Mapping If we take a complex number on the splane and substitute into a function an other complex number results eg substitut ing 4 3 into 2 1 yields 16 30 Contour Consider a collection of points called a contour A Contour A can be mapped ID: 26083 Download Pdf

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Mapping If we take a complex number on the splane and substitute into a function an other complex number results eg substitut ing 4 3 into 2 1 yields 16 30 Contour Consider a collection of points called a contour A Contour A can be mapped

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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 . If we take a complex number on the s-plane and substitute into a function ), an- other complex number results. e.g. substitut- ing = 4 + 3 into ) = + 2 + 1 yields 16 + 30. 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 to contour B.

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Assuming ) = )( )( If we assume

a clockwise direction for mapping the points on contour A, the contour B maps in a clockwise direction if ) has just one zero. If the zero is enclosed by contour A, then contour B enclose origin. Alternatively, the mapping is in a counterclock- wise direction if ) 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.

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Figure above; Examples of contour mapping.

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Consider the system in the Figure below. Figure above; closed

loop control system Letting ) = , H ) = We found ) = 1 + Note that 1 + ) =

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The poles of 1+ ) are the same as the poles of ), the open-looped system that are known. The zeros of 1 + are the same as the poles of ), the closed- looped system , that are unknown. Because stable systems have ) 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 half- plane to determine stability.

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We try to construct contour B via ) = which is the

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

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Each zero or pole of 1 + ) that is in- side contour A (the right half-plane), yields a rotation around ( , j 0) (clockwise for zero and counterclockwise for pole)

for the resul- tant Nyquist diagram. The total number of counterclockwise revolution, , around ( , j 0) is , where is the number of open- loop poles,and is the number of closed loop poles. Thus we determine that that the number of closed loop poles, , in the right half-plane equals the number of open-loop poles, , that are in the right half-plane minus the number of counterclockwise revolution, , around of the mapping, i.e. Use Nyquist criterion to determine stability If = 0 (open loop stable system), for a closed systems to be stable (i.e. = 0), we should have = 0. That is, the contour should

not enclose ( , j 0). This is as shown in next Fig- ure (a).

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On the other hand, another system with = 0 (open loop stable) has generated two clock- wise encirclement of ( , j 0), ( 2), as shown in Figure (b) below. Thus 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;

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Example: Apply the Nyquist criterion to deter- mine the stability of the following unit-feedback systems with ) = + 3 + 2)( + 2 + 25) ii ) = +

20 + 2)( + 7)( + 50) iii ) = 500( 2) + 2)( + 7)( + 50) Solution: For (i) and (ii), check polar plots in the previous lecture. For both systems we have = 0 (open loop stable system). The two nyquist plots does not enclose ( , j 0), (N=0) Thus = 0. Both systems (i) and (ii) are stable since there are no close-loop poles in the right half plane.

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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; −1.5 −1 −0.5 0.5 1.5 −1.5 −1 −0.5 0.5 1.5 0 dB −20 dB −10 dB

−6 dB −4 dB −2 dB 20 dB 10 dB 6 dB 4 dB 2 dB Nyquist Diagram Real Axis Imaginary Axis Figure above; The polar plots for ) = 500( 2) + 2)( + 7)( + 50) We have = 0 (open loop stable system), but 1, so System (iii) is unstable with one closed loop pole in the right half plane. 10

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