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


 Gregory Davidson
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1 Introduction to the Nyquist criterion The Nyquist criterion relates the stability of a closed system to the openloop frequency response and open loop pole location. Mapping. If we take a complex number on the splane 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 openlooped 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 halfplane, we apply the concept of contour to use the entire right halfplane 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 halfplane), 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 halfplane equals the number of openloop poles, P, that are in the right halfplane 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 closedloop 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 unitfeedback 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 closeloop poles in the right half plane. 9
10 For (iii), we run numg=500* [12];; 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
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