Control Systems I Lecture 7: Feedback and the Root Locus method Readings: Guzzella 9.1-3, 13.3 Emilio Frazzoli Institute for Dynamic Systems and Control D-MAVT ETH Zürich November 3, 2017 E. Frazzoli (ETH) Lecture 7: Control Systems I 3/11/2017 1 / 27
Tentative schedule # Date Topic 1 Sept. 22 Introduction, Signals and Systems 2 Sept. 29 Modeling, Linearization 3 Oct. 6 Analysis 1: Time response, Stability 4 Oct. 13 Analysis 2: Diagonalization, Modal coordinates. 5 Oct. 20 Transfer functions 1: Definition and properties 6 Oct. 27 Transfer functions 2: Poles and Zeros 7 Nov. 3 Analysis of feedback systems: internal stability, root locus 8 Nov. 10 Frequency response 9 Nov. 17 Analysis of feedback systems 2: the Nyquist condition 10 Nov. 24 Specifications for feedback systems 11 Dec. 1 Loop Shaping 12 Dec. 8 PID control 13 Dec. 15 Implementation issues 14 Dec. 22 Robustness E. Frazzoli (ETH) Lecture 7: Control Systems I 3/11/2017 2 / 27
Today s learning objectives Standard feedback control cofiguration and transfer function nomenclature Well-posedness, Internal vs. I/O stability The root locus method. E. Frazzoli (ETH) Lecture 7: Control Systems I 3/11/2017 3 / 27
Towards feedback control So far we have looked at how a given system, represented as a state-space model or as a transfer function, behaves given a certain input (and/or initial condition). Typically the system behavior may not be satisfactory (e.g., because it is unstable, or too slow, or too fast, or it oscillates too much, etc.), and one may want to change it. This can only be done by feedback control! The methods we will discuss next provide An analysis tool to understand how the closed-loop system (i.e., the system + feedback control) will behave for di erent choices of feedback control. A synthesis tool to design a good feedback control system. E. Frazzoli (ETH) Lecture 7: Control Systems I 3/11/2017 4 / 27
Standard feedback configuration r e C(s) u P(s) y Transfer functions make it very easy to compose several blocks (controller, plant, etc.) Imagine doing the same with the state-space model! (Open-)Loop gain: L(s) =P(s)C(s) Complementary sensitivity: (Closed-loop) transfer function from r to y T (s) = L(s) 1+L(s) Sensitivity: (Closed-loop) transfer function from r to e S(s) = 1 1+L(s) E. Frazzoli (ETH) Lecture 7: Control Systems I 3/11/2017 5 / 27
Concern #1: Well-posedness r e u y D 1 D 2 Assume both the plant and the controller are (static) gains. Then the denominator of the closed-loop transfer functions would be 1+D 2 D 1.IfD 2 D 1 = 1, the whole interconnections does not make sense it is not well posed. In general, never put in feedback blocks with a non-zero feed-through (i.e., non-zero D term). It is always a pain to deal with (Matlab complains of algebraic loops), and in certain conditions can make the system ill-posed. E. Frazzoli (ETH) Lecture 7: Control Systems I 3/11/2017 6 / 27
Concern #2: (Internal) Stability It may be tempting to just check that the interconnection is I/O stable, i.e., check that the poles of T (s) have negative real part. However, consider what happens if C(s) = 1 s 1 and P(s) = s 1 s+1 : The interconnection is I/O stable: T (s) = 1. s+2 The closed-loop transfer function from r to u is unstable: S(s)C(s) = s+1 (s 1)(s+2). While the system seems to be stable, internally the controller is blowing up. The pole-zero cancellation made the unstable controller mode unobservable in the interconnection. Internal stability requires that all closed-loop transfer functions between any two signals must be stable. Never be tempted to cancel an unstable pole with a non-minimum-phase zero or viceversa. E. Frazzoli (ETH) Lecture 7: Control Systems I 3/11/2017 7 / 27
