Topic # Feedback Control


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1 Topic # Feedback Control Stability in the Frequency Domain Nyquist Stability Theorem Examples Appendix (details) This is the basis of future robustness tests.
2 Fall Frequency Stability Tests Want tests on the loop transfer function L(s) = G c (s)g(s) that can be performed to establish stability of the closedloop system G cl (s) = G c(s)g(s) 1 + G c (s)g(s) Easy to determine using a root locus. How do this in the frequency domain? i.e., what is the simple equivalent of the statement does root locus go into RHP? Intuition: All points on the root locus have the properties that L(s) = ±180 and L(s) = 1 So at the point of neutral stability (i.e., imaginary axis crossing), we know that these conditions must hold for s = jω So for neutral stability in the Bode plot (assume stable plant), must have that L(jω) = ±180 and L(jω) = 1 So for most systems we would expect to see L(jω) < 1 at the frequencies ω π for which L(jω π ) = ±180 Note that L(jω) = ±180 and L(jω) = 1 corresponds to L(jω) = 1 + 0j
3 Fall Gain and Phase Margins Gain Margin: factor by which the gain is less than 1 at the frequencies ω π for which L(jω π ) = 180 GM = 20 log L(jω π ) Phase Margin: angle by which the system phase differs from 180 when the loop gain is 1. Let ω c be the frequency at which L(jω c ) = 1, and φ = L(jω c ) (typically less than zero), then P M = φ Typical stable system needs both GM > 0 and P M > 0 Gain Positive phase 1 1/GM γ φ 1 L(jω) Figure 1: Gain and Phase Margin for stable system in a polar plot
4 Fall Positive gain Negative phase 1 1/GM γ φ 1 1 γ φ Positive phase L(jω) L(jω) 1/GM Negative gain Stable system Unstable system Figure 2: Gain and Phase Margin in Polar plots + L db 0 _ ω c Positive gain Log ω + L db 0 _ Negative gain ω c Log ω 90 o 90 o L 180 o 270 o Positive phase Stable system Log ω L 180 o 270 o Negative phase Unstable system Log ω Figure 3: Gain and Phase Margin in Bode plots Can often predict closedloop stability looking at the GM and PM
5 Fall So the test for neutral stability is whether, at some frequency, the plot of L(jω) in the complex plane passes through the critical point s = 11 G c G (jω) Figure 4: Polar plot of a neutrally stable case This is good intuition, but we need to be careful because the previous statements are only valid if we assume that: Increasing gain leads to instability L(jω) = 1 at only 1 frequency which are reasonable assumptions, but not always valid. In particular, if L(s) is unstable, this prediction is a little more complicated, and it can be hard to do in a Bode diagram need more precise test. A more precise version must not only consider whether L(s) passes through 1, but how many times it encircles it. In the process, we must take into account the stability of L(s)
6 Fall Nyquist Stability Key pieces: an encirclement an accumulation of of 360 of phase by a vector (tail at s 0 ) as the tip traverses the contour c c encircles s 0 c s 0 We are interested in the plot of L(s) for a very specific set of values of s, called the Nyquist Path. R R C 2 Case shown assumes that L(s) has no imaginary axis poles, which is where much of the complexity of plotting occurs. Also note that if lim s L(s) = 0, then much of the plot of L(s) for values of s on the Nyquist Path will be at the origin. Nyquist Diagram: plot of L(s) as s moves around the Nyquist path C 2
7 Fall Steps: Construct Nyquist Path for particular L(s) Draw Nyquist Diagram Count # of encirclements of the critical point 1 Why do we care about the # of encirclements? Turns out that (see appendix) that if L(s) has any poles in the RHP, then the Nyquist diagram/plot must encircle the critical point 1 for the closedloop system to be stable. It is our job to ensure that we have enough encirclements how many do we need? Nyquist Stability Theorem: P = # poles of L(s) = G(s)G c (s) in the RHP Z = # closedloop poles in the RHP N = # clockwise encirclements of the Nyquist Diagram about the critical point 1. Can show that Z = N + P So for the closedloop system to be stable (i.e., no closedloop poles in the RHP), need Z 0 N = P Note that since P 0, then would expect CCW encirclements
8 Fall The whole issue with the Nyquist test boils down to developing a robust way to make accurate plots and count N. Good approach to find the # of crossing from a point s 0 is: Draw a line from s 0 Count # of times that line and the Nyquist plot cross N = #CW crossings CCW crossings G(s) 1 Observation: If the stability of the system is unclear from the Bode diagram, then always revert to the Nyquist plot.
