Linear control systems
Lecture #26
Contents of this lecture
1. Nyquist stability criterion.
 Minimum phase systems.
 Simplified Nyquist stability criterion.
 Relative stability measures for minimum phase systemss.
a. Gain margin.
b. Phase margin.
Brief description about lecture
Some Examples
Example 1 The characteristic equations of linear control systems follow. Apply the Nyquist criterion to
determine the value of K for stable system.
s 3  4Ks 2  ( K  5) s  10  0
Solution: The system can be changed to appropriate form
4s 2  s
1 K 3
0
s  5s  10
4s 2  s
has two RHP poles ( use Routh Hurwitz criteria for s 3  5s  10  0 ) so P0  P1  2 . The
3
s  5s  10
Nyquist plot for K=1 follows:
- for positive value of K:
Revised by: Ali Karimpour
30.1.2004
Linear control systems
Lecture #26
Z 1  2  2
K  (1 / 2.18)

stable
Z 1  2  0

Z 1  2
 K  (1 / 2.18)
- for negative value of K:
Z 1  P1  N1  Z 1  2  0

Z 1  2
unstable
So the system is stable for K  0.458
Example 2 The block diagram of a feedback control system follow.

unstable
Z 1  P1  N 1 

Z 1  0

C (s )
R(s )
G (s )
+
G (s )
+
-
+
+
Consider:
K
( s  5)( s  4)
a) Apply the Nyquist criterion to determine the range of K for stability.
b) Check the answer of part (a) by Routh Hurwitz.
G( s) 
Solution: The closed loop transfer function follows
C ( s)
G( s) 2
G( s) 2
K2



R( s) 1  G ( s)  G( s)  G( s) 2  (G ( s) 2 ) 1  2G( s) 2 ( s  5) 2 ( s  4) 2  2 K 2
The characteristic equation follows
2
( s  5) 2 ( s  4) 2  2 K 2  0

1 K 2
0
( s  5) 2 ( s  4) 2


The Nyquist plot for K 2  1 follows:
Revised by: Ali Karimpour
30.1.2004
--
Linear control systems
Lecture #26
- for positive value of K 2 :
Z 1  0  1
Z 1  P1  N 1 
Z 1  0  0

Z 1  1


Z 1  0
K 2  (1 / 0.005)

K 2  (1 / 0.005)


unstable
stable
2
- clearly K is not negative.
So the system is stable for K 2  200 or  10 2  K  10 2 .
b) The characteristic equation is:
( s  5) 2 ( s  4) 2  2K 2  s 4  18s 3  121s 2  360s  400  2K 2  0
The Routh Hurwitz table follows:
1
121
400  2 K 2
18
360
101
400  2 K
0
2
0
29160  36 K 2
0
101
400  2 K 2
So the system is stable for 400  2 K 2  0 or K 2  200 or  10 2  K  10 2 .
Example 3 The open loop transfer function of a unity-feedback (negative sign) system is:
K
G p ( s) 
( s  5) n
Sketch the Nyquist plot of G ( j ) for   0 to    and determine the range of K for the closed loop
system to be stable for:
a) n =2
b) n =3
c) n =4
Solution: Note that for all value of n the system is minimum phase.
Revised by: Ali Karimpour
30.1.2004
Linear control systems
Lecture #26
a) For n=2 the characteristic equation is 1 
K
 0 . So one must draw the simplified Nyquist plot of
( s  5) 2
G p ( j ) for   0 to    ( set K=1).
- for positive value of K the closed loop system is stable since,
Z 1  P1  N 1  Z 1  0  0 
Z 1  0
- for negative value of K:
Z 1  0  1

Z 1  1
 K  (1 / 0.04)

unstable
Z 1  P1  N1 
Z 1  0  0

Z 1  0
  (1 / 0.04)  K  0

stable
So the system is stable for K  25 .
K
 0 . So one must draw the simplified Nyquist plot of
b) For n=3 the characteristic equation is 1 
( s  5) 3
G p ( j ) for   0 to    (set K=1).
- for positive value of K:
Revised by: Ali Karimpour
30.1.2004
Linear control systems
Lecture #26
Z 1  P1  N 1 
Z 1  0  0

Z 1  0

K  (1 / 0.001)


K  (1 / 0.0004)

stable

(1 / 0.0004)  K

unstable

K  (1 / 0.0016)
stable
-Z 1  0  1

Z 1  1

(1 / 0.001)  K

unstable
- for negative value of K:
Z 1  0  1

Z 1  1
 K  (1 / 0.008)

unstable
Z 1  P1  N1 
Z 1  0  0

Z 1  0
  (1 / 0.008)  K  0

stable
So the system is stable for  125  K  1000 .
K
 0 . So one must draw the simplified Nyquist plot of
c) For n=3 the characteristic equation is 1 
( s  5) 4
G p ( j ) for   0 to    ( set K=1).
- for positive value of K:
Z 1  0  0

Z 1  0
Z 1  P1  N 1 
Z 1  0  1

Z 1  1
- for negative value of K:
Z 1  0  1

Z 1  1
Z 1  P1  N1 
Z 1  0  0

Z 1  0
So the system is stable for  625  K  2500 .
  (1 / 0.0016)  K  0


unstable
stable
Example 4 The characteristic equations of linear control systems follow. Apply the Nyquist criterion to
determine the value of K for stable system.
s 3  2s 2  20s  10 K  0
Solution: The system can be changed to appropriate form
10
1 K
0
2
s ( s  2 s  20)
10
Since
is minimum phase one can use simplified Nyquist plot. The Nyquist plot for K=1
2
s ( s  2 s  20)
follows:
Revised by: Ali Karimpour
30.1.2004
-
Linear control systems
Lecture #26
- for positive value of K:
Z 1  0  1

Z 1  1
Z 1  P1  N 1 
Z 1  0  0

Z 1  0
- for negative value of K:
Z 1  P1  N1  Z 1  0  1

Z 1  1
So the system is stable for 0  K  4 .

K  (1 / 0.25)

K  (1 / 0.25)


unstable
stable
unstable
Example 5 The open loop transfer function of a unity-feedback (negative sign) system with PD
controller is:
10( K p  K d s)
G( s) 
s2
Select the value of K p so that the parabolic error constant is 100. Find the equivalent open-loop transfer
function Geq (s) for stability analysis with K d as a gain factor. Sketch the Nyquist plot of Geq ( j ) for
  0 to    . Determine the range of K d for the stability by Nyquist criterion.
Solution: The parabolic error constant is 100 so:
K a  lim s 2 G(s)  10K p  100
s 0

K p  10
The characteristic equation is:
10(10  K d s)
1
0

s 2  10K d s  100  0
2
s
The equivalent open-loop transfer function Geq (s) for stability analysis with K d as a gain factor can be
find as follow:
10s
1 Kd 2
0
s  100
The Nyquist path and Nyquist plot are follows:
Revised by: Ali Karimpour
30.1.2004
Linear control systems
Lecture #26
j
..
.
*..
.
.
f
j10
-j10
j
g
e
d
c
b
a
.
s plane
h
k
.
Geq ( s ) 
10 s
s  100
.
c
2
..
.
b

k
a
g
h f 
.
d
*
.
e
- for positive value of K d the closed loop system is stable since,
Z 1  P1  N 1  Z 1  0  0 
- for negative value of K d the closed loop system is unstable since,
Z 1  P1  N1  Z 1  0  2 
So the system is stable for K d  0 .
Revised by: Ali Karimpour
Z 1  0
Z 1  2
30.1.2004