2. 1. Introduction to Nyquist stability criterion
The Nyquist stability criterion determines the stability
of a closed-loop system from its open-loop frequency
response and open-loop poles, and there is no need for
actually determining the closed-loop poles.
To determine the stability of a closed-loop system, one
must investigate the closed-loop characteristic equation
of the system.
3. )
(s
G
)
(s
H
We consider the following typical closed-loop system:
From the above diagram, we have
( )
( )
1 ( ) ( )
G s
s
G s H s
1 2
1 2
( ) ( )
( ) : , ( ) :
( ) ( )
N s N s
G s H s
D s D s
where
( ) 1 ( ) ( )
F s G s H s
We define
4. which is also called the characteristic equation of the
closed-loop system and will play an important role in
stability analysis. In fact, we have
1 2 1 2
1 2
( ) ( ) ( ) ( )
( )
( ) ( )
D s D s N s N s
F s
D s D s
Further, we assume that deg(N1N2)<deg(D1D2), which
implies that both the denominator and nominator
polynomials of F(s) have the same degree.
5. 2. Preliminary Study
Consider, for example, the following open-loop transfer
function:
2
( ) ( )
1
G s H s
s
1
( ) 1 ( ) ( ) ( )
1
s
F s G s H s F s
s
whose closed-loop characteristic equation is
which is analytic everywhere except at its singular
points. For each point analytic in s plane, F(s) maps
the point into F(s) plane. For example,
s=2+j:F(s)=2j.
Further, we have the following conclusion:
6. Conformal mapping: For any given continuous closed
path in the s plane that does not go through any singular
points, there corresponds a closed curve in the F(s) plane.
s-plane F(s) plane
j
s
[ ]
0
s
1
s
2
s
3
s
F s
[ ( )]
Im
Re
0
F
1
F s
( )
2
F s
( )
3
F s
( )
7. 0
-2 -1 1 2 3
s-plane
A B C
F
G
D
E
H
s: clockwise
1
-1 0 2 3
F(s)-Plane
A׳
C ׳
G ׳
D ׳
E ׳ H ׳
F: clockwise
2
( )
s
F s
s
Example. Consider
( ) ( 2) ( 0)
0
F s s s
F encircles the origin:
( 2) 0
( 0) 0
s
s
Case 1:
8. -2 1 0
s-plane
A B C
F
G
D
E
H
s: clockwise
1
-1 0
F(s) plane
A׳
C ׳
G ׳
D ׳
E ׳
F: clockwise
2
( )
s
F s
s
( ) ( 2) ( 0)
360
F s s s
F encircles the origin:
( 2) 360
( 0) 0
s
s
Case 2:
9. 2
( )
s
F s
s
-2 -1 0
s-plane
A B C
F
G
D
E
H
s: clockwise
1
-1 0
F(s) plane
A׳ C ׳
H ׳ D ׳
E ׳
B׳
F ׳
G ׳
F: counterclockwise
( ) ( 2) ( 0)
360
F s s s
F encircles the origin:
0
( 2) 0
( 0) 360
s
s
Case 3:
10. 2
( )
s
F s
s
Case 4:
It can be verified that when s encloses both the pole and
zero in the clockwise direction, the number of the F
encircles the origin is 1+1=0.
11. 1
-z
2
z
1
p
-
2
p
- 0
s 1
2
2
1
j
0
F
)
(s
F
u
jv
S
F
If, for example, F(s) has two zeros and one pole are
enclosed by s, it can be deduced from the above
analysis that F encircles the origin of F plane one time
in the clockwise direction as s traces out s.
12. Mapping theorem (The principle of argument)
Let F(s) be a ratio of two polynomials in s. Let P be the
number of poles and Z be the number of zeros of F(s)
that lie inside a closed contour s, which does not pass
through any poles and zeros of F(s).
Then the s in the s-plane is mapped into the F(s) plane
as a closed curve, F. The total number N of
counterclockwise encirclements of the origin of the
F(s), as s traces out s in the clockwise direction, is
equal to
N=PZ
13. 3.The Nyquist Stability Criterion
)
(s
G
)
(s
H
1 2 1 2
1 2
( ) ( ) ( ) ( )
( )
( ) ( )
=
D s D s N s N s
F s
D s D s
closed -loop characteristic equation
open -loop characteristic equation
For a stable system, all the zeros of F(s) must lie in the
left-half s-plane.
