Frequency Response Methods

Topic #3 16.30/31 Feedback Control Systems Frequency response methods • Analysis • Synthesis • Performance • Stability in the Frequency Domain • Nyquist Stability Theorem Fall 2010 16.30/31 3–2 FR: Introduction • Root locus methods have: • Advantages: ∗ Good indicator of transient response; ∗ Explicitly shows location of all closed-loop poles; ∗ Trade-offs in the design are fairly clear. • Disadvantages: ∗ Requires a transfer function model (poles and zeros); ∗ Difficult to infer all performance metrics; ∗ Hard to determine response to steady-state (sinusoids) ∗ Hard to infer stability margins • Frequency response methods are a good complement to the root locus techniques: • Can infer performance and stability from the same plot • Can use measured data rather than a transfer function model • Design process can be independent of the system order • Time delays are handled correctly • Graphical techniques (analysis and synthesis) are quite simple. September 15, 2010 Fall 2010 16.30/31 3–3 Frequency Response Function • Given a system with a transfer function G(s), we call the G(jω), ω ∈ [0, ∞) the frequency response function (FRF) G(jω) = |G(jω)|�G(jω) • The FRF can be used to find the steady-state response of a system to a sinusoidal input since, if e(t) G(s) y(t) and e(t) = sin 2t, |G(2j)| = 0.3, �G(2j) = −80◦ , then the steady-state output is y(t) = 0.3 sin(2t − 80◦) ⇒ The FRF clearly shows the magnitude (and phase) of the re­ sponse of a system to sinusoidal input • A variety of ways to display this: 1. Polar (Nyquist) plot – Re vs. Im of G(jω) in complex plane. • Hard to visualize, not useful for synthesis, but gives definitive tests for stability and is the basis of the robustness analysis. 2. Nichols Plot – |G(jω)| vs. �G(jω), which is very handy for sys­ tems with lightly damped poles. 3. Bode Plot – Log |G(jω)| and �G(jω) vs. Log frequency. • Simplest tool for visualization and synthesis • Typically plot 20log |G| which is given the symbol dB September 15, 2010 Fall 2010 16.30/31 3–4 • Use logarithmic since if � � � (s + 1)(s + 2) � � log |G(s)| = �� (s + 3)(s + 4) � = log |s + 1| + log |s + 2| − log |s + 3| − log |s + 4| and each of these factors can be calculated separately and then added to get the total FRF. • Can also split the phase plot since � (s + 1)(s + 2) = (s + 3)(s + 4) �(s + 1) + �(s + 2) −�(s + 3) − �(s + 4) • The keypoint in the sketching of the plots is that good straightline approximations exist and can be used to obtain a good prediction of the system response. September 15, 2010 Fall 2010 16.30/31 3–5 Bode Example • Draw Bode for G(s) = |G(jω)| = s+1 s/10 + 1 |jω + 1| |jω/10 + 1| log |G(jω)| = log[1 + (ω/1)2]1/2 − log[1 + (ω/10)2]1/2 • Approximation log[1 + (ω/ωi)2]1/2 ≈ � 0 ω � ωi log[ω/ωi] ω � ωi Two straightline approximations that intersect at ω ≡ ωi • Error at ωi obvious, but not huge and the straightline approximations are very easy to work with. Fig. 1: Frequency response basic approximation September 15, 2010 Fall 2010 16.30/31 3–6 • To form the composite sketch, • Arrange representation of transfer function so that DC gain of each element is unity (except for parts that have poles or zeros at the origin) – absorb the gain into the overall plant gain. • Draw all component sketches • Start at low frequency (DC) with the component that has the lowest frequency pole or zero (i.e. s=0) • Use this component to draw the sketch up to the frequency of the next pole/zero. • Change the slope of the sketch at this point to account for the new dynamics: -1 for pole, +1 for zero, -2 for double poles, . . . • Scale by overall DC gain Fig. 2: G(s) = 10(s + 1)/(s + 10) which is a lead. September 15, 2010 Fall 2010 16.30/31 3–7 • Since �G(jω) = �(1 + jω) − �(1 + jω/10), we can construct phase plot for complete system in a similar fashion • Know that �(1 + jω/ωi) = tan−1(ω/ωi) • Can use straightline approximations ⎧ ⎨ 0 ω/ωi ≤ 0.1 �(1 + jω/ωi) ≈ 90◦ ω/ωi ≥ 10 ⎩ ◦ 45 ω/ωi = 1 • Draw components using breakpoints that are at ωi/10 and 10ωi Fig. 3: