR

sKsssQ p

The MATLAB commands to generate the root locus diagram for 2.0=pK are >> Kp=0.2; >> Q=[1 3 2.16+Kp 0]; >> R=[2]; >> rlocus(R,Q) >> The corresponding root locus diagram indicating the minimum overshoot is

The following is the root locus diagram for 1=pK

373

Chapter 8

The root locus diagram can be drawn in this fashion for any value of the proportional gain. The following table gives, for several values of proportional gain, the minimum overshoot, corresponding gain, damping ratio, and frequency.

pK minη iK ζ ω 0 0 0 to 0.228 1 0.469 to 1.18

0.2 0 0 to 0.279 1 0.557 to 1.47 1.0 0.0396 0.58 0.928 1.08 1.5 1.42 0.695 0.804 1.46 2.0 3.24 0.455 0.737 1.85 5.0 11.9 0 0.561 2.68 10.0 22.4 0 0.43 3.49 25.0 38.9 0 0.288 5.21

374

Chapter 8

8.39 The transfer function of a third-order plant is

(a) 1053

2)( 23 ++++

=sss

ssG

(a) When placed in a feedback loop with a proportional controller the closed loop transfer function is

(b) )2(1053

)2(

105321

10532

)(

23

23

23

+++++

+=

++++

+

++++

=

sKssssK

ssssK

ssssK

sH

p

p

p

p

The root locus diagram is drawn for this system using

(d) 2)((c) 1053)( 23

+=+++=

ssRssssQ

The appropriate MATLAB commands to draw the root locus diagram are

>> Q=[1 3 5 10]; >> R=[1 2]; >> rlocus(R,Q) >>

The resulting root locus diagram is

(b) The closed-loop transfer function when the plant is placed in a feedback loop with a PI controller is

375

Chapter 8

(e) )2)((1053

)2)((

10532)(

1053

2)()(

234

23

23

++++++

++=

++++

++

++++

+=

sKsKsssssKsK

ssssKsKs

ssssKsK

sH

ip

ip

ip

ip

Substituting in Equation (e) leads to 25=pK

(f) 2)60(303

)2)(25()( 234

ii

i

KsKssssKs

sH+++++

++=

A root-locus diagram can be drawn for the system whose transfer function is Equation (f) with

(h) 2)((g) 60303)( 234

+=+++=

ssRsssssQ

The MATALB commands used to generate the root-locus plot are

>> Q=[1 3 30 60 0] >> R=[1 2] >> rlocus(R,Q) >> The resulting root-locus diagram is

The root-locus diagram shows that the system is stable for 2.21<iK . (c) The closed-loop transfer function for the system when the plant is placed in a feedback loop with a PID controller is

376

Chapter 8

(i) )2)((1053

)2)((

10532

10532)(

)(

2234

2

23

232

+++++++

+++=

++++

+

++++

++=

sKsKsKsssssKsKsK

sssss

ssssKsKsK

sH

ipd

ipd

ipd

Substitution of and 25=pK 100=iK in Equation (i) leads to

(j) 200160)230()3(

)10025()( 234

2

++++++++

=ssKsKs

ssKsH

dd

d

A root-locus diagram is constructed for this system using

(l) 2)((k) 200160303)(

23

334

sssRsssssQ

+=

++++=

The MATLAB commands used to generate the root-locus diagram are

>> Q=[1 3 30 160 200]; >> R=[1 3 0 0]; >> rlocus(R,Q) >>

- The resulting root-locus diagram is

The root-locus diagram shows that the system is stable for 29.2>dK

(d) The system with the proportional controller is stable for all values of the proportional gain, but has offset. The use of integral control eliminates the offset, but is destabilizing

377

Chapter 8

and increases overshoot. The addition of derivative control is stabilizing, but relative stability can be a problem. 8.40 The transfer function of a third-order plant is

(a) )4)(3)(2(

1)(+++

+=

sssssG

The closed-loop transfer function when this plant is used in a feedback loop with a PID controller is

(b) )24()26()9(

)()(

))((1()4)(3)(2())(1(

)4)(3)(2(1)(

)4)(3)(2(1)(

)(

234

2`3

2

2

2

2

iippdd

iippdd

ipd

ipd

ipd

ipd

KsKKsKKsKsKsKKsKKsK

KsKsKsssssKsKsKs

ssssKsKsKs

ssssKsKsK

sH

++++++++++

+++++=

+++++++

+++=

++++

+++

++++

++=

A MATLAB M file written to plot the root-locus curve, plot the impulsive response, and to plot the step response for input values of the controller gains is listed below % Program for Problem 8.40 % Design of a PID controller % Input controller gains disp('Please enter proportional gain') Kp=input('>>'); disp('Please enter integral gain ') Ki=input('>>'); disp('Please enter differential gain ') Kd=input('>>'); disp('Which parameter do you want to vary for root-locus curve?') disp('Enter 1 for proportional gain') disp('Enter 2 for integral gain') disp('Enter 3 for differential gain') i=input('>>') while i~=1&i~=2&i~=3 disp('Invalid enry, please reenter') disp('Enter 1 for proportional gain') disp('Enter 2 for integral gain') disp('Enter 3 for differential gain') i=input('>>'); end if i==1 Q=[1 9+Kd 26+Kd 24+Ki Ki]; R=[1 1 0]; elseif i==2

378

Chapter 8

Q=[1 9+Kd 26+Kd+Kp 24+Kp 0]; R=[1 1]; else Q=[1 9 26+Kp 24+Kp+Ki Ki]; R=[1 1 0 0]; end rlocus(R,Q) % Impulsive reponse and step response for input values N=[Kd Kd+Kp Kp+Ki Ki]; D=[1 9+Kd 26+Kd+Kp 24+Kp+Ki Ki]; figure impulse(N,D) figure step(N,D) % End of program Some output from the program follows. First consider 10 ,10 ,100 === idp KKK , but with the root-locus diagram drawn for the integral gain. The root locus plot is

Root Locus

Real Axis

Imag

inar

y Ax

is

-25 -20 -15 -10 -5 0 5-20

-15

-10

-5

0

5

10

15

20

System: sysGain: 237

Pole: -7.08 - 3.7iDamping: 0.887

Overshoot (%): 0.243Frequency (rad/sec): 7.99

System: sysGain: 2.15e+003

Pole: 0.0275 + 10.9iDamping: -0.00251

Overshoot (%): 101Frequency (rad/sec): 10.9

379

Chapter 8

The corresponding step response is

0 10 20 30 40 50 600

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1Step Response

Time (sec)

Ampl

itude

The figures show that the minimum overshoot occurs for a gain of 210 and that the settling time is almost 25 seconds for this design. Thus try a controller with a larger integral gain and a small differential gain. The next set is for Kp=100, Ki=100 and Kd=25 with the root-locus diagram drawn for varying Ki. 8.41 The transfer function of a third-order plant is

(a) )2)(1(

1)(++

=sss

sG

The closed-loop transfer function when the plant is placed in a feedback loop with a proportional controller is

(b) 23

1

)2)(1(1)(

23 Ksss

KssssH

+++=

+++=

The denominator of the transfer function of Equation (b) is of the form of Equation (6.32) with

(d) 1)((c) 23)( 23

=++=

sRssssQ

The MATLAB commands to draw the root-locus diagram for this system are >> Q=[1 3 2 0]; >> R=[1];

380

Chapter 8

>> rlocus(R,Q) >> rlocus(R,Q) >> The resulting root-locus diagram is

(b) The transfer function of the plant when placed in a feedback loop with a compensator is

(e) )())(2)(1(

)(

)2)(1(11

)2)(1(1

)(

asKbssssasK

sssbsasK

sssbsasK

sH

++++++

=

++++

+

++++

=

A maximum overshoot of 10 percent corresponds to a damping ratio of 0.591. Thus it is desired to design he compensator such that the dominant pole of the system has a damping ratio of 0.591 and a frequency of 1.5 r/s. From Equation (9.80) this implies the dominant pole is

(f) 21.1887.0 )591.0(1)5.1()5.1)(591.0( 2

jjs

±−=

−±−=

In order for this point to be on a branch of the root-locus diagram the angle criterion must be satisfied

(g) )12(21 πφφφφφ +=−−−− ++++ nbssssas where

381

Chapter 8

827.0113.121.1tan )21.1(113.12

478.1113.021.1tan )21.1(113.01

203.2887.021.1tan )21.1(887.0

12

11

=⎟⎠⎞

⎜⎝⎛=+=+

=⎟⎠⎞

⎜⎝⎛=+=+

=⎟⎠⎞

⎜⎝⎛−

=+−=

−+

−+

−

s

s

s

js

js

js

φ

φ

φ

Thus Equation (j) becomes

(h) 508.4)12( 827.0478.1203.2)12(

++=++++=− ++

ππφφ

nnbsas

Choose b=2.5 (the choice is arbitrary). Then

(i) 6436.0613.121.1tan 1 =⎟

⎠⎞

⎜⎝⎛= −

+Bsφ

Equation (h) becomes (j) 152.5)12( ++=+ πφ nas

For n=-1, Equation (j) leads to (k) 01.2152.5 =+−=+ πφ as

Equation (k) requires

(l) 319.0887.0

21.1)01.2tan(

=+−

=

aa

The magnitude criterion leads to

(m) 52.421.1568.0)319.021.1887.0()21.1887.0(

47.557.2 )21.1613.1)(21.1113.1)(21.1113.0)(21.1887.0()21.1887.0(

)21.1887.0()21.1887.0(

=+−=++−=+−

−=++++−=+−

+−+−

=

KjjjR

jjjjjjQ

jRjQK

Thus choosing a=0.319, b=2.5, and K=4.52, Equation (e) becomes

(n) 44.152.950.950.5

442.152.4

)319.0(52.4)5.2)(2)(1()319.0(52.4)(

234 +++++

=

++++++

=

sssss

sssssssH

The step responses without the compensator is determined using the MATLAB commands >> N=[1]; >> D=[1 3 2 0]; >> step(N,D) >>

382

Chapter 8

The resulting step response is

The step response of the system with the compensator is obtained using MATLAB by >> N=[4.52 1.44]; >> D=[1 5.5 9.5 9.52 1.44]; >> step(N,D) >> The resulting step response is

383

Chapter 8

8.42 The transfer function of the plant is

(a) )205(

2)( 2 ++=

ssssG

Use of the Ziegler-Nichols tuning rules require knowledge of the crossover frequency when a proportional controller is used. The closed-loop transfer function when the plant of Equation (a) is placed in a feedback loop with a proportional controller is

(b) 2)205(

2

)205(21

)205(2

)(

2

2

2

KsssK

sssK

sssK

sH

+++=

+++

++=

The root-locus diagram for the transfer function of Equation (b) is obtained by defining

(d) 2)((c) 205)( 23

=++=

sRssssQ

The MATLAB commands used to generate the root-locus diagram are >> Q=[1 5 20 0]; >> R=[2]; >> rlocus(R,Q) >> The resulting diagram is

384

Chapter 8

The root-locus diagram shows that the gain at the crossover frequency is and the crossover frequency is 4.5 r/s.