14. Therefore, we choose a contour s, say, Nyquist path, in
the s-plane that encloses the entire right-half s-plane. The
contour consists of the entire j axis ( varies from to
+) and a semicircular path of infinite radius in the right-
half s plane.
Nyquist contour
r
0
S
s
jw
s-plane
As a result, the Nyquist path
encloses all the zeros and poles of
F(s)=1+G(s)H(s) that have
positive real parts.
15. Now, let
P be the number of the unstable open-loop poles,
which is assumed to be known,
N be the number of encirclements of the origin of
F(s) in counterclockwise direction, which, by
drawing the curve, is also known, and
Z be the number of the unstable closed-loop poles,
which is unknown.
Then by applying the Mapping Theorem, we have
N=PZ.
If Z=0, the system is stable; if Z>0, the system is unstable
with the number of unstable poles being Z.
Question: How to obtain the curve of F(s) as s traces
out s in clockwise direction?
16. Nyquist contour
r
0
S
s
jw
s-plane
It is clear that when s traverses the semicircle of infinite
radius, F(s) remains a constant. That is, the encirclements
of the origin of F(s) only determines by F(j ) as varies
from to + provided that no zeros or poles lie on the
j axis.
lim ( )
lim [1 ( ) ( )] const.
s
s
F s
G s H s
By assumption,
17. Example.
Consider a unity feedback system whose open-loop
transfer function is as follows
2 1
( ) ( ) 1
1 1
s
G s F s G
s s
The system is stable.
1
P 1
N
0
N
P
Z
1
1
F(j)
0
j
18. Further, since
( ) 1
F s G
compare the two curves below:
1
1
F(j)
0
j
2
1
GH(j)
0
j
Clearly, the encirclement of the origin in F plane is
equivalent to the encirclement of 1 point in GH
plane!
19. Example.
Consider a unity feedback system whose open-loop
transfer function is as follows
0.5
( )
1
G s
s
The system is unstable.
1
P 0
N
1
N
P
Z
1
5
.
0
G(s)
0
20. Example.
Consider a single-loop control system, where
)
1
1
.
0
)(
1
(
100
)
(
)
(
s
s
s
H
s
G
0
,
0
N
P
0
N
P
Z
The system is stable. G(s)H(s)
1
0
Determine its stability.
21. Nyquist stability criterion
A system is stable if and only if from
N = PZ
we can obtain that Z=0, where P is the number of the
open-loop poles in the right-hand half s-plane and N is the
number of the encirclements of the point (1, j0) of
G(j)H(j) in the counterclockwise direction.
22. Because the plot of G(j )H(j) and the plot of
G(j )H(j) are symmetrical with each other about the
real axis. Hence, we can modify the stability criterion as
N = (P – Z)/2,
where only the polar plot of G(j)H(j) with varying
from 0 to + is necessary.
G(s)H(s)
0
-1
Re
Im
G(s)H(s)
0
-1
Re
Im
23. Example.
Consider a unity feedback system whose open-loop
transfer function is as follows
2
( )
1
G s
s
The system is stable.
1 ( )
2 2
(1 )
0
2
P Z
N
Z
Z
Investigate its stability by using Nyquist criterion.
Solution: Obviously, P=1 (nonminimum phase plant).
Now, drawing its Nyquist plot yields:
2
1
G(j)
0
j
24. Example.
Consider a unity feedback system whose open-loop
transfer function is as follows
1 2 3
( ) , 0, 0
( 1)( 1)( 1)
i
K
G s T K
T s T s T s
Using Nyquist plot, determine its stability.
Solution: Obviously, P=0. Now, drawing its Nyquist
plot yields:
25. 0
Im
Re
1
0
1
N
3
K
0
Im
Re
1
0
0
P
0
Im
Re
1
0
0
N
1
K 2
K
26. 4. Nyquist Criterion Based on Bode diagram
Again, we consider the open-loop transfer function with
Nyquist curve below:
1 2 3
( ) , 0, 0
( 1)( 1)( 1)
i
K
G s T K
T s T s T s
where P=0, N=1 and the
system is unstable. The key
is the point that the Nyquist
curve crosses the negative
real axis is less than 1
(abs(A)>1).
0
Im
Re
1
0
1
N
3
K
A
27. The Bode diagram is:
1
1
T
2
1
T 3
1
T
dB
[ 20]
[ 40]
[ 60]
20logK
w
w
0
180
0
270
c
w
0
0
G
The system is unstable because the phase angle plot
crosses the1800 line in the region of 20logG(j)>0.
