Modern Control System EKT 318

Transfer Function

Poles and Zeros

Block Diagram

Transfer Function

• A system can be represented, in s-domain, using the following block diagram.

Transfer Function

)(sG

Input Output

)(sX )(sY

For a linear, time-invariant system, the transfer function )(sG

is given by,

input. of transformLaplace theis )(

andoutput of transformLaplace theis )( where,

)(

)()(

sX

sY

sX

sYsG

Transfer Function (contd…)

• Consider the following linear time invariant (LTI) system

xbxbxbxb

yayayaya

mm

mm

nn

nn

1

)1(

1

)(

0

1

)1(

1

)(

0

....

....)( mn

output. theis andinput theis where, yx

With zero initial conditions, taking Laplace transform on both sides

mm

mm

nn

nn

bsbsbsbsX

asasasasY

1

1

10

1

1

10

...)(

...)(

Transfer Function (contd…)

nn

nn

mm

mm

asasasa

bsbsbsb

sX

sY

inputL

outputLsG

1

1

10

1

1

10

condition initial zero

...

...

)(

)(

][

][)(

Rearranging, we get

)()()( sXsGsY

Analysis of Transfer Function

• Consider the transfer function,

)(

)(

...

...

)(

)()(

1

1

10

1

1

10

sq

sp

asasasa

bsbsbsb

sX

sYsG

nn

nn

mm

mm

If the denominator polynomial )(sq is set to 0,

the resulting equation

0)( sq is called the characteristic equation

The roots of the characteristic equation are called poles.

called are 0)( of roots The sp zeros.

Poles and Zeros

• Suppose the following transfer function

)5.2)(5.1)(5.0(

)2)(1()(

sss

sssG

Note: This is only for illustration. Positive real poles lead instability.

Characteristic equation

2. and 1 are Zeros

2.5. and 1.5 0.5, are Poles

0)5.2)(5.1)(5.0( sss

poles

zeros

a pole–zero plot is a graphical representation of a transfer function in the complex plane which helps to convey certain properties of the system

The poles of a feedback system allow one to tell almost at a glance if a system is stable; zeros are often placed in the feedback network transfer function to cancel the natural poles of the amplifier, allowing the amplifier to retain a good frequency response and yet remain stable.

Example of Poles and Zeros

follows. as drepresente are they plane,-s In the

2. and 1 are Zeros

2.5. and 1.5 0.5, are Poles

0)5.2)(5.1)(5.0( sss

Poles and Zeros plot

Suppose, the following transfer function

follows. as drepresente are they plane,-s In the

2.- and 1- are poles and 3- is zero Clearly,

)2)(1(

)3()(

ss

ssG

Block Diagram representation

Transfer Function

)(sG

Input Output

)(sX )(sY

)(1 sG )(2 sG)(sX )()(1 sXsG )()()( 12 sXsGsG

Open-Loop System Block Diagram

Closed-Loop System Block Diagram

)(sG

)(sH

)(sE

+ -

)(sC)(sR

)(sB

[3] )()()(

[2] )()()(

[1] )()()(

sEsGsC

sHsCsB

sBsRsE

Our objective is to get )(

)(

sR

sC

Closed Loop Transfer function

function transfer loop-Closed)()(1

)(

)(

)(

)()()(1

)()(

)()()()(1)(

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

E(s)) the(replace (3), to(4)equation insert

[4] )()()()(

B(s)) the(replace (1), to(2)equation insert

[3] )()()(

[2] )()()(

[1] )()()(

sHsG

sG

sR

sC

sRsHsG

sGsC

sRsGsHsGsC

sHsCsRsGsC

sHsCsRsE

sEsGsC

sHsCsB

sBsRsE

Closed Loop Transfer function (contd…)

)(sG

)(sH

)(sE

+ -

)(sC)(sR

)(sB)(sR )(sC

)()(1

)(

sHsG

sG

Block Diagram Transformation

Block Diagram Reduction

Example 1:

H2

G4

G3G4

1. Moving a pickoff point behind a block

2. Eliminating feedback loop

3. Eliminating feedback loop

4. Eliminating feedback loop

G3G4

G2G3G4 1- G3G4H1

G1G2G3G4 1- G3G4H1+G2G3H2

)(sR )(sC

)()(1

)(

sHsG

sG

3. Eliminating feedback loop

4. Eliminating feedback loop

G2G3G4 1- G3G4H1

G1G2G3G4 1- G3G4H1+G2G3H2

)(sR )(sC

)()(1

)(

sHsG

sG

G2G3G4 1- G3G4H1

GS= H(s)=H2/G4

G2G3G4/1-G3G4H1

1+[(G2G3G3/1-G3G4h1)(H2/G4)]

SIGNAL FLOW GRAPH MODEL

17

Multiple subsystem can be represented in two ways;

(a) Block Diagram.