7.50=cK

(a) The Ziegler-Nichols tuning rule for a proportional controller is to choose

(c) 4.25 )7.50(5.0

5.0

==

= cp KK

(b) Application of the Ziegler-Nichols tuning rule for a PI controller gives

(e) 931.0 )5.4(3

4

34

(d) 8.22

45.0

=

=

=

=

=

πωπ

ci

cp

K

KK

(c) Application of the Ziegler-Nichols tuning rules for a PID controller leads to

(h) 175.0 4

(g) 698.0

(f) 4.30

6.0

=

=

=

=

=

=

cd

ci

cp

K

K

KK

ωπ

ωπ

385

Chapter 8

386

Chapter 9

9. State-Space Methods 9.1 Consider the differential equation

)a()(25163 tFxxxx =+++ &&&&&& Define

(d) (c) (b)

3

2

1

xyxyxy

&&&

===

Use of Equations (b)-(d) in Equation (a) gives

)e()(31625)(25163

3213

1233

tFyyyytFyyyy

+−−−==+++

&&

Equations (c), (d), and (e) are summarized in matrix form as

(f) )(

00

31625100010

3

2

1

3

2

1

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

tFyyy

yyy

&&&

Assuming the system output is 1)( ytx = the output is given as

[ ] [ ][ ] )g()(10001

3

2

1

tFyyy

x +⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

The state and input matrices are determined from Equation (f) as

(i) 100

(h) 31625

100010

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−=

B

A

The output matrix and the transmission matrix are determined from Equation (g) as [ ][ ] (k) 0

(j) 001==

DC

9.2 Consider a system governed by the differential equations

(b) )(325.0

(a) )(22.0

2212

1211

tFyydt

dy

tFyydtdy

=+−

=−+

The system inputs are and . The system outputs are and . Equations (a) and (b) are each divided by the coefficient of their derivative term leading to

)(1 tF )(2 tF )(1 ty )(2 ty

387

Chapter 9

(d) )(4124

(c) )(5510

2212

1211

tFyydt

dy

tFyydtdy

+−=

++−=

Using and as state variables, Equations (c) and (d) can be summarized in a matrix form as

1y 2y

(e) )()(

4005

124510

2

1

2

1

2

1⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−=⎥

⎦

⎤⎢⎣

⎡tFtF

yy

yy&&

The output relations are written as

(f) )()(

0000

1001

2

1

2

1

2

1⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡tFtF

yy

yy

The state matrix and input matrix are determined from Equation (e) as

(h) 4005

(g) 124510

⎥⎦

⎤⎢⎣

⎡=

⎥⎦

⎤⎢⎣

⎡−

−=

B

A

The output matrix and transmission matrix are determined from Equation (f) as

(j) 0000

(i) 1001

⎥⎦

⎤⎢⎣

⎡=

⎥⎦

⎤⎢⎣

⎡=

D

C

9.3 Consider a system governed by the differential equations

(b) 04223

(a) )(2432

2121

2121

=+−+

=−++

yydt

dydtdy

tFyydt

dydtdy

The system has one input and two outputs, and . Let and be the state variables. Equations (a) and (b) are summarized in matrix form as

)(tF )(1 ty )(2 ty )(1 ty )(2 ty

(c) )(01

)()(

4224

2332

2

1

2

1 tFtyty

yy

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡&&

Multiplying Equation (c) by the inverse of the coefficient matrix multiplying the vector of derivatives leads to

388

Chapter 9

(d) )(6.04.0

8.22.32.38.2

)(01

52

53

53

52

4224

52

53

53

52

2

1

2

1

2

1

2

1

tFyy

yy

tFyy

yy

⎥⎦

⎤⎢⎣

⎡−+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−=⎥

⎦

⎤⎢⎣

⎡

⎥⎦

⎤⎢⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−−=⎥

⎦

⎤⎢⎣

⎡

&&

&&

The relation between the input and output is [ ] (f) F(t)

01

001001

2

1

2

1⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡yy

yy

The state matrix and input matrix are determined from Equation (d) as

(i) 6.04.0

(h) 8.22.32.38.2

⎥⎦

⎤⎢⎣

⎡−=

⎥⎦

⎤⎢⎣

⎡−

−=

B

A

The output matrix and transmission matrix are determined from Equation (f) as

[ ] (k) 00

(j) 1001

=

⎥⎦

⎤⎢⎣

⎡=

D

C

9.4 The mathematical model is given in Equations (b), (d), and (f) in the solution to Example 3.27

(c) 01

(b) 011

(a) )(11

0332

32

02

0132

321

02

011

21

=++−+

=+−−+++

=−+++

∫

∫∫

∫∫

t

tt

tt

dtiC

RiRidtdi

LdtdiM

dtiC

dtiC

RiRidtdi

MdtdiL

dtdiM

tvdtiC

dtiC

Ridtdi

Mdtdi

L

The system has one input , , and three outputs and . Define the charges through the circuit branches by

)(tv )(),( 21 titi )(3 ti

(f)

(e)

(d)

33

22

11

dtdq

i

dtdq

i

dtdq

i

=

=

=

The state variables are the charges and the currents,

389

Chapter 9

)6( )5( )4( )3( )2( )1(

336

225

114

33

22

11

gyiygyiygyiygqygqygqy

&&&

======

===

Equations (a)-(c) are rewritten using the state variables as

(j) 01

(i) 011

(h) )(11

36565

2165654

21454

=++−+

=+−−+++

=−+++

yC

RyRyyLyM

yC

yC

RyRyyMyLyM

tvyC

yC

RyyMyL

&&

&&&

&&

Equations (g4), (g5), (g6), (h), (i), and (j) are summarized in a matrix form as

)(k )(

001000

0100

0011

00011100000010000001000

0000000

0000000100000010000001

6

5

4

3

2

1

6

5

4

3

2

1

tv

yyyyyy

RRC

RRCC

RCC

yyyyyy

LMMLM

ML

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

−−

−−=

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

&&&&&&

The inverse of the coefficient matrix on the left-hand side of Equation (k ) is determined using the symbolic capabilities of MATLAB as

( ) ( )

( )

(l)

222000

222000

222000

000100000010000001

0000000

0000000100000010000001

22

22

2222

2

222222

22

2

2222

22

1

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

−−

−−

−−

−−−

−−−

−−

=

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡−

MLLM

MLM

MLLM

MLM

MLL

MLM

MLLM

MLM

MLLML

LMMLM

ML

Pre-multiplying Equation (k) by the matrix on the right-hand side of Equation (l) leads to

390

Chapter 9

(m) )(

000

)()()()(

)()(

)()(100000010000001000

2

6

5

4

3

2

1

2

2

6

5

4

3

2

1

tv

LM

M

yyyyyy

LMLR

LMLR

LRM

LCLCLMM

LCLMM

MLRMLRMRCM

CLM

CLM

LMLMR

LMLMR

LR

LCM

LCML

LCML

yyyyyy

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

+

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

+−

+−−

++−

++−

+−

+

+−

+−−

++−=

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

α

α

αβ

αβ

αβ

ααβ

αα

αααααα

αααβ

ααβ

αβ

&&&&&&

where

(o) (n) 2

22

22

MLML

−=

−=

β

α

The state matrix and input matrix are determined from Equation (m) as

(q)

000

(p)

)()()()(

)()(

)()(100000010000001000

2

2

2

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

=

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

+−

+−−

++−

++−

+−

+

+−

+−−

++−=

α

α

αβ

αβ

αβ

ααβ

αα

αααααα

αααβ

ααβ

αβ

LM

M

LMLR

LMLR

LRM

LCLCLMM

LCLMM

MLRMLRMRCM

CLM

CLM

LMLMR

LMLMR

LR

LCM

LCML

LCML

B

A

391

Chapter 9

The relation between the output and state variables is

(r) )(000

100000010000001000

6

5

4

3

2

1

3

2

1

tv

yyyyyy

iii

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡+

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

The output matrix and transmission matrix are obtained from Equation (r) as

(t) 000

(s) 100000010000001000

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

D

C

9.5 Referring to the circuit diagram application of KCL at node leads to 1Q

(a) 0321 =−− iii

Assuming an ideal amplifier, and 03 =i 0

1=Qv . Thus

(b) 1

11 R

vi =

(c) 12

2 dtdvC

Rvi AA −−=

392

Chapter 9

Substitution of Equations (b) and (c) into Equation (a) leads to

(d)

0

1

1

2

121

1

CRv

CRv

dtdv

dtdv

CRv

Rv

AA

AA

−−=

=++

Application of KCL at node gives 2Q(e) 0654 =−− iii

Assuming an ideal amplifier, and 06 =i 22vvQ = . Thus

(f) 3

24 R

vvi A −=

( ) (g) 225 BvvdtdCi −=

Substitution of Equations (f) and (g) in Equation (e) leads to

( )

(h)

0

2

23

2

23

223

2

dtdv

CRv

CRv

dtdv

vvdtdC

Rvv

AB

BA

++−=

=−−−

Equations (d) and (h) provide a mathematical model for the system. However note that Equation (h) has a term proportional to the derivative of an input. Define state variables as

(j) (i)

22

1

vvyvy

B

A

µ+==

Use of these state variables in Equations (d) and (h) lead to

(l) 11

(k) 11

2223

123

22

111

112

1

vvCR

yCR

vy

vCR

yCR

y

&&&

&

++−=−

−−=

µ

Choosing 1−=µ , Equation (l) becomes

(m) 112

231

232 v

CRy

CRy +−=&

Equations (k) and (l) are summarized by

(n) 10

01

01

01

2

1

23

11

2

1

23

12

2

1⎥⎦

⎤⎢⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡−+⎥

⎦

⎤⎢⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−=⎥

⎦

⎤⎢⎣

⎡vv

CR

CRyy

CR

CRyy&&

The output from the system is . Thus the output matrix and transmission matrix are obtained from

Bv

393

Chapter 9

[ ] [ ] [ ] (o) 10102

1

2

1⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡=

vv

yy

vB

Substitution of numerical values Ω=Ω=Ω= 6000 , 4000 , 2000 321 RRR , F 3021 µ== CC leads to

(s) 1] 0[(r) 1] 0[

(q) 56.50067.16

(p) 056.5033.8

==

⎥⎦

⎤⎢⎣

⎡−=

⎥⎦

⎤⎢⎣

⎡−−

=

DC

B

A

9.6 The differential equations governing a mechanical system are

(a) )(

00

330352

023

000022022

300040002

3

2

1

3

2

1

3

2

1

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

−+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

−+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

tFxxx

xxx

xxx

&&&

&&&&&&

The state variables are defined by [ ] (b) 321321

Txxxxxx &&&=y The state matrix is as presented in Equation (9.25) where

(d) 00005.05.0011

000022022

300040002

)(c 110

75.025.15.0015.1

330352

023

300040002

1

1

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

−

−

−

−

d1

1

CM

KM

Thus

(e)

