1

Chapter 2

• We need to write differential equations representing the system or subsystem.

• Then write the Laplace transform of the system.• Then we will write the overall input-output relationship of the

interconnected components. That is the Transfer Function T(s), also called G(s).

• Section 2.2 in the text includes tables, in particular Table 2.2, p. 44, of various variables (elements) used in modeling physical systems and differential equations describing ideal elements of physical systems.

• It is expected that the student is already familiar with these concepts. In general the student will have access to this information for exams if needed.

• We will investigate making linear approximations of non-linear elements.

• We will briefly mention the Laplace Transform• Get our first exposure to poles and zeros and to the Characteristic

Equation• Discuss Partial Fraction Expansion• Introduce Final and Initial Value Theorems

Material We Will Cover Next

2

Control Systems are Often More Complex Than Single Loops

• One forward path

• Four transfer functions in the forward path G1, G2, G3, G4.

• Three Feedback Loops.

• Three Summing Junctions

3

Block Diagram = Flow GraphBesides Block diagrams, control systems are

often represented using Flow Graphs

R(s)C(s)G1 G21 1

-H1

G31 G4

H3

-H2

Block Diagram

Flow Graph

Note that two of the H transfer functions in the Flow Graph have negative signs. This is necessary since the summing nodes do not have any signs associated with them as in the block diagram.

4

Solve For The Transfer Function

Solution:C(s) = G1E (1)F = C(s)H (2)E = R(s)-F (3)E = R(s)- C(s)H (4)C(s) = G1[R(s)- C(s)H] (5)

C(s) +C(s)HG1 = G1(R(s)C(s)[1 + HG1] = G1(R(s)

C(s)/R(s)=G1/(1+G1H)

G1

H

_

+R(S) E C(s)

F

5

Solve For The Transfer Function

Solution:C(s) = A + B (1)A = G1E (2)B = G2E (3)F = HB (4)E = R(s)-F (5)

G1

G2

H

++

_

+R(S)

A

E C(s)B

F

Substitute (2) and (3) into (1); C(s) = G E G E (G G )E (6)Substitute (5) into (6); C(s) = (G G )[R(s) F] (7) Substitute (3) into (4); F = HG E (8)Substitute (5) into (8); F = HG [R(s) F] (9)

Solve (9) for F; F = HG R(s)1 G H

(10)

Substitute (10) into (7); C(s) = (G +G )[R(s)HG R(s)1 G H

] (11)

C(s) = (G G

1 2 1 2

1 2

2

2

2

2

1 22

2

1 2

)1 G H HG

1 G HR(s)

T(s) = C(s)R(s)

(G G )1 G H

2 2

2

1 2

2

6

Attempt to Solve For The Transfer Function

A B

C

D

E F

K

JM

7

Overall Transfer Function

(1) )(

kkM

sT

Mason’s Gain RuleThe overall Transfer function can be obtained from the formula:

• Mk is the transmittance of each forward path between a source and a sink node

• (2) 1 321 LLL

• ∑L1 is the sum of the transmittances of each closed path.

• ∑L2 is the sum of the product of the transmittances of all possible combinations of two non-touching loops.

• ∑L3 is the sum of the product of the transmittances of all possible combinations of three non-touching loops.

• ∆k is the cofactor of Mk. It is the determinant of the remaining sub-graph when the forward path which produces Mk is removed. Thus it does not include any loops which touch the forward path in question. It is equal to unity when the forward path touches all the loops in the graph of when the graph contains no loops.

8

Applying Mason’s Gain Rule

)(

kkM

sT

R(s)C(s)G1 G21 1

-H1

G31 G4

H3

-H2

214321243332121

432111

21432124333212121

2143212

2433321211

22143211

1

MT(s)

1-1

more.or or M no is There 1

HHGGGGHGGHGGHGG

GGGG

HHGGGGHGGHGGHGGLL

HHGGGGL

HGGHGGHGGL

GGGGM

9

Applying Mason’s Gain Rule

R(s) C(s)G1 G2 G3 G7

G6 G8G5G4

G9-H2

-H1

-H3

-H5-H4

M G G G G G G G G

M G G G G H G H G G G G H G G G H G H G H

G H G H G G G G H G G G H G H G H

G H G H G G G G H G G G

1 1 2 3 4 5 6 7 8

2 1 9 8 2 4 1 6 4 3 4 5 6 3 4 5 6 2 4 1 6 4

4 1 6 4 3 4 5 6 3 4 5 6 2 4 1 6 4

4 1 6 4 3 4 5 6 3 4 5

1

1

1

1

=

1

[ ]

