Control Systems I Lecture 6: Poles and Zeros Readings: Emilio Frazzoli Institute for Dynamic Systems and Control D-MAVT ETH Z¨urich October 27, 2017 E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/2017 1 / 33 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 coordi- nates. 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 6: Control Systems I 27/10/2017 2 / 33 Today's learning objectives Recognize different ways of writing transfer functions, and why. Graphical computation of g(s). Poles and their effects on the response. Zeros and their effects on the response. Zeros and derivative action Effects of non-minimum-phase zeros E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/2017 3 / 33 Recap We know that the time response of a causal LTI system with state-space model (A; B; C; D) is Z t y(t) = CeAt x(0) + CeA(t−τ)Bu(τ) dτ + Du(t): 0 Assuming all the eigenvalues of A have negative real part (i.e., the system is stable), the steady-state response to an input of the form u(t) = est is st yss(t) = G(s)e where G(s) = C(sI − A)−1B + D: E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/2017 4 / 33 Different ways to write transfer functions For causal, finite-dimensional SISO LTI systems, transfer functions take the form of proper rational functions, e.g., n n−1 n−2 N(s) bns + bn−1s + bn−2s + ::: + b0 g(s) = = n n−1 ; D(s) s + an−1s + ::: + a0 There are other forms that are convenient. The first one, called the partial fraction expansion, is useful to compute transient responses, and to assess how much different modes contribute to the response. It has the form: r1 r2 rn g(s) = + + ::: + + r0; s − p1 s − p2 s − pn where p1;:::; pn are the "poles", i.e., the roots of the characteristic polynomial det(sI − A), i.e., the eigenvalues of A. The numbers r0;:::; rn are called the \residues." There are many ways to compute the residues, we will look at one later today. E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/2017 5 / 33 Different ways to write transfer functions There are also slightly different ways to factorize the polynomials at the numerator and denominator: Root-locus form: This is useful to compute the value of g(s) \by hand", and to use control design techniques like the root locus (next week) (s − z1)(s − z2) ::: (s − zm) g(s) = krl (s − p1)(s − p2) ::: (s − pn) Bode form: This is useful to use control design techniques like the Bode plot (in 2-3 weeks) ( s + 1)( s + 1) ::: ( s + 1) g(s) = k −z1 −z2 −zm Bode ( s + 1)( s + 1) ::: ( s + 1) −p1 −p2 −pn In the above formulas, z1;:::; zn are called the \zeros" of g(s), and are the roots of the numerator. E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/2017 6 / 33 Example { Computing g(s) via factorization Consider an LTI system with transfer function s + 1 g(s) = 2 s3 + 4s2 + 6s + 4 What is the steady-state response to an input u(t) = sin(t) (i.e., s = j)? We know that yss(t) = jg(j)j sin(t + \g(j)) We could substitute s j in the transfer function and compute g(j), but that is not a nice/insightful calculation. Let's use graphical methods instead. Factorize the transfer function, and write in the root-locus form: with some creative eyeballing (or using matlab) we can find that the 3 poles of g(s) are at {−2; −1 ± jg. s + 1 g(s) = 2 : (s + 2)(s + 1 + j)(s + 1 − j) E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/2017 7 / 33 Computing the magnitude of g(s) Since ja · bj = jaj · jbj, we can write js + 1j jg(s)j = 2 js + 2j · js + 1 + jj · js + 1 − jj Graphically, js − pj is the length of the vector from p to s. E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/2017 8 / 33 Computing the magnitude of g(s) Im j Re From the graph, we find: p p 2 2 2 jg(j)j = 2p p = ≈ 0:5657: 5 · 5 · 1 5 E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/2017 9 / 33 Computing the phase (argument) of g(s) Since \(a · b) = (\a) + (\b), we can write \g(s) = \(2) + \(s + 1) − \(s + 2) − \(s + 1 + j) − \(s + 1 − j) Graphically, \(s − p) is angle formed by the vector from p to s with the real axis. E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/2017 10 / 33 Computing the phase of g(s) Im j Re From the graph, we find: ◦ ◦ ◦ \g(j) = 0 + 45 − arctan(1=2) − arctan(2) + 0 = −45 : E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/2017 11 / 33 Putting the results together The steady-state response of an LTI system with transfer function s + 1 g(s) = 2 s3 + 4s2 + 6s + 4 to an input of the form u(t) = sin(t) is given by ◦ yss(t) ≈ 0:5657 sin(t − 45 ): Linear Simulation Results 1 y 0.8 0.6 0.4 0.2 0 Amplitude -0.2 -0.4 -0.6 -0.8 -1 0 5 10 15 20 25 30 35 Time (seconds) E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/2017 12 / 33 What is the effect of poles and zeros? E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/2017 13 / 33 Transient response to special inputs In order to understand the effect of poles and zeros, it is useful to apply some other standard \test inputs" to a system to evaluate its transient behavior (vs. the steady-state one). Typical \test inputs' include: Unit impulse u(t) = δ(t). This is not really a function, but a mathematical construct such that R " 0 δ(t) dt = 1 for any " > 0. R t In particular, 0 f (τ)δ(τ) dτ = f (0), for any t > 0. Unit step input u(t) = 1, for t ≥ 0. (Note this is the same as u(t) = e0t .) Other less used test inputs include the \unit ramp" u(t) = t and higher order ramps. E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/2017 14 / 33 Impulse response Assume D = 0, x(0) = 0, and u(t) = δ(t). Then: Z t A(t−τ) At yimp(t) = Ce Bδ(τ) dτ = Ce B 0 . The impulse response is the same as the response to an initial condition x(0) = B. Remember: the initial-condition response is given by simple exponentials for real eigenvalues, and sinusoids with exponentially-changing magnitude for complex-conjugate eigenvalues. E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/2017 15 / 33 Unit step response Assume D = 0, x(0) = 0, u(t) = 1 = e0t , for t ≥ 0, and that A is invertible. Then: Z t A(t−τ) −1 −1 At ystep(t) = Ce B dτ = −CA B + CA e B 0 −1 The steady-state response is given by yss (t) = G(0) = −CA B. For a scalar system, the step response then is simply computed as at ystep(t) = yss(t)(1 − e ); i.e., the step response is the steady-state response minus the scaled impulse response. The impulse response totally defines the response of a system (it is in fact the inverse Laplace transform of the transfer function)! E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/2017 16 / 33 First-order system A system with state-space model (A = a; B = b; C = c; D = 0) The transfer function is r g(s) = s + a with r = bc. The response to an unit impulse (or to an initial condition x(0) = b) has the form y(t) = reat : E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/2017 17 / 33 Higher-order system If we write the partial fraction expansion of g(s), assuming no repeated poles, we get r r r g(s) = 1 + 2 + ::: + n : s − p1 s − p2 s − pn The response to an impulse will then be p1t p2t pnt y(t) = r1e + r2e + ::: rne : The effect of the poles is then clear: each pole pi generates a term of the form epi t in the impulse response (and step response, etc.) As we know, these are simple exponentials if the pole pi is real, and are sinusoids with exponentially-changing amplitude for complex-conjugate pole pairs. E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/2017 18 / 33 Response shapes as function of pole location Im Re Each pole pi = σi + j!i with residue ri determines a term of the impulse response. σi t Each term's magnitude is bounded by ri e and oscillates at frequency !i . E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/2017 19 / 33 Effects of zeros on the response How can we compute the residues ri ? A convenient approach is the \cover-up" method.