Modern Control System
EKT 308
 Transfer Function
 Poles and Zeros
Transfer Function
• A system can be represented, in s-domain, using the
following block diagram.
Input
X (s )
Transfer Function
G (s )
Output
Y (s )
For a linear, time-invariant system, the transfer function G (s )
is given by,
Y ( s)
G(s) 
X (s)
where, Y ( s ) is the Laplace transform of output and
X ( s ) is the Laplace transform of input.
Transfer Function (contd…)
• Consider the following linear time invariant (LTI) system
( n 1)
(n)

a0 y  a1 y  ....  an 1 y  an y
( m 1)
( m)
(n  m)

 b0 x  b1 x  ....  bm1 x  bm x
where, x is the input and y is the output.
With zero initial conditions, taking Laplace transform on both sides


Y ( s) a0 s n  a1s n1  ...  an1s  an 

X ( s) b0 s m  b1s m1  ...  bm1s  bm

Transfer Function (contd…)
L[output ]
G( s) 
L[input ] zero initialcondition
Y ( s) b0 s m  b1s m1  ...  bm 1s  bm


X ( s) a0 s n  a1s n 1  ...  an 1s  an
Rearranging, we get
Y (s)  G( s) X ( s)
Impulse Response
Suppose, input to a LTI system is unit impulse. We get ,
Y ( s )  G ( s ) X ( s )  G ( s ).L( (t ))
Y ( s)  G ( s)
Inverse Laplace Transform of this output gives the
impulse response of the system. I.e. impulse response
of the system is given by,
Impulse Response  L-1G(s)  g (t )
Given g(t), input-output relationship in t-domain is
given by the following convolution
t
y(t )   x( ) g (t   )d
0
Analisys of Transfer Function
• Consider the transfer function,
Y ( s) b0 s m  b1s m1  ...  bm1s  bm p( s)
G( s) 


n
n 1
X ( s) a0 s  a1s  ...  an 1s  an q( s)
If the denominator polynomial q (s )
is set to 0,
the resulting equation q ( s )  0 is called the characteristic equation
The roots of the characteristic equation are called poles.
The roots of p( s )  0 are called zeros.
Example of Poles and Zeros
• Suppose the following transfer function
( s  1)( s  2)
G( s) 
( s  0.5)( s  1.5)( s  2.5)
Note: This is only for illustration. Positive real poles lead instability.
Characteristic equation
( s  0.5)( s  1.5)( s  2.5)  0
Poles are 0.5, 1.5 and 2.5.
Zeros are 1 and 2.
Example of Poles and Zeros
Poles and Zeros plot
Suppose, the following transfer function
( s  3)
G ( s) 
( s  1)( s  2)
Clearly, zero is - 3 and poles are - 1 and - 2.
In the s - plane, they are represente d as follows.
Zeros are represented by circles (O) and poles by
cross (x).
Block Diagram representation
Input
X (s )
Transfer Function
G (s )
X (s )
G1 ( s)
G1 (s) X (s)
Output
Y (s )
G2 (s)
G2 (s)G1 (s) X (s)
Closed-Loop System Block Diagram
R(s)
+
E (s )
-
G (s )
C (s )
B(s)
H (s )
B( s )
Open loop transfer function 
 G( s) H ( s)
E ( s)
C ( s)
Feedforwar d transfer function 
 G( s)
E ( s)
Closed Loop Transfer function
C ( s)  G ( s) E ( s)
E ( s )  R( s )  B( s )  R( s )  H ( s )C ( s )
C ( s )  G ( s )R( s )  H ( s )C ( s )
C ( s )1  G ( s ) H ( s )  G ( s ) R( s )
C ( s)
G ( s)

 Closed - loop transfer function
R( s) 1  G ( s) H ( s)
G(s)
C ( s) 
R( s)
1  G ( s) H ( s)
Closed Loop Transfer function (contd…)
R(s)
+
E (s )
-
G (s )
C (s )
R(s)
B(s)
H (s )
G( s)
1  G( s) H ( s)
C (s )
Block Diagram
Input
X (s )
Transfer Function
G (s )
X (s )
G1 ( s)
G1 (s) X (s)
Output
Y (s )
G2 (s)
G2 (s)G1 (s) X (s)
Block Diagram Transformation
Block Diagram Transformation
Block Diagram Reduction
Moving a pickoff point
behind a block
Eliminating feedback loop
Eliminating feedback loop
Eliminating feedback loop
SIGNAL FLOW GRAPH MODEL
.
• Nodes which are connected by several directed
branches
• Graphical representation of a set of linear relation.
• Basic element is unidirectional path segment called a
branch.
• The branch relates the dependency of input/output
variable in a manner equivalent to a block of block
diagram
SIGNAL FLOW GRAPH MODEL
Y (s)
R( s)
• Variables are reperesented as nodes.
• Transmittence with directed branch.
• Source node: node that has only outgoing branches.
G( s) 
Y(s)
R(s)
G(s)
• Sink node: node that has only incoming branches.
As signal flow graph
D(s)
+
+
F(s) +
E(s)
D(s)
Y(s)
R(s) 1
E(s)
G(s) F(s)
G (s )
-1
B(s)
B(s)
H (s )
1
1
Y(s)
H(s)
Cascade connection
G1
n1

G2
n2
Gm 1
nm1
n3
G1G2 ....Gm
n1
nm
Gm
nm
Parallel connection
Two parallel branch
P
P+Q
u
y
Q
u
y
Mason rule
T (s) 
 Pk  k

where
  1   Li   Li L j   Li L j Lk  ....  Li L j ...Lz
 Li
 Total transmittence for every single loop
 Li L j
 Total transmittence for every 2 non-touching loops
 Li L j Lk  Total transmittence for every 3 non-touching loops
 Li L j ...Lz
Pk
 Total transmittence for every m non-touching loops
 Total transmittence for k paths from source to sink nodes.
 k  1   Li   Li L j   Li L j Lk  .....  Li L j ....Lz
*
*
*
*
*
*
*
where:
 k is obtained from  by removing the loops that
touch path Pk
*
*
Example:
Determine the transfer function of Y (s) R(s) the following block diagram.
+
R
+
Q
P
Y
R
-
-
P
Q
1
1
H
-
H
-1
L1  Q.H and . L2   P.Q.1
 Li  L1  L2  Q.H  P.Q
 Li L j  0 etc.
  1   Li   Li L j   Li L j Lk  .....  1  Q.H  P.Q
P1  1.P.Q.1
1  1
Transfer function
T (s) 
 Pk  k

Y
Y
P.Q

R 1  Q.H  P.Q
Example:
Determine Y (s) R(s)
R
+
+
A
-
-
Y
C
B
-
D
E
1
A
R
.
1
-1
D
-
-E
L1   A.D L2  1.B and L   A.1.B.C.E
3
 Li  L1  L2  L3   A.D  B  A.B.C.E
 Li L j  L1.L2   A.D.  B  A.D.B
 Li L j Lk  0
C
B
Y
1
Y