9 Fall Basic Robustness The number of encirclements is crucial, but for a stable system, and a stable controller, we would expect no encirclements. However, we might find that the plot of L(s) along the Nyquist path approaches 1 very closely So the system is stable, but ally because small changes in the system dynamics might change L(s) and thus the # of encirclements 1/GM PM 1 L(jω) Clearly see role of GM and P M now  they are measures of how close L(s) comes to the critical point the closer it comes, the easier it might be to change the # of encirclements and thus change the system stability  robustness issue. GM and P M are crude measures of the basic concern better measure is minimum distance from L(s) to the critical point d(s) = 1 + L(s) Note that d(s) = 1/S(s), so the inverse of the distance is the system Sensitivity transfer function.
10 Fall FR: Example 1 G=tf(1.25,conv([1.5 1],[ ])); figure(1);clf;rlocus(g);rr=rlocus(g,1);rr2=rlocus(2*g,1); % hold on;plot(rr+j*eps, rs );plot(rr2+j*eps, md );hold off; % print dpng r300 examp11.png % figure(2);clf;nyquist(g);hold on;nyquist(2*g);hold off;% print dpng r300 examp12.png % figure(3);clf;bode(g);hold on;bode(2*g);hold off;% grid on;print dpng r300 examp13.png % Figure 5: System G(s) = 1.25 (1.5s+1)(s s+1) for a gain of 1 and 2
11 Fall FR: Example 2 G=tf([2 1],conv([1.3],[12])); % figure(1);clf;rlocus(g);rr=rlocus(g,1);rr2=rlocus(2/3*g,1); % hold on;plot(rr+j*eps, rs );plot(rr2+j*eps, md );hold off; % print dpng r300 examp21.png % figure(2);clf;nyquist(g);hold on;nyquist(2/3*g);hold off;% print dpng r300 examp22.png % figure(3);clf;bode(g);hold on;bode(2/3*g);hold off;% grid on;print dpng r300 examp23.png % Figure 6: System G(s) = 2s+1 (s+0.3)(s 2) for a gain of 1 and 2/3 Not obvious what is going on in the Bode plot for this conditionally stable system Can see effect of raising and lowering the gain, but interpretation not as obvious as in the Nyquist plot.
12 Fall FR: Summary Nyquist test gives us the desired frequency domain stability test Corresponds to a test on the number of encirclements of the critical point For most systems that can be interpreted as needing the GM > 0 and P M > 0 Typically design to GM 6dB and P M Introduced S(s) as a basic measure of system robustness.
13 A51 Appendix Nyquist Stability Theorem N P Z # clockwise encirclements of Nyquist diagram about 1 # poles G c (s)g p (s) in the RHP # closedloop poles in the RHP Can show that Z = N + P Clearly, for stability, we need Z = 0 Î N = P Outline of proof: Proof based on undertanding of how functions (F(s)) map contours in the splane. Î Map of F(s) will only encircle the origin if the contour in the splane contains a pole or zero of F(s). ZERO 0 C F(s) CW Encirclement POLE C F(s) CCW Encirclement Î Can also have mixtures of poloes + zeros in c, [symbol] encircles the origin N z  N p
14 A52 How apply this? x Use F(s) = 1 + G c (s)g p (s) x x x x Use Nyquist path for c Plot F(s) along c + count # encirclements around the origin (N) Î N = Z  P note: (zeroes of F(s)) (poles of F(s)) Z = # zeroes of = closedloop poles of system F(s) P = # poles of = unstable poles of G c (s)g p (s) F(s) in c Usually plot G c (s)g p (s) and look at encirclements of s = 1 (same thing) Note: F(s) = 1 + Gc(s)G(s) therefore zeros of F(s) are the CLP poles poles of F(s) are the loop poles Î only care about the CLP and loop poles in the RHP.
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