00011005.05.075.025.15.0011015.1100000010000001000

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−

−−=A

The input matrix is given by Equation (9.26) where

(f) 333.000

100

300040002 1

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

−

− QM 1

394

Chapter 9

Thus the input vector is

(g)

333.000000

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=B

The output matrix is

(h) 000100000010000001

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=C

and the transition matrix is [ ] (i) 0=D

9.7 The integro-differential equations derived in Example 3.14 are

(b) 0)(1)(

(a) )()(1)(

01212223

22

02121211

11

=−+−++

=−+−++

∫

∫t

t

dtiiC

iiRiRdtdi

L

tvdtiiC

iiRiRdtdi

L

Substitution of given values into Equations (a) and (b) leads to

(d) 010435005.01041500

(c) )(104150010425002.0

02

62

2

01

61

02

62

01

61

=+++−−

=−−++

∫∫

∫∫tt

tti

dtixidtdi

dtixi

tvdtixidtixidtdi

The system has one input, , and two outputs and . The state-space variables are the charges and currents

)(tv )(1 ti )(2 ti

(e4) (e3) (e2) (e1)

224

113

22

11

yiyyiy

qyqy

&&

====

==

Use of Equations (e) in Equations (c) and (d) leads to

(g) 010435005.01041500

(f) )(104150010425002.0

26

4416

3

26

416

33

=+++−−

=−−++

yxyyyxy

tvyxyyxyy&

&

Equations (f) and (g) can be rewritten as

(i) 107103108108

(h) )(5105.71025.1102102

43

33

26

16

4

43

34

27

17

3

yxyxyxyxy

tvyxyxyxyxy

−+−=

++−+−=

&

&

Equations (e3), (e4), (h) and (i) are summarized in matrix form as

395

Chapter 9

(j) )(

0500

107103108108105.71025.1102102

10000100

3366

3477

4

3

2

1

tv

xxxxxxxx

yyyy

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

+

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

&&&&

The state matrix and input matrix are obtained for this state-space model using Equation (j) resulting in

(k)

107103108108105.71025.1102102

10000100

3366

3477

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−

=

xxxxxxxx

A

(l)

0500

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=B

The relation between the state variables and the output variables is

(m) )(00

10000100

4

3

2

1

2

1 tv

yyyy

ii

⎥⎦

⎤⎢⎣

⎡+

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡

The output matrix and transmission matrix are determined using Equation (m) as

)(o 00

(n) 10000100

⎥⎦

⎤⎢⎣

⎡=

⎥⎦

⎤⎢⎣

⎡=

D

C

9.8 The differential equations obtained in Example 2.28 are

(a)

00

222111222111

22112211

222

2111122

112221222

2111122

112221

⎥⎦

⎤⎢⎣

⎡+−+−

+++=

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡+−−+

+⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡+−−+

+⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

ycycykykycycykyk

xkkkkkkkkx

ccccccccx

Im

&λ&λλλ

&&

λλλλλλ

&&

λλλλλλ

&&&&

θθθ

Pre-multiplication by the inverse of the mass matrix leads to

396

Chapter 9

(b) y

2

1

2211

21

2

1

2211

21

2,21,2

2,11,1

2,21,2

2,11,1

⎥⎦

⎤⎢⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−+

⎥⎦

⎤⎢⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−+⎥

⎦

⎤⎢⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−⎥⎦

⎤⎢⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−=⎥⎦

⎤⎢⎣

⎡

yI

cI

cmc

mc

yy

Ik

Ik

mk

mk

x

Ik

Ik

mk

mk

x

Ic

Ic

mc

mc

x

&&

λλ

λλ&&

&&&&

θθθ

where

(d)

(c)

222

2111122

112221

2,21,2

2,11,1

222

2111122

112221

2,21,2

2,11,1

⎥⎦

⎤⎢⎣

⎡+−−+

=⎥⎦

⎤⎢⎣

⎡

⎥⎦

⎤⎢⎣

⎡+−−+

=⎥⎦

⎤⎢⎣

⎡

λλλλλλ

λλλλλλ

kkkkkkkk

kkkk

cccccccc

cccc

State variable are defined in the form of

(h) z

(g) (f) (e)

22124

21113

2

1

yy

yyxzz

xz

νµθ

νµθ

++=

++===

&&

Use of Equations (e)-(h) in Equation (b) leads to

(i) y

2

1

2211

21

2

1

2211

21

2

1

2,21,2

2,11,1

2

1

22

11

4

3

2,21,2

2,11,1

2

1

22

11

2

1

⎥⎦

⎤⎢⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−+⎥

⎦

⎤⎢⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

+⎥⎦

⎤⎢⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−⎟⎟⎠

⎞⎜⎜⎝

⎛⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−⎥

⎦

⎤⎢⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−=⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−⎥

⎦

⎤⎢⎣

⎡

yI

cI

cmc

mc

yy

Ik

Ik

mk

mk

zz

Ik

Ik

mk

mk

yy

zz

Ic

Ic

mc

mc

yy

zz

&&

λλλλ

&&

&&

νµνµ

νµνµ

The terms proportional to the derivatives of the input functions are eliminated from Equation (i) by choosing

(j) 2211

21

22

11

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−=⎥

⎦

⎤⎢⎣

⎡

Ic

Ic

mc

mc

λλνµνµ

Substitution of Equation (j) into Equation (i) leads to

397

Chapter 9

(k)

2

1

2,21,2

2,11,1

2

1

2,21,2

2,11,1

4

3

2,21,2

2,11,1

2

1

2211

21

2,21,2

2,11,1

2211

21

2

1

2,21,2

2,11,1

4

3

2,21,2

2,11,1

2

1

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−⎥⎦

⎤⎢⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−=

⎥⎦

⎤⎢⎣

⎡

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

+⎥⎦

⎤⎢⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−⎥⎦

⎤⎢⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−=⎥⎦

⎤⎢⎣

⎡

yy

zz

Ik

Ik

mk

mk

zz

Ic

Ic

mc

mc

yy

Ic

Ic

mc

mc

Ic

Ic

mc

mc

Ik

Ik

mk

mk

zz

Ik

Ik

mk

mk

zz

Ic

Ic

mc

mc

zz

αααα

λλλλ

&&

where

(l)

3222,221,222

2,2

2112,211,211

1,2

222,12

21,122,1

112,12

11,111,1

2211

21

2,21,2

2,11,1

2211

21

2,21,2

2,11,1

⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−−=

⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−=⎥

⎦

⎤⎢⎣

⎡

Icc

mIcc

Ik

Icc

mIcc

Ik

mIcc

mcc

mk

mIcc

mcc

mk

Ic

Ic

mc

mc

Ic

Ic

mc

mc

Ik

Ik

mk

mk

λλ

λλ

λ

λ

λλλλ

α

α

α

α

αααα

The state-space formulation of the equations is

(m) 0000

10000100

2

1

2,21,2

2,11,1

4

3

2

1

2,21,22,21,2

2,11,12,11,1

4

3

2

1

⎥⎦

⎤⎢⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

+

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−−−−=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

yy

zzzz

Ic

Ic

Ik

Ik

mc

mc

mk

mk

zzzz

αααα

&&&&

Assuming the output is the vector [ ]T θx , the output is determined from

(n) 0000

00100001

2

1

4

3

2

1

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡+

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡yy

zzzz

xθ

398

Chapter 9

9.9 The differential equations governing the motion of the system of Figure 9.9 are

(a) 0

2000

00

2

1

2

1

2

1⎥⎦

⎤⎢⎣

⎡ +=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡ kyycxx

kkkk

xxc

xx

mm &

&&

&&&&

Since the right-hand side of Equation (a) involves the derivative of the input, a state-space formulation similar to that of Example 9.6 is required. To this end define state variables as

(b) 24

23

12

11

yxzxz

yxzxz

υ

µ

+==

+==

&

&

Substitution into the differential equations of Equation (a) leads to

(c) 0

2000

00

3

1

4

2

4

2⎥⎦

⎤⎢⎣

⎡ +=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−+⎥

⎦

⎤⎢⎣

⎡−−

⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡−−

⎥⎦

⎤⎢⎣

⎡ kyyczz

kkkk

yzyzc

yzyz

mm &

&&&&

υµ

υµ

Setting mc /−=µ and 0=υ in Equation (c) leads to

(d) 0

2000

00

2

3

1

4

2

4

2

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡ ymck

zz

kkkk

zzc

zz

mm

&&

Equations (b) and (d) can be summarized as

(e) )(

0000

1000

020010

)(

00

00001000000001

001000020010

00001000000001

)(

00

001000020010

00001000000001

2

2

4

3

2

1

4

3

2

1

2

1

4

3

2

11

4

3

2

1

2

4

3

2

1

4

3

2

1

tymc

mk

mc

zzzz

mk

mk

mk

mc

mk

zzzz

tymck

c

m

m

zzzz

kk

kck

m

m

zzzz

tymck

c

zzzz

kk

kck

zzzz

m

m

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−+

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−

−−=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

+

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−+

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

&&&&

&&&&

&&&&

399

Chapter 9

The state matrix and input matrix are determined from Equation (e) as

(g)

00

(f)

001000

020010

2

2

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−

−

−−

−

=

mc

mk

mc

mk

mk

mk

mc

mk

B

A

The output and transition matrices for the model are

(i) 00

(h) 00100001

⎥⎦

⎤⎢⎣

⎡=

⎥⎦

⎤⎢⎣

⎡=

D

C

9.10 The state matrix derived in the solution of Problem 9.1 is

(a) 31625

100010

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−=A

The state transition matrix is ( ) (b) L)( 11 −− −=Φ AIst

To this end

(c) 31625

1001

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+−

−=−

ss

ss AI

The symbolic capabilities of MATLAB are used to invert the matrix in Equation (b). The MATLAB workspace to accomplish the inversion is >> syms s >> A=[s -1 0;0 s -1;25 16 s+3] A = [ s, -1, 0] [ 0, s, -1] [ 25, 16, s+3] >> A1=A^-1

400

Chapter 9

A1 = [ (s^2+3*s+16)/(s^3+3*s^2+16*s+25), (s+3)/(s^3+3*s^2+16*s+25), 1/(s^3+3*s^2+16*s+25)] [ -25/(s^3+3*s^2+16*s+25), s*(s+3)/(s^3+3*s^2+16*s+25), s/(s^3+3*s^2+16*s+25)] [ -25*s/(s^3+3*s^2+16*s+25), -(16*s+25)/(s^3+3*s^2+16*s+25), s^2/(s^3+3*s^2+16*s+25)] >> Thus

( ) (d) 251625)3(25

13163

251631

2

2

231

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+−−+−+++

+++=− −

sssss

sss

ssss AI

The poles of the transforms in the matrix of Equation (d) are obtained as

(g) 673.35971.0(f) 673.35971.0(e) 8059.1

3

2

1

jsjs

s

−−=+−=

−=

Partial fraction decompositions of the transforms in Equation (d) are of the form

