Chapter 3-3.doc

8/13/2019 Chapter 3-3.doc

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3.7 Response of higher order systems

The primary difference between first-order and second-order

linear models is that the characteristic roots of second-order

models can be complex numbers. This leads to types of behavior

not seen in first-order models and sometimes requires modified

algorithms in order to handle quantities with real and imaginary

parts. The second-order case contains all of the possible types of

behavior that can occur in linear systems. This is because a

characteristic equation with real coefficients must have roots

that fall into one of the three categories.

1. Real and distinct.2. Distinct but complex conugate pairs.

!. Repeated.

"nce we understand the behavior generated by each of the three

root cases# the superposition principle will allow us to apply the

results to a linear system of any order. $or example# the

response of a sixth-order system with a complex conugate pair

of roots# two real repeated roots# and two real distinct roots canbe analy%ed by combination of each response pattern will give

the total system response.

$or example# consider the third-order model

&'12

2

2!

!

! tfybdt

dyb

dt

ydb

dt

ydb o =+++ '!.(-1&

)ts characteristic equation is

*12

2

!

! =+++ obsbsbsb '!.(-2&

)f the three rootss1, s2# ands3are real and distinct# the free

response is an extension of the form given by '!.!-+

tststs eAeAeAty !21 !21&' ++= '!.(-!&

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)fs1ands2are complex conugates buts3is real# the free

response is an extension of '!.!-2+ namely#tsat eAbtBety !!&sin'&' ++=

'!.(-,&

wheres1= -a + ib# ands2= -a -ib. $inally# if all three roots are

repeated# then the extension of '!.!-!,& gives

tststs eAeAeAty !21 !21&' ++= '!.(-&

wheres1= s2= s3. These cover all the cases for the third-order

model. The forced response can be obtained using the trial

solution method using equation '!.-1 where the free-response

form is given by one of the preceding solutions# and the

particular solution form is obtained as outlined in /ection !..

The only situation requiring further algebraic manipulation is

the case# where any of the repeated roots are complex. This can

occur in only fourth-order models or higher. The free response

is an extension of the form given by '!.!-!, where there are

now two pairs of repeated roots#s1=s2= -a +ibands3=s4= -a

-ib. The free response is

tstststs etAeAetAeAty !!11 ,!21&' +++= '!.(-&

The algebra required to converty(t)into a useful form

containing sine functions is similar to the procedures used

following '!.!-1&.

The trial solution method can be used to solve a linear

differential equation of any order. 0owever# the algebra required

to solve for the solutions constants becomes tedious as the

systems order increases. $or example# the step response of a

third-order model requires four constants to be found# one from

the differential equation and three from the initial conditions

y'* dy/dt'* and d2y/dt2'*&. Two alternatives to trial solution

method are '1& the 3aplace transform method and '2&

simulation.

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3.8 The Routh-Hurwitz stability criterion

4erhaps the most important question to be answered by a

dynamic model concerns the stability of the system under study.

)f the proposed system is predicted to be unstable# the designer

immediately see5s to modify the system to ma5e it stable. This

will be especially true in the design of feedbac5 control systems.

Recall that the stability properties of a linear system are

determined by the roots of its characteristic equation. )f any root

lies in the right half of the complex plane# the system is

unstable. The Routh-0urwit% criterion tells us how many roots

lie in the right-half plane. )t sometimes gives enough

information to enable us to determine the location of all roots.

The second-order case

/ince the quadratic formula is relatively simple# it is not difficult

to use it to show that characteristic equation

*12

2 =++ obsbsb '!.+-1&

represents a stable system if and only if b2, b1, boall have thesame sign. )f bo6* and b2has the same sign as b1# the system is

neutrally stable# since one root iss6* and the other is negative.

)f b16* and bo/b27*# the system is again neutrally stable with

two purely imaginary roots. '"f course# if b26*# the system is no

longer second order.&

The third-order case

Third-order models occur quite frequently# and the Routh-0urwit% results are simply stated. 8onsider the equation

*12

2

!

