Poles and Zeros
• The dynamic behavior of a transfer function model can be
characterized by the numerical value of its poles and zeros.
Chapter 6
• Two equivalent general representation of a TF:
m
G s 

bi s
i
i0
n
 ai s

i
 s  zm 
p2   s  pn 
b m  s  z1   s  z 2 
a n  s  p1   s 
i0
where {zi} are the “zeros” and {pi} are the “poles”.
We will assume that there are no “pole-zero” cancellations.
That is, that no pole has the same numerical value as a zero.
Note that, for system to be physically realizable, n>m.
Example: 4 poles (denominator is 4th order
polynomial) & 0 zero (numerator is a const)
G (s) 
K
s ( 1 s  1)( 2 s  2   2 s  1)
s1  0; s 2  
2
1
1
2
; s3 , s 4  

2
 j
1
2
2
Y  s   G  s U  s 
If u  t   M S  t  , then
y  t   A0  A1t  B e

t
1
e


2
t

 C 1 sin


1
2
2
t  C 2 cos
1
2
2

t


Chapter 6
Effects of Poles on System
Response
(1)

2

1
1
 e

t
1

decays slow er than e

2
t
(2) R H P pole  unstable system
(3) com plex conjugate poles  oscillation
(4) origin  integrating elem ent
Example of Integrating
Element
A
dh
dt
 qi  q0
A sH  ( s )  Q i( s )  Q 0 ( s )
if Q 0 ( s )  0
then
H ( s )
Q i( s )

1
As
pure integrator (ramp) for step
change in qi
Cause of Zeros – Input
Dynamics
[E xam ple 1]
G s 
K  a s  1 
 1s  1
[E xam ple 2]
G s 
 1 y   y  K  a u   u 
 1
1 y  y  K 
a
K  a s  1 
 a s  1 s  1 

0 u   d   u 

t
Some Facts about Zeros
• Zeros do not affects the number and
locations of the poles, unless there is an
exact cancellation of a pole by a zero.
• The zeros exert a profound effect on the
coefficients of the response modes.
Example of 2nd-Order Overdamped
System with One (1) Zero
Y  s   G  s U  s  
K  a s  1 
M
 1 s  1   2 s  1 
s
w here  1   2
t
t

 a   1 1  a   2  2
y t   KM 1 
e

e

1   2
 2  1

and




y    KM
C ase (a):  a   1  overshoot
C ase (b):  1   a  0  sim ilar to 1st-order step re sponse
C ase (c):  a  0  inverse response
Chapter 6
Step Response of 2nd-Order
Overdamped System without Zeros
u t   MS t   U  s  
  1 G s 
 
=
M
s
K
 1 s  1   2 s  1 
1
 1 2
1   2
2  1 2
y t   L
1
 G  s  U  s 
 t /
 t /

 1e 1   2 e 2 
 KM 1 





1
2

Further Analysis of Inverse
Response
Y  s   G  s U  s  
K  a s  1 
M
 1 s  1   2 s  1 
s
w here  1   2  0,  a  0, K M  0
K M  a s  1 
 dy 
L 

sY
s

 

 1 s  1   2 s  1 
 dt 
B y initial value theorem
dy
dt
t 0
a
K M  a s  1 
 dy 
 lim s L 

s

s 
 1 s  1   2 s  1 
 dt 
KM
 1 2
0
Chapter 6
Common Properties of Overshoot
and Inverse Responses
O vershoot or inverse response can be exp ected
w henever there are tw o physical effects that act
on the sam e output in opposite w ays and w ith
different tim e scales, i.e.
(1) sgn  K 1    sgn  K 2 
(2)  1   2
Another Example
Y  s   G  s U  s  

y  t   K M 1 

y 0 
a
1
K  a s  1  M
 1 s  1 
t

 a  1
1 
e
1 

K M (jum p ) and
s



y    KM
C ase (a):  a   1  0  decr easing
C ase (b):  1   a  0  increasing
C ase (c):  a  0   1  increasing
Time Delays
Time delays occur due to:
1. Fluid flow in a pipe
Chapter 6
2. Transport of solid material (e.g., conveyor belt)
3. Chemical analysis
-
Sampling line delay
-
Time required to do the analysis (e.g., on-line gas
chromatograph)
Mathematical description:
A time delay, θ, between an input u and an output y results in the
following expression:
0

y t   
u  t  θ 
for t  θ
for t  θ

Y s
U s
e
 s
Chapter 6
Implication of Time Delay
The presence of time delay in a process
means that we cannot factor the
transfer function in terms of simple
poles and zeros!
Polynomial Approximation of
Time Delays
T alo r series ex p an sio n :
 s
2
e
 s
 1 s 
 s
2
3