( ) ( ) )(h 8.1319.1

1806.1

12

1321 CCCAI +

+++

+=− − s

ssss

MATLAB is used to help determine the partial fraction decompositions. For example the elements in the third row and third column of each matrix are obtained with the following worksheet >> D=[1 3 16 25] D = 1 3 16 25 >> N=[1 0 0] N = 1 0 0 >> [r,p,k]=residue(N,D) r = 0.3909 + 0.2912i 0.3909 - 0.2912i 0.2182

401

Chapter 9

p = -0.5971 + 3.6725i -0.5971 - 3.6725i -1.8059 k = [] >> C3=r(1)*(s-p(2))+r(2)*(s-p(1)) C3 = (7042123938002753/18014398509481984+655836290881283/2251799813685248*i)*(s+2688890991237795/4503599627370496+8269686015043673/2251799813685248*i)+(7042123938002753/18014398509481984-655836290881283/2251799813685248*i)*(s+2688890991237795/4503599627370496-8269686015043673/2251799813685248*i) >> C3A=vpa(simplify(C3),3) C3A = .782*s-1.67 >> The resulting matrices are

(i) 2182.0605.3673.11208.01443.0673.1

0669.00797.09261.0

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−=1C

(j) 728.0605.3673.11208.014.1673.10669.00797.00739.0

2

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−−=C

(k) 67.18.13023.1

926.011.1023.10409.005.176.1

3

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−=C

The state transition matrix is obtained by inverting the transforms of Equation (h) leading to

( ) (l) )67.3sin(595.0273.0)67.3cos()( 595.0595.0806.1 teteet ttt −−− −++=Φ 2321 CCCC

402

Chapter 9

9.11 The state matrix derived in the solution of Problem 9.2 is

(a) 124510⎥⎦

⎤⎢⎣

⎡−

−=A

The state transition matrix is ( ) (b) L)( 11 −− −=Φ AIst

To this end

)(c 124510⎥⎦

⎤⎢⎣

⎡+−−+

=−s

ss AI

and

( ) )(d 104

51210022

12

1⎥⎦

⎤⎢⎣

⎡+

+++

=− −

ss

sss AI

The following MATLAB workspace is used to determine the partial fraction decompositions for each element of the matrix in Equation (d) >> clear >> D=[1 22 100] D = 1 22 100 >> N11=[1 12] N11 = 1 12 >> [r11,p11,k11]=residue(N11,D) r11 = 0.3909 0.6091 p11 = -15.5826 -6.4174 k11 = []

403

Chapter 9

>> N12=[5] N12 = 5 >> [r12,p12,k12]=residue(N12,D) r12 = -0.5455 0.5455 p12 = -15.5826 -6.4174 k12 = [] >> N21=[4] N21 = 4 >> [r21,p21,k21]=residue(N21,D) r21 = -0.4364 0.4364 p21 = -15.5826 -6.4174 k21 = [] >> N22=[1 10]

404

Chapter 9

N22 = 1 10 >> [r22,p22,k22]=residue(N22,D) r22 = 0.6091 0.3909 p22 = -15.5826 -6.4174 k22 = [] >> From the MATLAB workspace it is clear that the poles of the transforms are -15.58 and -6.42. Use of the MATLAB workspace leads to

( ) )(e 391.0436.0546.0609.0

42.61

609.0436.0546.0391.0

58.1511

⎥⎦

⎤⎢⎣

⎡+

+⎥⎦

⎤⎢⎣

⎡−

−+

=− −

sss AI

Inversion of Equation (e) leads to

(f) 391.0436.0546.0609.0

609.436.0546.0391.0

)( 42.658.15 tt eet −−⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡−

−=Φ

9.12 The state matrix obtained during the solution of Problem 9.5 is

(a) 056.5033.8⎥⎦

⎤⎢⎣

⎡−−

=A

The state transition matrix is defined as ( ) (b) L)( 11 −− −=Φ AIst

To this end

( ) (c) 833.056.5

0)833.0(

1

56.50833.0

1⎥⎦

⎤⎢⎣

⎡+−+

=−

⎥⎦

⎤⎢⎣

⎡ +=−

−

ss

sss

ss

s

AI

AI

Inversion of the transforms in Equation (c) leads to

405

Chapter 9

( ) (d) 11675.00

833.0

833.0

⎥⎦

⎤⎢⎣

⎡

−=Φ

−

−

t

t

ee

9.13 The state transition matrix is defined as

( ) (a) L)( 11 −− −=Φ AIst The state matrix derived during the solution of Problem 9.9 is

(b)

001000

020010

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−

−

−−

−

=

mk

mk

mk

mc

mk

A

Thus

(c)

0100

02001

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−

−+=−

smk

mk

smk

mcs

mks

s AI

MATLAB’s symbolic capabilities are used to determined ( )

d)(

)2()2()()2()2()(

)()2()())((

)(1

22

22

222

22

1

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−+−+−−+−+−−++−

−−−−−−−−+

=

− −

kcsmsmskcsmskkmskmskskcsmsmkcsmsmskmkmsk

kmskmskmsmskmskkmmkskmskmskcs

sD

s AI

where

(e) 3)( 22342 kcksmkscmssmsD +−−+= It is not possible to proceed further without numerical values for system parameters. 9.14 The state transition matrix derived during the solution of Problem 9.10 is

( ) (a) )67.3sin(595.0273.0)67.3cos()( 595.0595.0806.1 teteet ttt −−− −++=Φ 2321 CCCCwhere

(b) 2182.0605.3673.11208.01443.0673.1

0669.00797.09261.0

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−=1C

(c) 728.0605.3673.11208.014.1673.10669.00797.00739.0

2

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−−=C

406

Chapter 9

)(d 67.18.13023.1

926.011.1023.10409.005.176.1

3

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−=C

The initial condition vector is given as

(e) 001

)0(⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=y

The free response is obtained using Equation (9.45) as )(f )0()()( ytty Φ=

Substitution of Equations (a)-(e) into Equation (f) leads to

(g) )67.3sin(001

728.0605.3673.11208.014.1673.10669.00797.00739.0

595.067.18.13023.1

926.011.1023.10409.005.176.1

273.0

)67.3cos(001

728.0605.3673.11208.014.1673.10669.00797.00739.0

001

2182.0605.3673.11208.01443.0673.1

0669.00797.09261.0)(

595.0

595.0

806.1

te

te

et

t

t

t

−

−

−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡×

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−−−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−−+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−=Φ

Simplification of Equation (g) leads to

(h) )67.3sin(551.0551.0

469.0)67.3cos(

673.1673.10739.0

673.1673.1

926.0

)()()(

595.0595.081.1 teteetxtxtx

tt =−−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

&&&

9.15 The state transition matrix derived in the solution to Problem 9.11 is

(a) 391.0436.0546.0609.0

609.436.0546.0391.0

)( 42.658.15 tt eet −−⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡−

−=Φ

The initial condition vector is given as

(b) 5.0

0)0( ⎥

⎦

⎤⎢⎣

⎡=y

The free response is obtained using Equation (9.45) (c) )0()()( ytty Φ=

407

Chapter 9

Substitution of Equations (a) and (b) into Equation (c) leads to

(d) 196.0273.0

305.0273.0

5.0

0391.0436.0546.0609.0

5.00

609.436.0546.0391.0

)(

42.658.15

42.658.15

tt

tt

ee

eet

−−

−−

⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡−=

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−=y

9.16 Application of Kirchoff’s laws to the circuit lead to the following equations

(b) 0

(a) )0(1

212

011

21

=+−

−=++− ∫

RidtdiL

dtdiL

Cqdti

CRi

dtdi

Ldtdi

Lt

Define the state variables as

)(23

112

11

ciydtdqiy

qy

=

==

=

Equations (a) and (b) are written in terms of the state variables as

(e)

0

(d) )0(1

)0(1

323

323

2132

1232

yLRyy

RyyLyLLCqy

LRy

LCyy

Cqy

CRyyLyL

−=−

=+−

−−−=−

−=++−

&&

&&

&&

&&

Equations (c)-(e) are summarized in matrix form as

(f) )0(

0

10

00

01010

110110

001

3

2

1

3

2

1

qLC

yyy

LR

LR

LCyyy

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−+⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−−=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

&&&

However the formulation of Equation (f) does not work as the matrix on the left hand side of Equation (f) is singular. This is to be expected as there are only two energy storage devices in the system, thus only two states are available. Notice the subtraction of Equation (e) from Equation (d)

(g) )0()(1)(2 23123 LCqyy

LRy

LCyy −−+−=− &&

Equation (g) suggests that the state variables are

(h) 122

11

iiyqy−=

=

408

Chapter 9

Note that Equation (d) can be formulated in terms of the state variables of Equation (h) as

(i) )0(1112 LC

qyLRy

LCy −−= &&

Thus the state-space formulation of the system using the state variables of Equation (h) is

(j) )0(

211

221

01

10

12

1

2

1

LCq

yy

LR

LC

LCyy

LR

⎥⎥⎦

⎤

⎢⎢⎣

⎡−

−+⎥

⎦

⎤⎢⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−=⎥

⎦

⎤⎢⎣

⎡⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

&&

Substituting given values,

)(k 5.0

0015.07.1661033.8

500.01051.7

5.0

17.1661033.8

01067.11013.333

2

16

4

2

1

2

16

7

2

1

φ

φ

⎥⎦

⎤⎢⎣

⎡−

−+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

−−

=⎥⎦

⎤⎢⎣

⎡

⎥⎦

⎤⎢⎣

⎡−−

+⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

−=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

yy

xx

yy

yy

xx

yy

&&

&&

where )/()0( LCq=φ is taken as the input. The state matrix and input matrix are defined in Equation (k). The output matrix and transition matrix are

[ ] (l) ][ 01 0DC == (b) The state transition matrix is

( ) (m) L)( 11 −− −=Φ AIst To this end

[ ]

(o) 1034.81053.7)(

(n) 1051.71033.8

5.07.166)(

1

7.1661033.85.01051.7

652

461

6

4

xxssD

xsxs

sDs

sxxs

s

+−=

⎥⎦

⎤⎢⎣

⎡−−−−

=−

⎥⎦

⎤⎢⎣

⎡

−−

=−

−AI

AI

Inversion of Equation (n) leads to

(p) 11.111

6669.01040.71049.11.111

6669.09993.0)( 0111.0

41051.7

4

4 ttx ex

ex

t ⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡−−−

=Φ−

−

The system response is

[ ]

(q) 332.0332.0

)()(

0

)(0111.0)(1051.71

0

4

∫

∫

−− −=

−Φ=

tttx

t

deey

dt

τφ

τττ

ττ

Buy

9.17 The differential equations governing the motion of the system are

(a) 00

105105105105

3000003000

2

177

77

2

1⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

−−

+⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡xx

xxxx

xx&&&&

409

Chapter 9

When premultiplied by the inverse of the mass matrix, Equation (a) becomes

(b) 67.167.167.167.1

102

14

2

1⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−+⎥

⎦

⎤⎢⎣

⎡xx

xx&&&&

The state vector is defined as

(c)