[ 6 2 8 5

4 1 6 4 4 1 8 5 6 4 8 5 8 5 4 5 6 2 8 5 3 4 5 6 3

1 6 4 8 5

4 1 6 4 3 4 5 6 3 4 5 6 2 8 5

4 1 6 4 4 1 8 5 6 4 8 5 8 5 4 5 6

1

H G H

G H G H G H G H G H G H G H G G G H G H G G G G H

H G H G H

G H G H G G G G H G G G H G H

G H G H G H G H G H G H G H G G G

]

+ G

+

4

H G H G G G G H

H G H G H

T sC s

R s

M M G G G G G G G G G G G G H G H G G G G H G G G H G H G H

2 8 5 3 4 5 6 3

1 6 4 8 5

1 2 2 1 2 3 4 5 6 7 8 1 9 8 4 1 6 4 3 4 5 6 3 4 5 6 2 4 1 6 4

+ G4

1( )( )

( )

[1 ]

10

Using Mason’s Gain RuleSolve for the Transfer Function – C/R

1 1

-1

G4

-1-1

G2G3G1

1

1A CBR

11

Applying Mason’s Gain Rule1 1

-1

G4

-1-1

G2G3G1

1

1A CBR

M

M

M

M

M

a)

4 1

2

3 3 4 3 1

4 1

5

1 1 2 3

2 1

4

5 3 4 1

1 2 1 2 3 4 1 2 1 2 1 2 3 4 1 2

1 2 3 4 1 3 4 1 1 1 3 4

1

1

1

1

1 1

1

1 1

1 1 1

1

G G G G

G

G G G

G

G G G

G G G G G G G G G G G G G G G G

C

M

G G G G G G G G G G G G

G G

( )

( ) ( )( )

( 2 1 2 3 4 1 2

3 4 1 2 3 4

1 2 1 2 3 4 1 2

1

1

G G G G G G

G G G G G G

G G G G G G G G

)

12

Derive Transfer Function for Previous Control System Example

)22)(22(

12

84

12

R(s)

C(s))(

then,, 3select weSuppose

2)24(

4

5.024

41

244

)(2)(11

)(2)(1

R(s)

C(s))(

Rule sMason' Applying

2

2

2

2

jsjssssT

K

Kss

K

ssK

ssK

HsGsG

sGsGsT

a

a

a

a

a

0.5H(s) 64s

KG2(s)

1)(1

2b

sas

KsG a

H = 0.5

-

+R(s) Error

Feedback

ManipulatedVariable

C(s)aKsG )(1

24s

4)(2

2

ssG

Note: The denominator, set equal to zero, is called the Characteristic Equation

13

Spring Mass System

f Ma f t My t by t Ky tF s Y s Ms Y s bs Y s K Y s Ms bs K

T s Y s

F s bs K

Differential Equation( ) = ( ) + + )

= Transfer Function 1

Ms2

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

( ) ( )

( )

2 2

Input = f(t)Process

Output = y(t)

Block Diagram

12Ms bs K

Y(s)F(s)Block Diagram 1

2

/ M

s sbM

KM

Y(s)F(s)

As s

n

n n

2

2 2 2 Y(s)F(s)General Quadratic Form

Natural Frequency Damping Ratio System Constant

2

n n n

n

K

M

b

M M

K

M

b

M M M K M K

b

KM

A

A

2 2

2

1

1 1 1

2

( )

M

b

fy

K

14

More on the Characteristic Equation

Characteristic Equation (C.E.)

If Roots = Are the same If Roots are real but different If Roots are complex conjugates

Also many text books refer to

Roots of the C.E. are: ss

1

1

s s

j

n n

n n n n

n

ss

2 22

111

0

22 2

2

2

1 1

1

,,

12Ms bs K

Y(s)F(s)Block Diagram 1

2

/ M

s sbM

KM

Y(s)F(s)

As s

n

n n

2

2 2 2 Y(s)F(s)General Quadratic Form

15

Characteristic Equation

• The denominator polynomial of the transfer function when set equal to zero is called the characteristic equation C.E..

• The roots of the C.E. are the poles of the system.