2!
3
3!
1 /1 P ad e ap p ro x im atio n :
e
 s

1 s / 2
1 s / 2
2 /2 P ad e ap p ro x im atio n :
1   s / 2   s / 12
2
e
 s

2
1   s / 2   s / 12
2
2

Chapter 6
Approximation of nth-Order
Systems
A n-th order system w ith n equal tim e co nstants:
Gn  s  
K


s

1


n

n
n t n 1


 nt / 
M


1

y  n, t   L G n  s 
  K M 1  e 
s 
i!

i0

y ,t   KMS t  
 lim G n  s   e
n 
 s


i



Chapter 6
Approximation of Higher-Order Transfer
Functions
In this section, we present a general approach for
approximating high-order transfer function models with
lower-order models that have similar dynamic and steady-state
characteristics.
Previously we showed that the transfer function for a time
delay can be expressed as a Taylor series expansion. For small
values of s,
(A)
e
θ0s
 1  θ0s
(zero)
An alternative first-order approximation is
(B )
e
θ0s
1

e
θ0s

1
1  θ0s
(pole)
Skogestad’s “Half Rule”
Chapter 6
1. Largest neglected time constant
•
One half of its value is added to the existing time delay (if
any) .
•
The other half is added to the smallest retained time
constant.
2. Time constants that are smaller than those in item 1.
•
Use (B)
3. RHP zeros.
•
Use (A)
Example 6.4
Consider a transfer function:
Chapter 6
G s 
K   0.1 s  1 
 5 s  1   3 s  1   0.5 s  1 
Derive an approximate first-order-plus-time-delay (FOPDT)
model,
G s 
Ke
 θs
τs  1
using two methods:
(a) The Taylor series expansions (A) and (B).
(b) Skogestad’s half rule
Compare the normalized responses of G(s) and the approximate
models for a unit step input.
Solution
(a) The dominant time constant (5) is retained. Applying
the approximations in (A) and (B) gives:
Chapter 6
 0.1s  1  e
 0.1 s
and
1
3s  1
e
1
3s
0.5 s  1
e
 0.5 s
Substitution into G(s) gives the Taylor series approximation,
GT S  s  
Ke
 0.1 s  3 s  0.5 s
e
5s  1
e

Ke
 3.6 s
5s  1
(b) To use Skogestad’s method, we note that the largest neglected
time constant in G(s) has a value of three.
• According to “half rule” (Rule 1), half of this value is added to
the next largest time constant to generate a new time constant
Chapter 6
τ  5  0.5(3)  6.5.
• Rule 1: The other half provides a new time delay of 0.5(3) = 1.5.
• The approximation of the RHP zero in Rule 3 provides an
additional time delay of 0.1.
• Approximating the smallest time constant of 0.5 in G(s) by
Rule 2 produces an additional time delay of 0.5.
• Thus the total time delay is, θ  1.5  0.1  0.5  2.1
• Therefore
G Sk  s  
Ke
 2.1 s
6.5 s  1
Chapter 6
Example
G s 
K 1  s  e
s
 1 2 s  1   3 s  1   0.2 s  1   0.05 s  1 

(a) G 1  s  
(b) G 2  s  
Ke
 s
(FO P D T )
 s 1
Ke
 s
 1 s  1   2 s  1 
(S O P D T )
Part (a)
  1
3
 0 .2  0 .0 5  1  3 .7 5
2
  12 
3
 1 3 .5
2
G1  s  
Ke
 3 .7 5 s
1 3 .5 s  1
Part (b)
  1
0.2
 0.05  1  2.15
2
 1  12
2  3
G2  s  
0.2
 3.1
2
Ke
 2.15 s
12 s  1   3.1 s  1 