How to determine closed-loop stability? Assuming the feedback interconnection is well-posed and internally stable, then all that remains to do is to design C(s) in such a way that all the poles of T (s) have negative real part. In principle one could just pick a trial design for C(s), go to matlab, and check what the closed-loop response (e.g., T (s)) looks like. However, it is desired to find a systematic way to choose C(s), while doing as little calculations as possible. The first automatic control engineers were working with paper, pencil, and possibly a slide-rule. All of classical control can be summarized in exploit the knowledge of the loop gain L(s) to figure out the properties of the closed-loop transfer functions T (s) and S(s) with the least e ort possible. E. Frazzoli (ETH) Lecture 7: Control Systems I 3/11/2017 8 / 27
Classical methods for feedback control Remember: the main objective is to assess/design the properties of the closed-loop system by exploiting the knowledge of the open-loop system, and avoiding complex calculations. We have three main methods: Root Locus Quick assessment of control design feasibility. The insights are correct and clear. Can only be used for finite-dimensional systems (e.g. systems with a finite number of poles/zeros) Di cult to do sophisticated design. Hard to represent uncertainty. Nyquist plot The most authoritative closed-loop stability test. It can always be used (finite or infinite-dimensional systems) Easy to represent uncertainty. Di cult to draw and to use for sophisticated design. Bode plots Potentially misleading results unless the system is open-loop stable and minimum-phase. Easy to represent uncertainty. Easy to draw, this is the tool of choice for sophisticated design. E. Frazzoli (ETH) Lecture 7: Control Systems I 3/11/2017 9 / 27
Evan s Root Locus method Invented in the late 40s by Walter R. Evans. Useful to study how the roots of a polynomial (i.e., the poles of a system) change as a function of a scalar parameter, e.g., the gain. r y k L(s) Let us write the loop gain in the root locus form The sensitivity function is kl(s) =k N(s) D(s) = k (s z 1)(s z 2 )...(s z m ) (s p 1 )(s p 2 )...(s p n ) S(s) = 1 1+kL(s) = D(s) D(s)+kN(s) The closed-loop poles are the solutions of the characteristic equation D(s)+kN(s) =0. E. Frazzoli (ETH) Lecture 7: Control Systems I 3/11/2017 10 / 27
The root locus rules What can we say about the closed-loop poles? Since the degree of D(s)+kN(s) is the same as the degree of D(s), the number of closed-loop poles is the same as the number of open-loop poles. For k! 0, D(s)+kN(s) D(s), and the closed-loop poles approach the open-loop poles. For k!1, and the degree of N(s) is the same as the degree of D(s), then 1 k D(s)+N(s) N(s), and the closed-loop poles approach the open-loop zeros. If the degree of N(s) is smaller, then the excess closed-loop poles go to infinity (we will look into this more). The closed-loop poles need to be symmetric wrt the real axis (i.e., either real, or complex-conjugate pairs). E. Frazzoli (ETH) Lecture 7: Control Systems I 3/11/2017 11 / 27
The angle and magnitude rules Let us rewrite the closed-loop characteristic equation as N(s) D(s) = 1 k The angle rule Take the argument on both sides: \(s z 1 )+\(s z 2 )+...+ \(s z m ) \(s p 1 ) \(s p 2 )... \(s p n )= 180 (±q 360 )ifk > 0 0 (±q 360 )ifk < 0 The magnitude rule Take the argument on both sides: s z 1 s z 2... s z m s p 1 s p 2... s p n = 1 k E. Frazzoli (ETH) Lecture 7: Control Systems I 3/11/2017 12 / 27
Graphical interpretation All points on the complex plane that could potentially be a closed-loop pole (i.e., the root locus) have to satisfy the angle condition which is essentially THE rule for sketching the root locus. Im s \s p 2 \s z 1 Re \s p 1 The sum of the angles (counted from the real axis) from each zero to s, minus the sum of the angles from each pole to s must be equal to 180 (for positive k) or0 (for negative k), ± an integer multiple of 360. E. Frazzoli (ETH) Lecture 7: Control Systems I 3/11/2017 13 / 27
