Control Systems I. Lecture 6: Poles and Zeros. Readings: Emilio Frazzoli. Institute for Dynamic Systems and Control D-MAVT ETH Zürich

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1 Control Systems I Lecture 6: Poles and Zeros Readings: Emilio Frazzoli Institute for Dynamic Systems and Control D-MAVT ETH Zürich October 27, 2017 E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/ / 33

2 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 6: Control Systems I 27/10/ / 33

3 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/ / 33

4 Recap We know that the time response of a causal LTI system with state-space model (A, B, C, D) is t y(t) = Ce At x(0) + Ce A(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) = e st is y ss (t) = G(s)e st where G(s) = C(sI A) 1 B + D. E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/ / 33

5 Different ways to write transfer functions For causal, finite-dimensional SISO LTI systems, transfer functions take the form of proper rational functions, e.g., g(s) = N(s) D(s) = b ns n + b n 1 s n 1 + b n 2 s n b 0 s n + a n 1 s n a 0, 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: g(s) = r 1 + r r n + r 0, s p 1 s p 2 s p n where p 1,..., p n are the poles, i.e., the roots of the characteristic polynomial det(si A), i.e., the eigenvalues of A. The numbers r 0,..., r n 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/ / 33

6 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) g(s) = k rl (s z 1)(s z 2)... (s z m) (s p 1)(s p 2)... (s p n) Bode form: This is useful to use control design techniques like the Bode plot (in 2-3 weeks) ( s z g(s) = k 1 + 1)( s z 2 + 1)... ( s Bode ( s p 1 + 1)( s p 2 z m + 1) + 1)... ( s p n + 1) In the above formulas, z 1,..., z n are called the zeros of g(s), and are the roots of the numerator. E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/ / 33

7 Example Computing g(s) via factorization Consider an LTI system with transfer function s + 1 g(s) = 2 s 3 + 4s 2 + 6s + 4 What is the steady-state response to an input u(t) = sin(t) (i.e., s = j)? We know that y ss (t) = g(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 ± j}. s + 1 g(s) = 2 (s + 2)(s j)(s + 1 j). E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/ / 33

8 Computing the magnitude of g(s) Since a b = a b, we can write s + 1 g(s) = 2 s + 2 s j s + 1 j Graphically, s p is the length of the vector from p to s. E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/ / 33

9 Computing the magnitude of g(s) Im j Re From the graph, we find: 2 g(j) = 2 = E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/ / 33

10 Computing the phase (argument) of g(s) Since (a b) = ( a) + ( b), we can write g(s) = (2) + (s + 1) (s + 2) (s 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/ / 33

11 Computing the phase of g(s) Im j Re From the graph, we find: g(j) = arctan(1/2) arctan(2) + 0 = 45. E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/ / 33

12 Putting the results together The steady-state response of an LTI system with transfer function s + 1 g(s) = 2 s 3 + 4s 2 + 6s + 4 to an input of the form u(t) = sin(t) is given by y ss (t) sin(t 45 ). 1 Linear Simulation Results 0.8 y Amplitude Time (seconds) E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/ / 33

13 What is the effect of poles and zeros? E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/ / 33

14 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 ε δ(t) dt = 1 for any ε > 0. 0 In particular, t f (τ)δ(τ) dτ = f (0), for any t > 0. 0 Unit step input u(t) = 1, for t 0. (Note this is the same as u(t) = e 0t.) 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/ / 33

15 Impulse response Assume D = 0, x(0) = 0, and u(t) = δ(t). Then:. y imp (t) = t 0 Ce A(t τ) Bδ(τ) dτ = Ce At B 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/ / 33

16 Unit step response Assume D = 0, x(0) = 0, u(t) = 1 = e 0t, for t 0, and that A is invertible. Then: y step (t) = t 0 Ce A(t τ) B dτ = CA 1 B + CA 1 e At B The steady-state response is given by y ss (t) = G(0) = CA 1 B. For a scalar system, the step response then is simply computed as y step (t) = y ss (t)(1 e at ), 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/ / 33

17 First-order system A system with state-space model (A = a, B = b, C = c, D = 0) The transfer function is with r = bc. g(s) = r s + a The response to an unit impulse (or to an initial condition x(0) = b) has the form y(t) = re at. E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/ / 33