(b) Signal Flow Graph.

The block diagram.

The signal flow graph,

18

Signal flow graph consists only branches which represent

system and nodes which represents signals.

Variables are represented as nodes.

Transmittance with directed branch.

Source node: node that has only outgoing branches.

Sink node: node that has only incoming branches.

Cont’d…

19

Cont’d…

Parallel connection,

Signal-flow graph components: (a) system; (b) signal; (c)

interconnection of systems and signals

21

Example:

Step 1:

Step 2:

Mason rule

kkP

sT )(

where

zjikjijii LLLLLLLLL .......1 iL Total transmittence for every single loop

jiLL Total transmittence for every 2 non-touching loops

kji LLL Total transmittence for every 3 non-touching loops

zji LLL ... Total transmittence for every m non-touching loops

*********.........1 zjikjijiik LLLLLLLLL

where:

*iL Total transmittance for every single non-touching loop of ks’ paths

**ji LL Total transmittence for every 2 non-touching loop of ks’ paths

***kji LLL Total transmittance for every 3 non-touching loop of ks’ paths.

Total transmittance for every n non-touching loop of ks’ paths.

***.... zji LLL

R(s) C(s) G1(s) G2(s) G3(s) G4(s) G5(s)

V1(s) V2(s) V3(s) V4(s)

V5(s) V6(s)

H1(s)

G6(s) G8(s)

H2(s)

H4(s)

G7(s)

kP Total transmittance for k paths from source to sink nodes Forward Path

1. The forward-path gains:

Pk=G1(s)G2(s)G3(s)G4(s)G5(s)

Mason rule

kkP

sT )(

zjikjijii LLLLLLLLL .......1 iL Total transmittence for every single loop

jiLL Total transmittence for every 2 non-touching loops

kji LLL Total transmittence for every 3 non-touching loops

zji LLL ... Total transmittence for every m non-touching loops

R(s) C(s) G1(s) G2(s) G3(s) G4(s) G5(s)

V1(s) V2(s) V3(s) V4(s)

V5(s) V6(s)

H1(s)

G6(s) G8(s)

H2(s)

H4(s)

G7(s)

The loop gains: i. G2(s)H1(s)

ii. G4(s)H2(s)

iii. G7(s)H4(s)

iv. G2(s)G3(s)G4(s)G5(s)G6(s) G7(s) G8(s)

iL i + ii + iii + iv

jiLL (i.ii) + (i.iii) + (ii.iii) Non touching loop Non touching loop

Non touching loop kji LLL i.iii.iii

R(s) C(s) G1(s) G2(s) G3(s) G4(s) G5(s)

V1(s) V2(s) V3(s) V4(s)

V5(s) V6(s)

H1(s)

G6(s) G8(s)

H2(s)

H4(s)

G7(s)

The loop gains: i. G2(s)H1(s)

ii. G4(s)H2(s)

iii. G7(s)H4(s)

iv. G2(s)G3(s)G4(s)G5(s)G6(s) G7(s) G8(s)

iL iii Non touching loop Non touching loop

Non touching loop at k path

*********.........1 zjikjijiik LLLLLLLLL

where:

*iL Total transmittance for every single non-touching loop of ks’ paths

**ji LL Total transmittence for every 2 non-touching loop of ks’ paths

***kji LLL Total transmittance for every 3 non-touching loop of ks’ paths.

Total transmittance for every n non-touching loop of ks’ paths.

***.... zji LLL

5 steps in Mason Rule

1. Calculate forward path gains

2. Identify every loop gains

3. Identify the nontouching loops

4. Identify

5. Calculate the Transfers Function using Mason Rule :

,k

kkP

sT )(

R QP Y

11

-1

-H

Example:

Determine the transfer function of )()( sRsY the following block diagram.

P Q

H

R Y + +

- -

0 jiLL

QPHQLLLLLL kjijii ..1.....1

1...1

:Gain Path Forward .1

1 QPP

101 k

5. Transfer function QPHQ

QP

R

Y

..1

.

kkP

sT )(

and . HQL .1 1..2 QPL 2. Loop Gain:

3. Nontouching loop :

, .4 k

Example:

Determine )()( sRsY

.

A B

D

C

E

R Y

- - -

+ +

R Y

1

A

-D

1 B

-1

C

-E

Y1

DAL .1

BL .12

ECBAL ...1.3

ECBABDALi ....

BDABDALL ji ....

BDAECBABDA ..)....(1

1...1.

:Gain Path Forward .1

1 CBAP

2. Loop Gain:

3. Nontouching loop : DAL .1

BL .12

, .4 k zjikjijii LLLLLLLLL .......1

ECBABDALLLLi ....321

BDABDALLLL ji ..... 21

5. Transfer function

BDAECBABDA

CBA

R

Y

......1

..

kkP

sT )(

1k