2

1

2

1

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

xxxx

&&

y

The state-space formulation of Equation (c) is

(d)

001067.11067.1001067.11067.110000100

4

3

2

1

44

44

4

3

2

1

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

yyyy

xxxx

yyyy

&&&&

The state matrix for the system is

(e)

001067.11067.1001067.11067.110000100

44

44

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

=

xxxx

A

The state transition matrix is

( ) (f) L)( 11 −− −=Φ AIst which is determined from the following MATLAB work session >> c=1.67e4 c = 16700 >> A=[0 0 1 0;0 0 0 1;-c c 0 0;c -c 0 0] A = 0 0 1 0 0 0 0 1 -16700 16700 0 0 16700 -16700 0 0 >> syms s >> I=[1 0 0 0;0 1 0 0;0 0 1 0;0 0 0 1]

410

Chapter 9

I = 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 >> B=(s*I-A)^-1 B = [ 1/s*(s^2+16700)/(s^2+33400), 16700/s/(s^2+33400),

(s^2+16700)/s^2/(s^2+33400), 16700/s^2/(s^2+33400)] [ 16700/s/(s^2+33400), 1/s*(s^2+16700)/(s^2+33400), 16700/s^2/(s^2+33400),

(s^2+16700)/s^2/(s^2+33400)] [ -16700/(s^2+33400), 16700/(s^2+33400), 1/s*(s^2+16700)/(s^2+33400),

16700/s/(s^2+33400)] [ 16700/(s^2+33400), -16700/(s^2+33400), 16700/s/(s^2+33400),

1/s*(s^2+16700)/(s^2+33400)] >> PHI=ilaplace(B) PHI = [ 1/2+1/2*cos(10*334^(1/2)*t), -1/2*cos(10*334^(1/2)*t)+1/2,

1/6680*334^(1/2)*sin(10*334^(1/2)*t)+1/2*t, 1/2*t-1/6680*334^(1/2)*sin(10*334^(1/2)*t)]

[ -1/2*cos(10*334^(1/2)*t)+1/2, 1/2+1/2*cos(10*334^(1/2)*t), 1/2*t-1/6680*334^(1/2)*sin(10*334^(1/2)*t), 1/6680*334^(1/2)*sin(10*334^(1/2)*t)+1/2*t]

[ -5*334^(1/2)*sin(10*334^(1/2)*t), 5*334^(1/2)*sin(10*334^(1/2)*t), 1/2+1/2*cos(10*334^(1/2)*t), -1/2*cos(10*334^(1/2)*t)+1/2]

[ 5*334^(1/2)*sin(10*334^(1/2)*t), -5*334^(1/2)*sin(10*334^(1/2)*t), -1/2*cos(10*334^(1/2)*t)+1/2, 1/2+1/2*cos(10*334^(1/2)*t)]

>> PHI1=vpa(PHI) PHI1 = [

.50000000000000000000000000000000+.50000000000000000000000000000000*cos(182.75666882497065256033071913464*t), -.50000000000000000000000000000000*cos(182.75666882497065256033071913464*t)+.50000000000000000000000000000000, .27358782758229139604839928014167e-

411

Chapter 9

2*sin(182.75666882497065256033071913464*t)+.50000000000000000000000000000000*t, .50000000000000000000000000000000*t-.27358782758229139604839928014167e-2*sin(182.75666882497065256033071913464*t)]

[ -.50000000000000000000000000000000*cos(182.75666882497065256033071913464*t)+.50000000000000000000000000000000, .50000000000000000000000000000000+.50000000000000000000000000000000*cos(182.75666882497065256033071913464*t), .50000000000000000000000000000000*t-.27358782758229139604839928014167e-2*sin(182.75666882497065256033071913464*t), .27358782758229139604839928014167e-2*sin(182.75666882497065256033071913464*t)+.50000000000000000000000000000000*t]

[ -91.378334412485326280165359567320*sin(182.75666882497065256033071913464*t), 91.378334412485326280165359567320*sin(182.75666882497065256033071913464*t), .50000000000000000000000000000000+.50000000000000000000000000000000*cos(182.75666882497065256033071913464*t), -.50000000000000000000000000000000*cos(182.75666882497065256033071913464*t)+.50000000000000000000000000000000]

[ 91.378334412485326280165359567320*sin(182.75666882497065256033071913464*t), -91.378334412485326280165359567320*sin(182.75666882497065256033071913464*t), -.50000000000000000000000000000000*cos(182.75666882497065256033071913464*t)+.50000000000000000000000000000000, .50000000000000000000000000000000+.50000000000000000000000000000000*cos(182.75666882497065256033071913464*t)]

>> PHI1=vpa(PHI,3) PHI1 = [ .500+.500*cos(183.*t), -.500*cos(183.*t)+.500, .274e-2*sin(183.*t)+.500*t,

.500*t-.274e-2*sin(183.*t)] [ -.500*cos(183.*t)+.500, .500+.500*cos(183.*t), .500*t-.274e-2*sin(183.*t),

.274e-2*sin(183.*t)+.500*t] [ -91.5*sin(183.*t), 91.5*sin(183.*t), .500+.500*cos(183.*t), -

.500*cos(183.*t)+.500] [ 91.5*sin(183.*t), -91.5*sin(183.*t), -.500*cos(183.*t)+.500,

.500+.500*cos(183.*t)]

412

Chapter 9

>>

)3.183cos(5.05.0)3.183cos(5.05.0)3.183sin(5.91)3.183sin(5.91)3.183cos(5.05.0)3.183cos(5.05.0)3.183sin(5.91)3.183sin(5.91

)3.183sin(0274.05.0)3.183sin(0274.05.0)3.183cos(5.05.0)3.183cos(5.05.0)3.183sin(0274.05.0)3.183sin(0274.05.0)3.183cos(5.05.0)3.183cos(5.05.0

tttttttt

tttttttttttt

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

+−−++−−+−+−−+−+

=Φ

The initial condition vector is

(h)

0800

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=0y

The system response is determined using Equation (9.34)

(i)

)3.183cos(44)3.183cos(44

)3.183sin(2192.04)3.183sin(2192.04

0800

)3.183cos(5.05.0)3.183cos(5.05.0)3.183sin(5.91)3.183sin(5.91)3.183cos(5.05.0)3.183cos(5.05.0)3.183sin(5.91)3.183sin(5.91

)3.183sin(0274.05.0)3.183sin(0274.05.0)3.183cos(5.05.0)3.183cos(5.05.0)3.183sin(0274.05.0)3.183sin(0274.05.0)3.183cos(5.05.0)3.183cos(5.05.0

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−+−

−+

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

+−−++−−+−+−−+−+

=

Φ=

tt

tttt

X

tttttttt

tttttttttttt

0yy

Thus

(k) )3.183sin(2192.04)((j) )3.183sin(2192.04)(

2

1

tttxtttx

−=+=

9.18 The differential equations governing the system after the coupling engages the bumper are

(a) 00

105105105102.5

000102.1

3000003000

2

177

77

2

14

2

1⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

−−

+⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡xx

xxxx

xxx

xx

&&

&&&&

Equation (a) may be rewritten as

413

Chapter 9

(b) 1067.11067.11067.11073.1

0004

2

144

44

2

1

2

1⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

−−

−⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−=⎥

⎦

⎤⎢⎣

⎡xx

xxxx

xx

xx

&&

&&&&

Defining state variables as

)c(

2

1

2

1

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

xxxx

&&

y

The state-space formulation of the system is

(d)

001067.11067.1041067.11073.110000100

44

44

4

3

2

1

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

xxxx

yyyy

&&&&

Equation (d) is valid after the bumper is engaged, after m 101 =x . For the appropriate state-space model is that of Equation (d) of the solution of Problem 9.17

m 101 <x

)e(

001067.11067.1001067.11067.110000100

4

3

2

1

44

44

4

3

2

1

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

yyyy

xxxx

yyyy

&&&&

Equations (d) and (e) are integrated using the MATLAB program ode45. The initial conditions as formulated in Equation (h) of the solution of Problem 9.17 are

)f(

0800

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=0y

The program applies the initial conditions to Equation (e) and solves Equation (e) until . After this time it solves Equation (d). m 101 =x

The MATLAB m file for this simulation is shown below % Problem 9.18 clear % Specify initial conditions y0=[0 0 8 0]; % Specify final time for integration tf=3; % Call ode45 [t,y]=ode45(@bumper,[0 tf],y0) % Plots results figure plot(t,y(:,1),'-',t,y(:,2),'.')

414

Chapter 9

xlabel('t (s)') ylabel('x (m)') title('Dispalcements of railroad cars of Problem 9.18') legend('x_1','x_2') figure plot(t,y(:,3),'-',t,y(:,4),'.') xlabel('t (s)') ylabel('v (m/s)') title('Velocities of railroad cars of Problem 9.18') legend('v_1','v_2') The required function bumper is % Function for Problem 9.18 function dy=bumper(t,y) if y(1)<10 dy1=y(3); dy2=y(4); dy3=-1.67e4*y(1)+1.67e4*y(2); dy4=1.67e4*y(1)-1.67e4*y(2); else dy1=y(3); dy2=y(4); dy3=-1.73e4*y(1)+1.67e4*y(2)-4*y(3); dy4=1.67e4*y(1)-1.67e4*y(2); end dy=[dy1;dy2;dy3;dy4] The generated plots are

415

Chapter 9

The details are difficult to determine in these plots. The following is a Figure acquired using the zoom in feature on the plot of displacements

416

Chapter 9

9.19 The differential equations governing the motion of the system after the cars engage the bumper are

(b) 01051053000

(a) 102.11021051053000

27

17

2

6.11

41

62

71

71

=+−

−+−+

xxxxx

xxxxxxxxx&&

&&&

Dividing by the mass and rearranging, Equations (a) and (b) are written as

(d) 1067.11067.1

(c) 41067.61067.11067.1

24

14

2

6.111

22

41

41

xxxxx

xxxxxxxx

−=

−−+−=

&&

&&&

Define state variable by

)e(

2

1

2

1

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

xxxx

&&

y

Equations (c) and (d) are written using state variables as

(i) 1067.11067.1

(h) 41067.61067.11067.1

(g) (f)

24

14

4

6.131

22

41

43

42

31

yxyxy

yyxyxyxy

yyyy

−=

−−+−=

==

&

&

&&

Equations (f)-(i) are valid after the bumper is engaged, after m 101 =x . For the appropriate state-space model is that of Equation (d) of the solution of Problem 9.17

m 101 <x

417

Chapter 9

)e(

001067.11067.1001067.11067.110000100

4

3

2

1

44

44

4

3

2

1

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

yyyy

xxxx

yyyy

&&&&

Equations (f)-(i) and (e) are integrated using the MATLAB program ode45. The initial conditions as formulated in Equation (h) of the solution of Problem 9.17 are