All points on the real axis are on the root locus. All points on the real axis to the left of an even number of poles/zeros (or none) are on the negative k root locus All points on the real axis to the right of an odd number of poles/zeros are on the positive k root locus. When two branches come together on the real axis, there will be breakaway or break-in points. E. Frazzoli (ETH) Lecture 7: Control Systems I 3/11/2017 14 / 27
Asymptotes So what happens when k!1and there are more open-loop poles than zeros? We can see, e.g., from the magnitude condition, that the excess closed-loop poles will have to go to to infinity. Since this is the complex plane, we need to identify in which direction they go towards infinity. This is were we use the angle rule again. If we zoom out su ciently far, the contributions from all the finite open-loop poles and zeros will all be approximately equal to \s, andthe angle rule is approximated by (m n)\s = \ k ± q360. In other words, as k!1, the excess poles will go to infinity along asymptotes at angles of \s = \ k ± q360 = \ k ± q360 m n n m These asymptotes meet in a center of mass lying on the real axis at P n i=1 s com = p P m i j=1 z j n m (note that poles are positive unit masses, and zeros are negative unit masses in this analogy) E. Frazzoli (ETH) Lecture 7: Control Systems I 3/11/2017 15 / 27
Examples r k 1 s 1 y Im Re E. Frazzoli (ETH) Lecture 7: Control Systems I 3/11/2017 16 / 27
Examples r k 1 (s 1)(s 2) y Im Re E. Frazzoli (ETH) Lecture 7: Control Systems I 3/11/2017 17 / 27
Examples r k 1 s 2 +s+1 y Im Re E. Frazzoli (ETH) Lecture 7: Control Systems I 3/11/2017 18 / 27
Examples r k s+1 s 2 +s+1 y Im Re E. Frazzoli (ETH) Lecture 7: Control Systems I 3/11/2017 19 / 27
Examples r k 1 (s+1)(s 2 +s+1) y Im Re E. Frazzoli (ETH) Lecture 7: Control Systems I 3/11/2017 20 / 27
Examples r k s+1 s 2 +s+1 y Im Re E. Frazzoli (ETH) Lecture 7: Control Systems I 3/11/2017 21 / 27
Examples r k 1 (s+1)(s 2 +s+1) y Im Re E. Frazzoli (ETH) Lecture 7: Control Systems I 3/11/2017 22 / 27
Stabilizing an inverted pendulum The equations of motion for an inverted pendulum (see lecture 3) can be written as ml 2 = mgl + u; The transfer function is given by G(s) = 1 s 2 g/l. How does the pendulum s behavior change if I attach a spring to the pendulum? E. Frazzoli (ETH) Lecture 7: Control Systems I 3/11/2017 23 / 27
Proportional control of an inverted pendulum r k 1 s 2 g/l y Im Re E. Frazzoli (ETH) Lecture 7: Control Systems I 3/11/2017 24 / 27
Proportional-derivative control of an inverted pendulum r k p + k d s 1 s 2 g/l y Im Re E. Frazzoli (ETH) Lecture 7: Control Systems I 3/11/2017 25 / 27
Root locus summary (for now) Great tool for back-of-the-envelope control design, quick check for closed-loop stability. Qualitative sketches are typically enough. There are many detailed rules for drawing the root locus in a very precise way: if you really need to do that, just use matlab or other methods. Closed-loop poles start from the open loop poles, and are repelled by them. Closed-loop poles are attracted by zeros (or go to infinity). Here you see an obvious explanation why non-minimum-phase zeros are in general to be avoided. Remember that the root locus must be symmetric wrt the real axis. E. Frazzoli (ETH) Lecture 7: Control Systems I 3/11/2017 26 / 27
Today s learning objectives Standard feedback control cofiguration and transfer function nomenclature Well-posedness, Internal vs. I/O stability The root locus method. E. Frazzoli (ETH) Lecture 7: Control Systems I 3/11/2017 27 / 27