1. 7-4 Nyquist Stability Criterion

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. We consider the following typical closed-loop system:
where
From the above diagram, we have
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
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:
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)=2j.
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
Angular displacement in the counterclockwise direction is positive. Angular displacement in the clockwise direction is negative. We assume that a rotation of an angle is positive if it rotates in the counterclockwise direction. In mathematics counter-clockwise rotations are considered “positive

7. F encircles the origin:
Example. Consider
Case 1: -2 -1 1 2 3
s-plane A B C F G D E H
s: clockwise 1 -1 2 3
F(s)-Plane A׳
C ׳
G ׳
D ׳
E ׳
H ׳
F: clockwise
F encircles the origin:

8. F encircles the origin:
Case 2: 1 -1
F(s) plane A׳
C ׳
G ׳
D ׳
E ׳
F: clockwise -2 1
s-plane A B C F G D E H
s: clockwise
F encircles the origin:

9. F encircles the origin:
Case 3: -2 -1
s-plane A B C F G D E H
s: clockwise 1 -1
F(s) plane A׳
C ׳
H ׳
D ׳
E ׳ B׳
F ׳
G ׳
F: counterclockwise
F encircles the origin:

10. 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. 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=PZ
Or in other words, N=Z-P, the total number N of clockwise encirclements of the origin of the F(s), as s traces out Gama s in the clockwise direction is equal to N=Z-P. (Negative values of N imply counterclockwise encirclements.)

13. 3.The Nyquist Stability Criterion
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
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=PZ.
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. By assumption,
Nyquist contour
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.

17. Example.
Consider a unity feedback system whose open-loop transfer function is as follows
F(j)
The system is stable.

18. compare the two curves below:
Further, since
compare the two curves below:
GH(j)
F(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
G(s)
The system is unstable.

20. Consider a single-loop control system, where
Example.
Consider a single-loop control system, where
G(s)H(s)
Determine its stability.
The system is stable.

21. Nyquist stability criterion
A system is stable if and only if from
N = PZ
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) -1 Re Im
G(s)H(s) -1 Re Im

23. Investigate its stability by using Nyquist criterion.
Example.
Consider a unity feedback system whose open-loop transfer function is as follows
Investigate its stability by using Nyquist criterion.
Solution: Obviously, P=1 (nonminimum phase plant). Now, drawing its Nyquist plot yields:
G(j)
The system is stable.

24. Example.
Consider a unity feedback system whose open-loop transfer function is as follows
Using Nyquist plot, determine its stability.
Solution: Obviously, P=0. Now, drawing its Nyquist plot yields:

26. 4. Nyquist Criterion Based on Bode diagram
Again, we consider the open-loop transfer function with Nyquist curve below:
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).

27. The Bode diagram is:
N=1
The system is unstable because the phase angle plot crosses the1800 line in the region of 20logG(j)>0.

28. We use N to denote the number of crosses of 1800 line in the region of 20logG(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 20logG(j)>0 if each across makes the phase angle larger than 1800.
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.

29. Example. Both the Nyquist plot and Bode diagram of an open-loop minimum phase transfer function are shown below. Determine its stability.

30. 5. Special case when G(s)H(s) involves Poles and Zeros on the j Axis
Consider the following open-loop transfer function:
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):
where  varies from 00 to 900 as  varies from 0 to 0+.

31. s-plane
j0+
=0
A modified contour s :
The modified s implies that we regard 1/s as a stable pole.
Then, for the controlled plant with integral factor 1/s, we have
where 1/s rotates from 00 to 900 with an infinite radius as  varies from 0 to 0+.

32. j0+
=0 j
The integral factor 1/s rotates from 00 to 900 with an infinite radius as  varies from 0 to 0+.

33. N = (P – Z)/2 Z=0, the system is stable.
s-plane -1
GH-plane C B’ B C’
The Nyquist curve starts from real axis when =0 and rotates 900 as =0+.
N = (P – Z)/2 Z=0, the system is stable.

35. Example. Let the open-loop transfer function be
Determine its stability by using Nyquist Theorem.
Solution: When =0,
When =0+,
Hence, when >0+,

36. P=0, N=1 N=(PZ)/2  Z=2. The system is unstable.

37. P=0, N=1, N+=0  N+N=(P Z)/2  Z=2. The system is unstable.

38. 6. Stability Analysis Example. Let the open-loop transfer function be
Determine its stability by using Nyquist Theorem.
Solution: When =0,
When =0+
Hence, when >0+,

39. If T1>T2, the Nyquist curve is shown below
If T1>T2, the Nyquist curve is shown below. Since P=0, N=1 N=(P Z)/2  Z=2. The system is unstable.

41. If T2>T1, the Nyquist curve is shown below
If T2>T1, the Nyquist curve is shown below. Since P=0, N=0 N=(P Z)/2  Z=0. The system is stable.

42. P=0, N=1/2, N+=1/2  N+N=(P Z)/2  Z=0. The system is stable.

43. Example. Consider a unity feedback system whose open-loop transfer function is as follows
Investigate its stability by using Nyquist criterion.
Solution: Obviously, P=1 (nonminimum phase plant). Now, drawing its Nyquist plot yields:
The system is unstable.

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.

45. Example. Consider a single-loop control system, where open-loop transfer function is
Determine the range of K so that the system keeps stable.

46. Example. Consider a single-loop control system, where open-loop transfer function is
Discuss its stability by using Nyquist stability criterion.