)f(

0800

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=0y

The program applies the initial conditions to Equation (e) and solves Equation (e) until . After this time it solves Equation (d). m 101 =x

The MATLAB m file for this simulation is shown below % Problem 9.19 clear % Specify initial conditions y0=[0 0 8 0]; % Specify final time for integration tf=3; % Call ode45 [t,y]=ode45(@bumpern,[0 tf],y0) % Plots results figure plot(t,y(:,1),'-',t,y(:,2),'.') xlabel('t (s)') ylabel('x (m)') title('Dispalcements of railroad cars of Problem 9.19') legend('x_1','x_2') figure plot(t,y(:,3),'-',t,y(:,4),'.') xlabel('t (s)') ylabel('v (m/s)') title('Velocities of railroad cars of Problem 9.19') legend('v_1','v_2') The required function bumpern is % Function for Problem 9.19 function dy=bumpern(t,y) if y(1)<10 dy1=y(3); dy2=y(4); dy3=-1.67e4*y(1)+1.67e4*y(2);

418

Chapter 9

dy4=1.67e4*y(1)-1.67e4*y(2); else dy1=y(3); dy2=y(4); dy3=-1.67e4*y(1)+1.67e4*y(2)-4*y(3)^1.6-6.67e2*y(1); dy4=1.67e4*y(1)-1.67e4*y(2); end dy=[dy1;dy2;dy3;dy4]; The plots generated from running this simulation are

419

Chapter 9

Zooming in a part of the figure showing displacements reveals

420

Chapter 9

9.20 The state transition matrix derived in the solution of Problem 9.10 is

( ) (a) )67.3sin(595.0273.0)67.3cos()( 595.0595.0806.1 teteet ttt −−− −++=Φ 2321 CCCC where

)(b 2182.0605.3673.11208.01443.0673.1

0669.00797.09261.0

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−=1C

(c) 728.0605.3673.11208.014.1673.10669.00797.00739.0

2

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−−=C

(d) 67.18.13023.1

926.011.1023.10409.005.176.1

3

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−=C

The input vector determined in the solution of Problem 9.1 is

(e) 100

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=B

The input is given as [ ] [ ])20sin(5.0)( ttF ==u . The response is determined using Equation (9.54)

( ) [ ] (f) )20sin(5.0100

)](67.3sin[595.0273.0

)](67.3cos[

)()(

)(595.0

0

)(595.0)(806.1

0

τττ

τ

τττ

τ

ττ

dte

tee

dt

t

ttt

t

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−−

+−+=

−Φ=

−−

−−−−∫

∫

23

21

CC

CC

Buy

Substitution of Equations (b), (c), and (d) into Equation (f) leads to

(g) 5742.0

0148.00220.0

)20sin()](67.3sin[5.0

728.01208.00669.0

)20sin()](67.3cos[5.02182.01208.0669.0

)20sin(5.0

0

)(59.0

0

)(59.0

0

)(806.1

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−−+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−=

∫

∫∫

−−

−=−=

tt

tt

tt

dte

dtede

τττ

τττττ

τ

ττy

Trigonometric identities are used to rewrite Equation (g) as

421

Chapter 9

∫

∫

∫

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−+

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−=

−

−

−

tt

tt

tt

dette

dette

dee

0

59.059.0

0

59.059.0

0

80.180.1

)20sin()67.3sin()67.3sin(728.0

1208.00669.0

)67.3cos(5742.0

0148.00220.0

5.0

)20sin()67.3cos()67.3sin(5742.0

0148.00220.0

)67.3cos(728.0

1208.00669.0

5.0

2182.01208.0669.0

)20sin(5.0

τττ

τττ

ττ

τ

τ

τy

Integration leads to

422

Chapter 9

(j) 33.16)33.16cos(33.16)33.16sin(59.0

)67.3sin(5742.0

0148.00220.0

)67.3cos(728.01208.00669.0

1026.9

67.23)67.23cos(67.23

)67.23sin(59.0)67.3sin(5742.0

0148.00220.0

)67.3cos(728.0

1208.00669.0

1046.4

33.16)33.16cos(33.16)33.16sin(59.0

)67.3sin(5742.0

0148.00220.0

)67.3cos(728.0

1208.00669.0

1026.9

67.23)67.23cos(67.23)67.23sin(59.0

)67.3sin(5742.0

0148.00220.0

)67.3cos(728.0

1208.00669.0

1046.4

20)20cos(20)20sin(80.12182.01208.0669.0

1017.5

59.0

4

59.0

4

59.0

4

59.0

4

80.15

t

t

t

t

t

ett

ttx

et

tttx

ett

ttx

ett

ttx

ettx

−

−

−

−

−

−

−

−

−−

−+

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

+−+

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−+

+−

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−+

+−

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−+

+−⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−=y

Further simplification to Equation (j) can be made. 9.21 The state transition matrix derived in the solution of Problem 9.10 is

( ) (a) )67.3sin(595.0273.0)67.3cos()( 595.0595.0806.1 teteet ttt −−− −++=Φ 2321 CCCC where

(b) 2182.0605.3673.11208.01443.0673.1

0669.00797.09261.0

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−=1C

(c) 728.0605.3673.11208.014.1673.10669.00797.00739.0

2

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−−=C

423

Chapter 9

(d) 67.18.13023.1

926.011.1023.10409.005.176.1

3

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−=C

The input vector determined in the solution of Problem 9.1 is

(e) 100

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=B

The input is given as [ ] [ ]))3()((5.0)( −−== tututFu . The response is determined using Equation (9.54)

( ) [ ] (f) )3()((5.0100

)](67.3sin[595.0273.0

)](67.3cos[

)()(

)(595.0

0

)(595.0)(806.1

0

ττττ

τ

τττ

τ

ττ

duute

tee

dt

t

ttt

t

−−⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−−

+−+=

−Φ=

−−

−−−−∫

∫

23

21

CC

CC

Buy

Substitution of Equations (b), (c), and (d) into Equation (f) leads to

(g) 5742.0

0148.00220.0

)]3()()][(67.3sin[5.0

728.01208.00669.0

)]3()()][(67.3cos[5.0

2182.01208.0669.0

)]3()([5.0

0

)(59.0

0

)(59.0

0

)(806.1

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−−−−+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−−−=

∫

∫

∫

−−

−=

−=

tt

tt

tt

duute

duute

duue

ττττ

ττττ

τττ

τ

τ

τy

Note that

( ) )(h )()3()]3()()[(300∫∫∫ −−=−−ttt

dftudfduuf ττττττττ

424

Chapter 9

Use of Equation (h) to evaluate Equation (g) leads to

( )[ ]

[ ]

[ ]

[ ]

[ ](i) ))]3(67.3cos(67.3))3(67.3sin(59.0[

)67.3cos(67.3)67.3sin(59.0)[3(

67.3)67.3cos(67.3)67.3sin(59.05742.0

0148.00220.0

0362.0

))]3(67.3sin(67.3))3(67.3cos(59.0[ )67.3sin(67.3)67.3cos(59.0)[3(

59.0)67.3sin(67.3)67.3cos(59.0728.0

1208.00669.0

0362.0

)3(112182.01208.0669.0

2688.0

)3(59.0

59.0

59.0

)3(59.0

59.0

59.0

)3(860.1860.1

−+−−

−+−−

+−+−⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−+

−+−−

−+−−

+++−⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−+

−−−−⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−=

−−

−

−

−−

−

−

−−−

ttettetu

tte

ttettetu

tte

tuee

t

t

t

t

t

t

tty

9.22 The state transition matrix derived in the solution of Problem 9.11 is

(a) 391.0436.0546.0609.0

609.436.0546.0391.0

)( 42.658.15 tt eet −−⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡−

−=Φ

The input matrix derived in the solution of Problem 9.2 is

)(b 4005⎥⎦

⎤⎢⎣

⎡=B

The input vector for this problem is given as

(c) 5.02.0

)( 2.0 tet −⎥⎦

⎤⎢⎣

⎡=u

The convolution integral solution for the forced response is obtained using Equation (9.54) assuming initial conditions of zero

∫ −Φ=t

dtt0

)(d )()()( τττ Buy

Substitution of Equations (a)-(c) into Equation (d) leads to

(e) 218.1701.1

782.0701.0

21

391.0436.0546.0609.0

609.436.0546.0391.0

5.02.0

4005

391.0436.0546.0609.0

609.436.0546.0391.0

)(

0

22.642.6

0

38.1558.15

0

2.0)(42.6)(58.15

0

2.0)(42.6)(58.15

∫∫

∫

∫

⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡−=

⎥⎦

⎤⎢⎣

⎡

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡−

−=

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡−

−=

−−

−−−−−

−−−−−

tt

tt

ttt

ttt

deedee

deee

deeet

ττ

τ

τ

ττ

τττ

τττy

425

Chapter 9

Simplification of Equation (e) leads to

( ) ( )

(f) 196.0274.0

0508.00456.0

247.0229.0

196.0274.0

0508.0456.

)(

42.658.152.0

42.62.058.152.0

ttt

tttt

eee

eeeet

−−−

−−−−

⎥⎦

⎤⎢⎣

⎡−⎥

⎦

⎤⎢⎣

⎡−−⎥

⎦

⎤⎢⎣

⎡=

−⎥⎦

⎤⎢⎣

⎡+−⎥

⎦

⎤⎢⎣

⎡−=y

9.23 The state transition matrix obtained during the solution of Problem 9.12 is

( ) (a) 11675.00

833.0

833.0

⎥⎦

⎤⎢⎣

⎡

−=Φ

−

−

t

t

ee

The input matrix obtained during the solution of Problem (9.5) is

(b) 56.50067.16

⎥⎦

⎤⎢⎣

⎡−=B

The input for this problem is

(c) 0

)800sin(120⎥⎦

⎤⎢⎣

⎡=

tu

The convolution integral solution for the circuit response is

( )

( )

( ) (d) )800sin(11350

)800sin(2000

0)800sin(2000

11675.00

0)800sin(120

56.50067.16

11675.00

)()(

0)(833.0

)(833.0

0)(833.0

)(833.0

0)(833.0

)(833.0

0

∫

∫

∫

∫

⎥⎦

⎤⎢⎣

⎡

−−−

=

⎥⎦

⎤⎢⎣

⎡−⎥⎦

⎤⎢⎣

⎡

−=

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−⎥⎦

⎤⎢⎣

⎡

−=

−Φ=

−−

−−

−−

−−

−−

−−

t

t

t

t

t

t

t

t

t

t

de

e

de

e

de

e

dt

ττ

τ

ττ

ττ

τττ

τ

τ

τ

τ

τ

τ

Buy

Evaluation of the integrals in Equation (d) leads to

(f) 6875.16875.1)800sin(00176.0)(

(e) 5.2)800cos(5.2)800sin(0260.0)(833.0

2

833.01

t

t

etty

ettty−

−

−+−=

−+−=

The output relationship obtained from the solution of Problem 9.5 is

[ ] [ ] [ ] (g) 10102

1

2

1⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡=

vv

yy

vB

Thus (h) 6875.16875.1)800sin(00176.0)( 833.0 t

B ettv −−+−=

426

Chapter 9

9.24 The differential equations modeling the system of Figure 9.24 are

(c) 0525.0125.05.2

(b) 0125.0325.02.02.1

(a) 2.02.05.1

323

3212

211

=+−

=−+−

=−+

hhdtdh

hhhdt

dh

qhhdtdh

i

Defining the state variables as

)(d 33

22

11

hyhyhy

===

The state equations become

(e) )(00667.0

210.050.001042.2708.167.