Phase plot for (s + 1) September 15, 2010 Fall 2010 16.30/31 3–8 • Then add them up starting from zero frequency and changing the slope as ω → ∞ Fig. 4: Phase plot G(s) = 10(s + 1)/(s + 10) which is a “lead”. September 15, 2010 Fall 2010 16.30/31 3–9 Frequency Stability Tests • Want tests on the loop transfer function L(s) = Gc(s)G(s) that can be performed to establish stability of the closed-loop system Gcl (s) = Gc(s)G(s) 1 + Gc(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 ωπ for which �L(jωπ ) = ±180◦ • Note that �L(jω) = ±180◦ and |L(jω)| = 1 corresponds to L(jω) = −1 + 0j September 15, 2010 Fall 2010 16.30/31 3–10 Gain and Phase Margins • Gain Margin: factor by which the gain is less than 1 at the frequen­ cies ωπ for which �L(jωπ ) = 180◦ GM = −20 log |L(jωπ )| • Phase Margin: angle by which the system phase differs from 180◦ when the loop gain is 1. • Let ωc be the frequency at which |L(jωc)| = 1, and φ = �L(jωc) (typically less than zero), then P M = 180◦ + φ • Typical stable system needs both GM > 0 and P M > 0 Gain margin 1/GM -1 � Positive phase margin 1 � L(j�) Image by MIT OpenCourseWare. Fig. 5: Gain and Phase Margin for stable system in a polar plot September 15, 2010 Fall 2010 16.30/31 3–11 Im Positive gain margin Im Negative phase margin 1/GM -1 � 1 1 � Re Re � � Positive phase margin 1/GM L(j�) L(j�) Negative gain margin Unstable system Stable system Image by MIT OpenCourseWare. Fig. 6: Gain and Phase Margin in Polar plots Positive gain margin + |L| db 0 _ Log � + |L| db 0 Log � _ �c -90o L Negative gain margin �c -90o -180o -270o Positive phase margin Stable system Log � L -180o -270o Log � Negative phase margin Unstable system Image by MIT OpenCourseWare. Fig. 7: Gain and Phase Margin in Bode plots • Can often predict closed-loop stability looking at the GM and PM September 15, 2010 Fall 2010 16.30/31 3–12 • 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 = −1 -1 � GcG(jω) Fig. 8: 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 com­ plicated, 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) September 15, 2010 Fall 2010 16.30/31 3–13 Nyquist Stability • Key pieces: an encirclement – an accumulation of of 360◦ of phase by a vector (tail at s0) as the tip traverses the contour c ⇒ c encircles s0 Im c Re s0 Image by MIT OpenCourseWare. • We are interested in the plot of L(s) for a very specific set of values of s, called the Nyquist Path. Im R R Re C2 Image by MIT OpenCourseWare. • 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 lims→∞ L(s) = 0, then much of the plot of L(s) for values of s on the Nyquist Path is at the origin. • Nyquist Diagram: plot of L(s) as s moves around the Nyquist path C2 September 15, 2010 Fall 2010 16.30/31 3–14 • 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 closed-loop 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)Gc(s) in the RHP • Z = # closed-loop 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 closed-loop system to be stable (i.e., no closed-loop poles in the RHP), need Z�0 ⇒ N = −P • Note that since P ≥ 0, then would expect CCW encirclements September 15, 2010 Fall 2010 16.30/31 3–15 • 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 s0 is: ∗ Draw a line from s0 ∗ Count # of times that line and the Nyquist plot cross N = #CWcrossings − #CCWcrossings Im G(s) -1 Re Image by MIT OpenCourseWare. • Observation: If the stability of the system is unclear from the Bode diagram, then always revert to the Nyquist plot. September 15, 2010 Fall 2010 16.30/31 3–16 FR: Summary • Bode diagrams are easy to draw • Will see that control design is relatively straight forward as well • Can be a bit complicated to determine stability, but this is a relatively minor problem and it is easily handled using Nyquist plots • Usually only necessary to do one of Bode/Root Locus analysis, but they do provide different perspectives, so I tend to look at both in sisotool. • 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 ∼ 30◦ − 60◦ • Introduced S(s) as a basic measure of system robustness. September 15, 2010