N=1
28. The Nyquist Theorem can be interpreted as:
Theorem: The system is stable if and only if from
N+ N = (P – Z)/2
we can obtain that Z=0.
We use N to denote the number of crosses of 1800 line
in the region of 20logG(j)>0 if each across makes
the phase angle less than 1800. Similarly, we use N+ to
denote the number of crosses of 1800 line in the region
of 20logG(j)>0 if each across makes the phase angle
larger than 1800.
29. Example. Both the Nyquist plot and Bode diagram of
an open-loop minimum phase transfer function are
shown below. Determine its stability.
)
(
GH
L lg
20
GH
1
2
3
1
2
3
0
1
)
(
)
(
0
180
0 0
2 2
P Z Z
N N Z
30. 5. Special case when G(s)H(s) involves Poles
and Zeros on the j Axis
Consider the following open-loop transfer function:
( ) ( )
( 1)
K
G s H s
s Ts
which does not satisfy the Nyquist Theorem because s is
a singular point. Therefore, the contour s must be
modified. A way of modifying the contour near the
origin is to use a quarter-circle with an infini-tesimal
radius (>0):
j
s e q
e
where varies from 00 to 900 as varies from 0 to 0+.
31. A modified contour s :
The modified s implies
that we regard 1/s as a
stable pole.
0 0
1 1
, :0 90
j
e
s
q
q
e
-
= × - ® -
where 1/s rotates from 00 to 900 with an infinite radius
as varies from 0 to 0+.
1
T
Radius
Re
j
s-plane
j0+
=0
Then, for the controlled plant with integral factor 1/s, we
have
32. The integral factor 1/s rotates from 00 to 900 with an
infinite radius as varies from 0 to 0+.
1
T
Radius
r
Radius
Re
j
j0+
=0
j
34. L(dB)
f w
( )
135
90
10
180
20dB/ dec
1 1 3
T
40dB/ dec
0
0
35. Example. Let the open-loop transfer function be
2
( ) ( ) , 0, 0
( 1)
K
G s H s T K
s Ts
Determine its stability by using Nyquist Theorem.
Solution: When =0,
2 0
2 2
( ) ( ) , 0, 0 ;
( 1)
j
K K
G s H s e
s T s
q
w q
e
0
2 180
q
Hence, when >0+,
0 1
( ) ( ) 180 tan
G j H j T
w w w
When =0+,
36. 0
Im
Re
1
0
0
P
0
P=0, N=1 N=(PZ)/2 Z=2. The system is
unstable.
38. 6. Stability Analysis
Example. Let the open-loop transfer function be
2
1 2
2
1
( 1)
( ) ( ) , 0, 0, 0
( 1)
K T s
G s H s T T K
s T s
Determine its stability by using Nyquist Theorem.
Solution: When =0,
2 0
2
2 2
1
( 1)
( ) ( ) , 0, 0 ;
( 1)
j
K T s K
G s H s e
s T s
q
w q
e
0
0 , 2 180
w q
Hence, when >0+,
0 1 1
1 2
( ) ( ) 180 tan tan
G j H j T T
w w w w
When =0+
39. If T1>T2, the Nyquist curve is shown below. Since P=0,
N=1 N=(P Z)/2 Z=2. The system is unstable.
0
Re
Im
0
-1 0
43. Example. Consider a unity feedback system whose
open-loop transfer function is as follows
( )
( 1)
K
G s
s T s
The system is unstable.
1 ( )
2 2
(1 )
2
2
P Z
N
Z
Z
Investigate its stability by using Nyquist criterion.
Solution: Obviously, P=1 (nonminimum phase plant).
Now, drawing its Nyquist plot yields:
-1
j
0
0
44. 7. Conditionally Stable Systems
A conditionally stable system is stable for the value
of the open-loop gain lying between critical values, but
it is unstable if the open-loop gain is either increased or
decreased sufficiently.
0
Im
Re
1
0
0
P
45. Example. Consider a single-loop control system,
where open-loop transfer function is
)
4
2
(
)
( 2
s
s
s
K
s
G )
(s
G
Determine the range of K so that the system keeps
stable.
2
w
n
0 0 0
( ) 90 90 180
w
n
G j
( 2) 1
2 4
K
G j
8
K
r
46. Example. Consider a single-loop control system,
where open-loop transfer function is
1 2
3
( 1)( 1)
( )
K s s
G s
s
t t
Discuss its stability by using Nyquist stability
criterion.