01333.1333.

3

2

1

3

2

1

tqhhh

hhh

i

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

&&&

The state-transition matrix is )(f )(L 1 1AI −− −=Φ s

where

)(g 210.050.00

1042.2708.167.01333.1333.

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+−−+−

−+=−

ss

ss AI

Inversion of Equation (g) leads to )(h )( 0289.0196.0391.0 ttt eeet −−− ++=Φ 321 CCC

where

(i) 125.0183.0118.0380.0661.0428.0

196.0341.0238.0

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

−=1C

(j) 825.00956.0256.0201.00271.00719.0425.00573.0124.0

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−−=2C

(k) 0498.00860.0138.0179.0312.0500.0229.0399.0637.0

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=3C

427

Chapter 9

Noting that the system response is )(2.0)( tutqi =

[ ] (l) ]1[955.0460.3408.4

]1[0261.00734.0127.0

10604.0

229.0122.0

138.0500.0637.0

256.00719.0124.0

118.0428.0

238.02.0

))(2.0(001

)(

0289.0196.0391.0

0

)(0289.0)(196.0)(391.0

0

ttt

tttt

t

eee

deee

dut

−−−

−−−−−−

−⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡+−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−+−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−Φ=

∫

∫

τ

τττ

τττ

y

9.25 The mathematical model for the system is

)(a )200sin(1010420)200sin(10010420010

4

5

txxxxtxxxx

=++

=++

&&&

&&&

Define state-space variables as

(b) 2

1

xyxy&=

=

The state-space formulation for the system is

)(c )(10

2010410

2

14

2

1 tFyy

xyy

⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−−

=⎥⎦

⎤⎢⎣

⎡&&

The state transition matrix is calculated from

( ) (e) 104

12010420

1

(d) 201041

4421

4

⎥⎦

⎤⎢⎣

⎡−+

++=−

⎥⎦

⎤⎢⎣

⎡+−

=−

−

sxs

xsss

sxs

s

AI

AI

Inversion of Equation (e) leads to

(f) )75.199sin(0501.0)75.199cos()75.199sin(25.200

)75.199sin(0050.0)75.199sin(0501.0)75.199cos(101010

101010

⎥⎦

⎤⎢⎣

⎡

−−+

=Φ

−−−

−−−

tetetetetete

ttt

ttt

The system response is determined from

∫

∫

⎥⎦

⎤⎢⎣

⎡

−−−−

=

−Φ=

−−

−−t

t

t

t

dtte

te

dt

0)(10

)(10

0

(g) )200sin(100)(75.199sin[0501.0)](75.199cos[

)](75.199sin[0050.0

)()(

ττττ

τ

τττ

τ

τ

Buy

The system response is

428

Chapter 9

[ ]

(h) )75.199sin()200sin()75.199cos(

)75.199cos()200sin()75.199sin(5.0

)]75.199sin()75.199cos()75.199cos()75.199sin()200sin(5.0

)200sin(100)](75.199sin[0050.0)(

0

10

0

1010

0

1010

0

)(10

⎭⎬⎫

−

⎩⎨⎧

=

−=

−=

∫

∫

∫

∫

−

−

−−

t

tt

tt

tt

det

dete

dttee

dtetx

τττ

τττ

ττττ

τττ

τ

τ

τ

τ

Evaluation of integrals in Equation (h) leads to

[ ]

)(i ]10)75.399cos(10)75.399sin(75.399)[75.199cos(1056.1 ]25.0)25.0cos(10)25.0sin(25.0)[75.199cos(0025.0

75.399)75.399cos()75.399()75.399sin(10)75.199sin(1056.1 ]25.)25.0cos(25.0)25.0sin(10)[75.199sin(0025.0)(

106

10

106

10

t

t

t

t

etttxettt

etttxettttx

−−

−

−−

−

−+−

−++

+−+

+−=

9.26 If the mass of the system varies with time, the appropriate form of Newton’s law is

∑ = (a) )( xmdtdF &

Application of Equation (a) to the particle leads to )(b )()()( tFkxxcmxtm =+++ &&&&

where

)(d (c) 10)(

1.0

1.0

t

t

emetm

−

−

−=

=

&

Defining state-space variables of

(e) 2

1

xyxy&=

=

the state-space formulation for the system is

(f) )100sin(10

0201.010410

1.02

11.01.04

2

1⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−−

=⎥⎦

⎤⎢⎣

⎡tey

yeexy

yttt&

&

A MATLAB M file which numerically obtains the solution of Equation (f) is % Problem 9.26 % Initial conditions clear y0=[0 0]; tf=1; % Nuerical integration using ode45 [t,y]=ode45(@xxy,[0 tf],y0); plot(t,y(:,1)) xlabel('t') ylabel('x') title('Response of mechanical system losing mass')

429

Chapter 9

Execution of the file requires a user defined program of derivatives such as % Functions defined for Problem 9.26 function dy=xxy(t,y) dy1=y(2); dy2=-4e4*exp(0.1*t)*y(1)+(0.1-20*exp(0.1*t))*y(2)... +10*exp(0.1*t)*sin(100*t); dy=[dy1;dy2]; The resulting plot is

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-4

-3

-2

-1

0

1

2

3

4

5x 10

-4

t

x

Response of mechanical system losing mass

9.27 The state-space formulation for the system is

(a) 2064010

2

1

2

1⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−−

=⎥⎦

⎤⎢⎣

⎡zz

zz&&

The problem is modeled as that of the free response due to initial condition of

(c) 0)0((b) 0128.0)0(

2

1

==

zz

A MATLAB program using ode45 to numerically integrate the differential equation follows % Problem 9.27 % Initial conditions clear z0=[0.0128 0]; tf=1; % Nuerical integration using ode45 [t,z]=ode45(@zxy,[0 tf],z0); plot(t,z(:,1)) xlabel('t') ylabel('x') title('Response of simplified syspension system')

430

Chapter 9

The required file zxy.m which supplies the vector of derivatives for use in ode45 is % Functions defined for Problem 9.27 function dy=zxy(t,y) dy1=y(2); dy2=-640*y(1)-20*y(2); dy=[dy1;dy2]; The resulting plot of the system response is

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-4

-2

0

2

4

6

8

10

12

14x 10-3

t

x

Response of simplified syspension system

9.28 The state-space formulation of the nonlinear pendulum problem is

(d) )sin(100(c) (b)

(a)

12

21

2

1

yyyy

y

y

−===

=

&&

&θ

θ

The initial conditions are of the form

(f) 0)0((e) )0(

2

01

==

yy θ

The MATLAB program to numerically determine the response of the nonlinear pendulum is % Problem 9.28 % Initial conditions clear disp('Please input initial angular displacement in degrees') theta0=input('>>') theta1=theta0*180/pi;

431

Chapter 9

z0=[theta1 0]; tf=4; % Nuerical integration using ode45 [t,z]=ode45(@theta,[0 tf],z0); plot(t,z(:,1)) xlabel('t') ylabel('\theta') str=['Response of nonlinear pendulum with \theta_0='... ,num2str(theta0),' degrees'] title(str) figure plot(z(:,1),z(:,2)) xlabel('\theta') ylabel('\omega') str=['State plane for nonlinear pendulum with \theta_0='... ,num2str(theta0),' degrees']; title(str) The function file supplying the required derivatives is % Functions defined for Problem 9.28 function dy=theta(t,y) dy1=y(2); dy2=-100*sin(y(1)); dy=[dy1;dy2];

0 0.5 1 1.5 2 2.5 3 3.5 4570.5

571

571.5

572

572.5

573

t

θ

Response of nonlinear pendulum with θ0=10 degrees

432

Chapter 9

570.5 571 571.5 572 572.5 573-15

-10

-5

0

5

10

15

θ

ω

State plane for nonlinear pendulum with θ0=10 degrees

0 0.5 1 1.5 2 2.5 3 3.5 42573.5

2574

2574.5

2575

2575.5

2576

2576.5

2577

2577.5

2578

2578.5

t

θ

Response of nonlinear pendulum with θ0=45 degrees

433

Chapter 9

2573.5 2574 2574.5 2575 2575.5 2576 2576.5 2577 2577.5 2578 2578.5-20

-15

-10

-5

0

5

10

15

20

θ

ω

State plane for nonlinear pendulum with θ0=45 degrees

It is clear from running the file that the period decreases as the initial angle increases. 9.29 A MATLAB program which uses the state-space formulation to model the CSTR of Example 4.17 is % Problem 9.29 % CSTR of Example 4.17 % Numerical simulation clear y0=[0 0 0] tf=44 [t,y]=ode45(@lin,[0 tf],y0) plot(t,y(:,1)) figure plot(t,y(:,2)) figure plot(t,y(:,3)) % Function for Problem 9.29 function dy=lin(t,y) V=1e-3 q=1e-6; alpha=1e-3; lambda=1e-3; R=480; Ts=25; Q=1; rho=760;

434

Chapter 9

cp=.06; E=5; B=exp(-E/(R*Ts)); cai=1e-4; dy1=-1/V*(q*y(2)+alpha*V*B*(y(1)+E/R/Ts^2*y(3)*cai)); dy2=-1/V*(-alpha*V*B*(y(1)+E/R/Ts^2*cai*y(3))+q*y(2)); dy3=-1/(rho*V*cp)*(rho*q*cp*y(3)-lambda*alpha*V*B*(y(1)+E/R/Ts^2*y(3)*cai))+Q; dy=[dy1;dy2;dy3]; 9.30 The state matrix and the input vector obtained in the solution of Problem 9.1 are

(b) 100

(a) 31625

100010

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−=

B

A

Noting that the state variables are defined such that xy =1 the transfer function

