The Nyquist criterion relates the stability of a closed system to the open-loop frequency response and open loop pole location.

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Introduction to the Nyquist criterion The Nyquist criterion relates the stability of a closed system to the open-loop frequency response and open loop pole location. Mapping.

1 Introduction to the Nyquist criterion The Nyquist criterion relates the stability of a closed system to the open-loop frequency response and open loop pole location. Mapping. If we take a complex number on the s-plane and substitute into a function F(s), another complex number results. e.g. substituting s = 4 + j3 into F(s) = s 2 + 2s + 1 yields 16 + j30. Contour. Consider a collection of points, called a contour A. Contour A can be mapped into Contour B, as shown in the next Figure. Figure above; Mapping contour A through F(s) to contour B. 1

2 Assuming F(s) = (s z 1)(s z 2 ) (s p 1 )(s p 2 ) If we assume a clockwise direction for mapping the points on contour A, the contour B maps in a clockwise direction if F(s) has just one zero. If the zero is enclosed by contour A, then contour B enclose origin. Alternatively, the mapping is in a counterclockwise direction if F(s) has just one pole, and if the pole is enclosed by contour A, then contour B enclose origin. If there is the one pole and one zero is enclosed by contour A, then contour B does not enclose origin. 2

3 Figure above; Examples of contour mapping. 3

4 Consider the system in the Figure below. Figure above; closed loop control system Letting G(s) = N G D G, H(s) = N H D H, We found T(s) = Note that G(s) 1 + G(s)H(s) = N G D H D G D H + N G N H 1 + G(s)H(s) = D GD H + N G N H D G D H 4

5 The poles of 1+G(s)H(s) are the same as the poles of G(s)H(s), the open-looped system, that are known. The zeros of 1 + G(s)H(s) are the same as the poles of T(s), the closedlooped system, that are unknown. Because stable systems have T(s) with poles only in the left half-plane, we apply the concept of contour to use the entire right half-plane as contour A, as shown in the Figure below. Figure above; Contour enclosing right halfplane to determine stability. 5

6 We try to construct contour B via F(s) = G(s)H(s) which is the same as that of 1 + G(s)H(s), except that it is shifted to the right by (1, j0). The mapping is called the Nyquist diagram of G(s)H(s). Assuming that A starts from origin, A is a path traveling up the jω axis, from j0 to j, then a semicircular arc, with radius, followed by a path traveling up the jω axis, from j to origin. So substituting s = jω, with ω changing from 0 to, we obtain part of contour B, which is exactly the polar plot of G(s)H(s). 6

7 Each zero or pole of 1 + G(s)H(s) that is inside contour A (the right half-plane), yields a rotation around ( 1, j0) (clockwise for zero and counterclockwise for pole) for the resultant Nyquist diagram. The total number of counterclockwise revolution, N, around ( 1, j0) is N = P Z, where P is the number of openloop poles,and Z is the number of closed loop poles. Thus we determine that that the number of closed loop poles, Z, in the right half-plane equals the number of open-loop poles, P, that are in the right half-plane minus the number of counterclockwise revolution, N, around 1 of the mapping, i.e. Z = P N. Use Nyquist criterion to determine stability If P = 0 (open loop stable system), for a closed systems to be stable (i.e. Z = 0), we should have N = 0. That is, the contour should not enclose ( 1, j0). This is as shown in next Figure (a). 7

8 On the other hand, another system with P = 0 (open loop stable) has generated two clockwise encirclement of ( 1, j0), (N = 2), as shown in Figure (b) below. Thus Z = P N = 2. The system is unstable with two closed-loop poles in the right hand plane. Figure above; Mapping examples: (a) contour does not enclose closed loop poles; (b) contour does enclose closed loop poles; 8

9 Example: Apply the Nyquist criterion to determine the stability of the following unit-feedback systems with (i) G(s) = s + 3 (s + 2)(s 2 + 2s + 25). (ii) G(s) = (iii) G(s) = s + 20 (s + 2)(s + 7)(s + 50). 500(s 2) (s + 2)(s + 7)(s + 50). Solution: For (i) and (ii), check polar plots in the previous lecture. For both systems we have P = 0 (open loop stable system). The two nyquist plots does not enclose ( 1, j0), (N=0) Thus Z = P N = 0. Both systems (i) and (ii) are stable since there are no close-loop poles in the right half plane. 9

10 For (iii), we run numg=500* [1-2];; deng=conv([1 2],[1 7]); deng=conv(deng,[1 50]); G=tf(numg,deng); nyquist(g); grid on; Nyquist Diagram db 0 db 2 db 1 4 db 4 db 6 db 6 db Imaginary Axis db 20 db 10 db 20 db Real Axis Figure above; The polar plots for G(s) = 500(s 2) (s + 2)(s + 7)(s + 50). We have P = 0 (open loop stable system), but N = 1, so System (iii) is unstable with one closed loop pole in the right half plane. 10

Linear Control Systems Lecture #3 - Frequency Domain Analysis. Guillaume Drion Academic year

Linear Control Systems Lecture #3 - Frequency Domain Analysis Guillaume Drion Academic year 2018-2019 1 Goal and Outline Goal: To be able to analyze the stability and robustness of a closed-loop system