! =+++ obsbsbsb '!.+-2&

$or simplicity# assume that we have normali%ed the equation so

that b37*. The system is stable if and only if all of the following

conditions are satisfied

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!21

12 **#*

bbbb

bbb

o

o

>

>>>

'!.+-!&

The neutrally stable cases occur if '!.+-!& is satisfied except for

the following1. bo6* ' a root exists ats6*&

2. b2=bo6* 'roots are 1#* bis = &

!. b1b2-bob36* 'roots are !21 9# bbbs =

Example 3.13

: certain system has the characteristic equation

*2.; 2!

=+++ Ksss '!.+-,&

$ind the range ofK values for which the system will be stable.

/olution

$rom '!.+-!& it is necessary and sufficient that

K7*# and ;'2& -K7*

Thus# the system is stable if and only if *


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$rom '!.+-! this occurs when

K-2,7*

!'2& - 'K-2,&7*

"r

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Problems

!.1 : tan5 has the model 1!"dt

d"A += # withA6 ! ft2and6

(** sec9ft2. /uppose that the height h is initially ft when the

inflow rate !1is suddenly turned "? at a constant value of *.*ft!9sec. $ind "(t).

!.2 8onsider the model

&'1*2 t#ydt

dy=+

'1& )fy($)=* and #(t)=1 'a unit step what is the steady-

state value of the outputyss@ 0ow long does it ta5e

before ;+A of the difference betweeny($)andyssis

eliminated@

'2& Repeat 'a& withy($)6 and #(t)=1.

'!& Repeat 'a& withy($)and #(t)62*.

!.! $ind the free response and the step response of the

following models. )f the solution is oscillatory# it should be

expressed in the form of a sine function with a phase shift.

'1& 1&*'#*&*'#2+,22

===++dt

dxxx

dt

dx

dt

xd

'2& 1&*'#*&*'#212+22

===++dt

dxxx

dt

dx

dt

xd

'!& 1&*'#*&*'#2,,22

===++dt

dxxx

dt

dx

dt

xd

!., 4ut the models in 4roblem !.! into state-variable form.

!. 8onvert the following models into reduced form in terms

of the variablex1. The input isf.

'1&fxx

dt

dx

xdt

dx

+=

=

212

2

21

-!

'2&

fxxdt

dx

xxdt

dx

+=

+=

212

2

211

-!

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!. Determine ## % and d for the dominant root in each of

the following sets of characteristic roots.

'1& s6 -2# -!i

'2& s6-!# -22i

!.( $ind ## % and d for the models given in 4roblem !.!.

!.+ Biven the model

*-2&2'2

2

=+++ &dt

dx&

dt

xd

'a& $ind the values of the parameter r for which the system is

'1& /table'2& ?eutrally stable

'!& Cnstable

'b& $or the stable case# for what values of r is the system

'1& Cnderdamped@

'2& "verdamped@

!.; : certain system is described by the model

fxdtdx'

dtxd =++ ,2

2

/et the value of the damping constant c so that both of the

following specifications are satisfied. Bive priority to the

overshoot specification. )f both cannot be satisfied# state the

reason.

1. aximum percent overshoot as small as possible and no

greater than 2*A.2. 1**A rise time as small as possible and no greater than !.

!.1* $ind the forced response of the following models

'1& teydt

dy

=+2

'2& 22 tydt

dy=+

!.11 $ind the steady-state part of the ramp response of thefollowing model# wheref(t)6t. :t steady-state# what is the

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difference betweenf(t)andx(t)@ :pproximately how long will

it ta5e to reach steady-state@

fxdt

dx

dt

xd=++

+2

2

!.11 : certain system has the characteristic equation

s(s+1) + K =$

>e desire that all the roots lie to the left of the lines= -bto

guarantee a time constant no longer than 61/b. Cse the Routh-

0urwit% criterion to determine the value ofKand so that bothroots meet this requirement.

!.12 Determine the range ofKvalues that will give a time

constant of less than 192 for the following characteristic

equation

s3+ 1$s2+ 31s + K =$

!.1! : mercury-in-glass thermometer has a time constant of

1*s. )f it is suddenly ta5en from being at 2* o8 and plunged into

hot water at +* o8# what will be the temperature indicated by the

thermometer after 'a& 1*s# 'b& 2*s@

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