16.06 Principles of Automatic Control, Lecture 21

16.06 Principles of Automatic Control Lecture 21 The Nyquist Stability Criterion Can apply the argument principle to finding the stability of the closed loop system r + G(s) k y - The closed loop transfer function is T psq “ Y psq kGpsq “ R 1 ` kGpsq The closed loop poles of T psq are the roots of 0 “ 1 ` kGpsq That is, the closed loop poles of T psq are zeros of 0 “ 1 ` kGpsq. Note that the poles of 1 ` kGpsq are just the open loop poles of Gpsq. This suggests the following test for stability of the closed loop system: 1 Stability Test, Version1: Define the contour C1 as shown below: Im(s) Re(s) radius goes to ∞ C1 The contour encloses (in the limit) the entire right half plane. For this contour, plot the contour map 1 ` kGpsq The number of CW encirclements of the origin by 1 ` kGpC1 q is equal to Z ´ P , where Z is the number of closed loop poles in the right half plane, and P is the number of open loop poles in the right half plane. As an equation Z “N `P where Z - the number of closed loop unstable poles, N - the number of CW encirclements of 0, P - the number of unstable poles Stability Test, Version 2: Since the “1” term in 1 ` kGpsq just shifts the contour map of kGpsq by one unit to the right, it is often (usually) easier to plot kGpsq alone. This is known as the polar plot or Nyquist plot for the system. Note that for each encirclement of 0 by 1 ` kGpsq, there is one encirclement of ´1 by kGpsq. So the Nyquist Criterion, in the usual form, is 1. Plot kGpsq for ´j8 ď s ď j8. First evaluate kGpjωq for ω P r0, 8s and plot. Then reflect the image about the real axis and add to the previous image. Note that there no need to calculate kGpsq on the circular part of C1 if kGpsq Ñ 0 as s Ñ 8. 2 2. Evaluate the number of CW encirclements about ´1, and call that number N (see FPE for how to count encirclements). 3. Determine the number of unstable poles of Gpsq, P . 4. The number of unstable poles of the closed loop system is Z “N `P Finally, if k is unknown, we can instead plot Gpsq, and count encirclements of the point ´1{k. This is useful for determining the range of gains for which the closed loop system is stable, as in root locus. Examples G(s) + 1 (s+1)2 k - Root locus: Im(s) k>0 k<0 -1 Re(s) 3 Magnitude, dB Bode plot: 0 10 −40 dB/dec −2 10 −3 −2 10 −1 10 0 10 1 10 2 10 10 0 Phase (deg) −50 −100 −150 −200 −250 −300 −3 10 −2 −1 10 0 1 10 10 Frequency, ω (rad/sec) 2 10 10 Nyquist plot: 0.8 0.6 ω=-1 Imaginary Axis 0.4 0.2 ω=-∞ ω=+∞ 0 -1 ω=0 1 −0.2 −0.4 ω=1 −0.6 −0.8 −1 −0.8 −0.6 −0.4 −0.2 0 Real Axis 0.2 0.4 0.6 0.8 1 Note that the Nyquist plot does not encircle ´1, and therefore the number of unstable closed loop poles is 4 Z “N ` P “0 ` 0 (no unstable open loop poles) “0, for k “ 1. However, we can conclude more than that. The number of encirclements of ´1{k is zero for 1 ă 0 or k 1 ñ ą 0 or k ñ 0 ă k ă 8 or ´ ´ 1 ă ´1 k 0 ą k ą ´1 Therefore, the system is stable for k ą ´1. For k ă ´1, N “ 1, so there is one unstable pole. Example: + 1 ą1 k 1 (s+1)3 k - 5 The Nyquist plot is: 0.8 0.6 Imaginary Axis 0.4 -1 8 0.2 1 0 ω=0 ω=√3 −0.2 −0.4 −0.6 −0.8 −1 −0.8 −0.6 −0.4 −0.2 0 Real Axis 0.2 0.4 0.6 For ´1{k ă ´1{8 p0 ă k ă 8q, system is stable. For ´1{k ą 1 p0 ą k ą ´1q, system is stable. For ´1{8 ă ´1{k ă 0, pk ą 8q, system has 2 unstable poles. For 0 ă ´1{k ă 1 pk ă ´1q, system has one unstable pole. Of course, this agrees with our Routh and root locus analysis. 6 0.8 1