Post on 12-Aug-2019
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