)()()(

sFsXsG = is obtained from

(c) )( 1

AI −=

sV

sG

where

(e) 25163 1625)3(s

3162510

01(d) 1

316110

010

23

2

+++=

+++=

+−

−=−

=

+−

−=

sssss

ss

ss

ss

AI

V1

Thus

(f) 25163

1)( 23 +++=

ssssG

435

Chapter 9

9.31 The state-space formulation obtained during the solution of Example 9.8 is

(a) )()(

10

010000

10000100

2

1

2

1

4

3

2

1

2

2

1

2

2

2

1

2

2

2

1

21

2

2

1

21

4

3

2

1

⎥⎦

⎤⎢⎣

⎡

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

+

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

+−

+−=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

tFtF

m

myyyy

mc

mc

mk

mk

mc

mcc

mk

mkk

yyyy

&&&&

The transfer function )()()(

1

11,1 sF

sYsG = is determined as

(b) )( ,1,1 AI

V 11

−=

ssG

where

(c) 1

10010

1

0

1100

0100

2

2

2

2

1

2

2

2

2

2

21

2

2

2

2

2

2

2

2

1

21

2

2

1,

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛++=

+−

−−

=

+−

−+

+−

−−

=

mcss

mk

m

mcs

mc

mks

m

mcs

mc

mk

mc

mccs

mk

m

s

11V

436

Chapter 9

(d) )(

1010

100010

2

2

1

21

21

22

21

212

21

22

21

22

1

21

2

2

21

22

2

2

1

21

2

2

2

2

1

2

2

2

2

2

1

21

2

2

1

2

2

2

2

2

1

21

2

2

2

2

1

2

2

2

1

2

2

2

1

21

2

2

1

21

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛ +−+

+−−

⎥⎦

⎤⎢⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛ +++−⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛ ++=

+−

−−+

−

−

+−

−+

+−

−

=

+−−

−+

+−+

−−

=−

mc

ssm

kkmm

kmm

kkks

mmck

mmc

sm

ccs

mk

mmck

mc

sm

ccsss

mc

smk

mk

mc

mk

mkk

s

mc

smc

mk

mc

mcc

smks

s

mc

smc

mk

mk

mc

mcc

smk

mkk

ss

s AI

The transfer function )()()(

1

21,2 sF

sYsG = is obtained from

(e) )( ,21,2 AI

V 1

−=

ssG

where AI −s is given in Equation (d) and

(f) 1

10001

1

0

1 1000

010

2

2

1

2

1

2

2

1

2

1

21

2

2

1

2

1

2

2

2

1

21

11

21,

⎥⎦

⎤⎢⎣

⎡−−−=

+−−

−−

−

+−−

−+

++

−−

=

mcs

mk

m

mcs

mc

mk

s

m

mcs

mc

mk

mc

mccs

mmkk

s

12V

437

Chapter 9

9.32 The transfer function of a second-order system is

(a) 256

32)( 2 +++

=ss

ssG

A state matrix for a state-space formulation of the system is

(b) 625

10⎥⎦

⎤⎢⎣

⎡−−

=A

The input vector is obtained by setting [ Tbb 21=B ]

(c) 3225

3225

32

12

2

1

+=−

+=

+=

sbsb

sbbs

s2V

Equation (c) must be valid for all s, thus

(e) 2

(d) 12.0253

2

1

=

−=−=

b

b

Thus the state-space formulation is

(f) )(212.0

62510

2

1

2

1 tfyy

yy

⎥⎦

⎤⎢⎣

⎡−+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−−

=⎥⎦

⎤⎢⎣

⎡&&

9.33 The transfer function of a fourth-order system is

(a) 5020103

523)( 234

2

++++++

=ssss

sssG

A state matrix that can be used in a state-space model for the system is

(b)

3102050100001000010

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−

=A

The input vector is obtained by requiring [ Tbbbb 4321 =B ]

(c) 523

10205000

1001

523

2

4

3

2

1

2

++=−

−

++=

ss

bbsbsbs

ss4V

438

Chapter 9

The determinant on the left-hand side of Equation (c) is evaluated by expanding about its first column

( ) ( )(d) 52350)5020()501020(

52350102020

5230

101

501020

01

2323

2132

34

32

213323

24

2

3

2

1

4

3

2

++=−−−+−−−+

++=++−−−−

++=−−

−−

ssbsbbsbbbsb

sssbsbbsbsbbsbs

ssbsbsb

bbsbs

s

Equating coefficients of like powers of s in Equation (d) leads to

(h) 0b(g) 04.)1.(103503501020(f) 0)1.(2025025020(e) 1.0550

4

11132

2223

33

=−=⇒−+=−⇒=−−−

=⇒−+=−⇒=−−−=⇒=−

bbbbbbbbb

bb

Thus a state-space formulation of the equations governing the response of the system is

(i) )(

00.100.04

3102050100001000010

4

3

2

1

4

3

2

1

tf

yyyy

yyyy

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−

+

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

&&&&

9.34 The closed-loop transfer function for the system is

(a) 3216

352

)1)(32()164()1)(32(

1641321

164132

)(

23

2

2

2

2

+++++

=

+++++++

=

⎟⎠⎞

⎜⎝⎛

+++

⎟⎠⎞

⎜⎝⎛ ++

⎟⎠⎞

⎜⎝⎛

+++

⎟⎠⎞

⎜⎝⎛ +

=

sssss

sssssss

sss

s

sss

ssH

The state matrix for the state-space formulation of this closed loop system is

(b) 6213

100010

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−=A

439

Chapter 9

The input vector is obtained by requiring [ Tbbb 321 =B ]

(c) 3522133

352213

01

352

2212

23

2

3

2

1

2

++=−−−

++=−

++=

sssbsbbsb

ssbbsbs

ss3V

Equation (c) must hold for all s, thus coefficients of like powers of s must be the same on both sides of the equation,

(f) 2(e) 3/161635213

)(d 133

3

1121

22

==⇒−=−⇒=−−

−=⇒=−

bbbbb

bb

Thus the state-space formulation for this system is

2s+3

sTransfer Fcn1

s+1

s +4s+162

Transfer Fcn

Y

To WorkspaceStep

)(h )(21

3/26

6213100010

3

2

1

3

2

1

tayyy

yyy

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

&&&&&&

SIMULINK models are constructed using both the transfer function formulation as well as the state-space formulation. The SIMULINK transfer function model is shown below

440

Chapter 9

A customized plot of the system response is

0 5 10 15 20 250

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

t

x

SIMULINK simulation of Problem 9.34 using transfer function model

When developing the state-space model note that since the input vector is determined using 3V the transfer function corresponds to that of and thus the appropriate output vector is

3y]100[ =C . The SIMULINK model developed using the state-space formulation

is

X

To WorkspaceStep

x' = Ax+Bu y = Cx+Du

A=[0 1 0;0 0 1;-3 -21 -6]B=[16/3;-1;2]

C=[0 0 1]D=[0]

441

Chapter 9

A customized plot of the output is

0 5 10 15 20 250

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

t

x

SIMULINK simulation of PRoblem 9.34 using state-space formulation

9.35 From the block diagram shown

( ) (b) )()()(21

2)(

(a) )(1

3)(

⎥⎦⎤

⎢⎣⎡ +−⎟⎠⎞

⎜⎝⎛

+=

+=

sCsBsAss

sB

sBs

sC

Equation (b) is used to obtain

(c) )()(24

2)( 2 ⎥⎦⎤

⎢⎣⎡ +

++= sCsA

sssssB

442

Chapter 9

Substitution of Equation (c) into Equation (a) leads to

( )

(d) )(42

12)(

)(2)(6

644

)()(2)(6

)1(4

)()(24

2)()(3

1

23

223

2

2

sAsss

sC

sAs

sCs

ssssss

sCsAs

sCs

sss

sCsAsss

ssBsCs

+−+=

=−+++++

+=+++

⎥⎦⎤

⎢⎣⎡ +

++==⎟

⎠⎞

⎜⎝⎛ +

Thus the transfer function for the system is

(e) 42

12

)()()(

23 +−+=

=

sss

sAsCsH

The state matrix for a state-space formulation of the closed loop is

(f) 214

100010

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−=A

The input vector is obtained from [ Tbbb 321 =B ]

(g) 124)4(

1244

1214

01

12

2212

3

2122

3

3

2

1

=+−+

=−++

=−

−

=

bsbbsb

sbsbbsb

bbsbs

3V

Equation (g) requires

(j) 0

(i) 43

404

(h) 3124

3

2121

22

=

==⇒=−

=⇒=

b

bbbb

bb

Thus the state-space formulation for the closed-loop system is

(f) )(0

0.753

214

100010

3

2

1

3

2

1

tayyy

yyy

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

&&&

443

Chapter 9

9.36 The Simulink model for the system is A simulation of the model is run with a step input. The response is illustrated below. Note that since the sign on the summing junction in one of the feedback loops is positive the system is unstable.

9.37 The state-space model for the system of Figure P9.24 is obtained in the solution of Problem 9.24 as

(a) )(00667.0

210.050.001042.2708.167.

01333.1333.

3

2

1

3

2

1

tqhhh

hhh

i

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

&&&

444

Chapter 9

H

x' = Ax+Bu y = Cx+Du

A=[-0.133 0.133 0;0.167 -0.2708 0.1042;0 0.05 -0.210B=[0.667;0;0]

C=[1 1 1]D=[0;0;0]

0.2u(t)

The SIMULINK model using the state-space formulation is illustrated below A customized graph drawn using the simulation data follows

0 1 2 3 4 5 6 7 8 9 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

t (s)

h (m

)

SIMULINK simulation of three-tank liquid level problem

h1(t)

h2(t)

h3(t)

9.38 The differential equations obtained during the solution of Example 6.41 are

(b) )(1005.11035.11099.24.04x10

(a) 01099.21048.11073.5

5434

343

tuxxxdt

d

xxdt

dx

wow

woo

=+−

=−+

θθθ

θθθ

Equations (a) and (b) are rewritten as

(d) )(60.23342.00740.0d

(c) 522.058.2

0w

00

tudt

dtd

w

w

+−=

+−=

θθθ

θθθ

445

Chapter 9

Define state variables as

(f) (e)

2

1

w

o

yy

θθ

==

The state-space formulation of the problem is

(g) )(60.20

3342.00740.0522.058.2

2

1 tuyy

⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡−

−=⎥

⎦

⎤⎢⎣

⎡&&

The steps in obtaining the state transition matrix are

( ) (h) 58.20740.0

522.03342.0924.091.2

1

3342.00740.0522.058.2

21

⎥⎦

⎤⎢⎣

⎡+

+

++=−

⎥⎦

⎤⎢⎣

⎡+−−+

=−

−

ss

sss

ss

s

AI

AI

The state transition matrix is obtained by inverting the transforms of Equation (h)

(i) 9947.00325.02295.00072.0

0053.00325.02295.09928.0 318.059.2 tt ee −−

⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡−

−=Φ

The response is obtained from

(j) )(586.2

5967.00138.05967.0

)(60.20

9947.00325.02295.00072.0

0053.00325.02295.09928.0

)()(

0

)(318.0)(59.2

0

)(318.0)(59.2

0

∫

∫

∫

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡−=

⎥⎦

⎤⎢⎣

⎡

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡−

−=

−Φ=

−−−−

−−−−

ttt

ttt

t

duee

duee

dt

ττ

ττ

τττ

ττ

ττ

Buy

Evaluation of the integrals of Equation (j) leads to

(l) )1(13.8)1(0053.0)(

(k) )1(867.1)1(2304.0)(318.059.2

318.059.2

ttw

tto

eet

eet−−

−−

−+−=

−+−−=

θ

θ

446