• The roots of the transfer function’s numerator are the zeros of the system.

• Poles and zeros are critical frequencies

• At the poles, the transfer function becomes infinite

• At the zeros the transfer function becomes zero.

• The complex frequency s-plane plot of the poles and zeros graphically portrays the character of the natural transient response of the system.

0)22)(22(842 jsjsss

)43)(43(

)5(4)(

jsjs

ssT

jω

x

x

4

-4

-6 -5 -4 -3 -2 -1

4

16

Root Location vs Stability of the 2nd order C.E.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.70

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1Settling Time ~ Ts

Time (sec)

Am

plitu

de

jω

X

Double Root

• Any time the roots are equal and real, the system is critically damped, no overshoot.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.70

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1Settling Time ~ Ts

Time (sec)

Am

plitu

de

• Any time the roots are unequal and real, the system is overdamped, no overshoot.

jω

X X

17

Root Location vs Stability of the 2nd order C.E.Continued

0 0.1 0.2 0.3 0.4 0.5 0.6 0.70

0.2

0.4

0.6

0.8

1

1.2

1.4Settling Time 2 ~ Ts

Time (sec)

Am

plitu

de

• Anytime the roots are complex conjugates, the system experiences damped oscillation

jω

X

X

jω

X

X

0 0.1 0.2 0.3 0.4 0.5 0.6 0.70

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2Settling Time ~ Ts

Time (sec)

Am

plitu

de

• If the poles lie on the imaginary axis the system is marginally stable. It oscillates.

18

Root Location vs Stability of the 2nd order C.E.Continued

• Anytime the roots are in the RHP the system is unstable and diverges

jω

X

X

jω

XX

0 5 10 15 20 25-100

-80

-60

-40

-20

0

20

40

60

80Unit Step Response

Time (sec)

Am

plitu

de

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.80

1

2

3

4

5

6

7

8

9Unit Step Response #2

Time (sec)

Am

plit

ud

e

19

Deriving a Transfer Function for a Control System ComponentConsider a DC Torque Motor

ea

Ra La

em

ө, ω

J , b

Ia

J = Motor Inertia, could also include load inertiab = FrictionRa = Motor winding resistanceLa = Motor inductanceem = Back emfKT = Motor Torque Sensitivity – A motor parameterKm = Relates back emf to shaft speed, em = Kmω

(2) )()(or ehen constant t Assusme ; :emfBack

(1) )()( :Developed Torque

m ssKsEKKe

sIKsT

mmmBm

aTm

20

Deriving a Transfer Function for a DC Torque MotorContinued

(3))()()(Ifor Solving )(])[()(or eILR

amaa LsR

sEsEssELsRsIsEIe

a

mamaaaaa

ea

Ra La

em

ө, ω

J , b

Ia

KVL around loop:

Motor and Load Inertia

(4)22

)()(for solving )()()(or

bsJs

sTssbssTsJsbJT m

mm

Equations (1), (2), (3), and (4) are called Equilibrium Equations

21

Constructing Block Diagram for a DC Torque MotorFrom Equilibrium Equations

(3) )()(

)(

(2) )()(

(4) )(

)(

(1) )()(

2

LsR

SEsEsI

ssKsE

bsJs

sTs

sIKsT

a

maa

mm

m

aTm

KT

Tm(s)Ia(s)

KT

Tm(s)Ia(s)

bsJs 2

1 )(s

KT

Tm(s)Ia(s)

bsJs 2

1 )(s

KmsEm(s)

KT

Tm(s)Ia(s)

bsJs 2

1 )(s

KmsEm(s)

+

-

Ea(s)

LsRa 1

Ea - Em

22

Constructing DC Motor Transfer Function from Block Diagram Using Mason’s Rule

])([))((

1

1))((

)(

)(

))((1

1 ))((

11

2

2

211

2

1221

mtaaaa

t

ma

T

a

T

a

ma

T

a

TT

a

KKbRsbLJRsJLs

K

sKBsJsLsR

KBsJsLsR

KM

sE

s

sKBsJsLsR

K

BsJsLsR

K

BsJsK

LsRM

KT

Tm(s)Ia(s)

bsJs 2

1 )(s

KmsEm(s)

+

-

Ea(s)

LsRa 1

Ea - Em

])([)(

)(2

mtaaaa

t

a KKbRsbLJRsJLs

K

sE

s