18 Higher-order system If we write the partial fraction expansion of g(s), assuming no repeated poles, we get g(s) = r 1 + r r n. s p 1 s p 2 s p n The response to an impulse will then be y(t) = r 1 e p1t + r 2 e p2t +... r n e pnt. The effect of the poles is then clear: each pole p i generates a term of the form e p i t in the impulse response (and step response, etc.) As we know, these are simple exponentials if the pole p i is real, and are sinusoids with exponentially-changing amplitude for complex-conjugate pole pairs. E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/ / 33

19 Response shapes as function of pole location Im Re Each pole p i = σ i + jω i with residue r i determines a term of the impulse response. Each term s magnitude is bounded by r i e σ i t and oscillates at frequency ω i. E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/ / 33

20 Effects of zeros on the response How can we compute the residues r i? A convenient approach is the cover-up method. For a non-repeated pole p i this takes the form r i = lim s pi (s p i )g(s) which in practice means remove the factor (s p i ) from the denominator and compute g(p i ) only considering the other terms. While the exponents in the terms of the response only depend on the poles p i, the residues are affected by the zeros z i. E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/ / 33

21 Example Consider g(s) = 1 (s + 1)(s j)(s + 1 j). Im Re Using the cover-up method we get g(s) = 1 s /2 s j + 1/2 s + 1 j. E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/ / 33

22 Example - impulse response Impulse Response p1 p2,p3 combined 0.6 Amplitude Time (seconds) E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/ / 33

23 Example adding a zero near a pole Consider g(s) = s ε (s + 1)(s j)(s + 1 j). Im Re Using the cover-up method we get g(s) ε s /2j s j + 1/2j s + 1 j. E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/ / 33

24 Example - impulse response 0.35 Impulse Response Amplitude Time (seconds) A zero can reduce the residue (i.e., the effect) of a nearby pole. E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/ / 33

25 Pole-zero cancellation What if a zero matches a pole exactly? g(s) = s + 1 (s + 1)(s j)(s + 1 j) = 1 (s j)(s + 1 j). One of the poles has been cancelled by the zero. Effectively its residue is zero, i.e., g(s) = 0 s /2j s j + 1/2j s + 1 j. Recall from the modal (diagonal) form that the residue is also given by r i = b i c i ; if the residue is zero, the i-th mode is either uncontrollable, unobservable, or both. This is ok if the i-th mode (i.e., p i ) is stable, but a big problem if it is unstable. Avoid unstable pole-zero cancellation! E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/ / 33

26 More effects of zeros... E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/ / 33

27 Integrator u(t) y(t) = t u(τ) dτ If the input is u(t) = e st, then the output will be y(t) = 1 s est. Hence, the transfer function of an integrator is G(s) = 1 s. You can verify from the state-space model: ẋ(t) = u(t), y(t) = x(t) A = 0, B = 1, C = 1, D = 0, and G(s) = C(sI A) 1 B + D = s 1. E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/ / 33

28 Differentiator u(t) d dt y(t) = du(t) dt If the input is u(t) = e st, then the output will be y(t) = se st. Hence, the transfer function of an integrator is G(s) = s. Note that you cannot obtain this from a state-space model, because a differentiator is not a causal operator! E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/ / 33

29 Zeros as derivative action If we have a transfer function g(s) = (s + z) g(s), we can decompose it into. g(s) = z g(s) + s g(s) If the impulse response of g(s) is given by ỹ(t), and the impulse response of g(s) is y(t), then remembering that s is the transfer function of a differentiator, we can write y(t) = zỹ(t) + ỹ(t). In other words, the zero is effectively adding a derivative term to the output. This typically has an anticipatory effect. E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/ / 33

30 With and without a zero / derivative 1 Impulse Response 0.35 Impulse Response 0.8 p1 p2,p3 combined Amplitude 0.4 Amplitude Time (seconds) Time (seconds) E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/ / 33

31 Non-minimum-phase zeros We know that poles with positive real part result in an unstable system. (The output diverges over time.) What happens when zeros have positive real part? The stability of the system is preserved (since the growth/decay of the terms in the response is not affected by the zeros only the respective residues) However, a zero in the right half plane effectively means a negative derivative action. This is the opposite of anticipatory indeed the output will tend to move in the wrong direction initially. These are called non-minimum phase zeros and are typically very bad news for control engineers, they make our work much harder. (Typically the presence of non-minimum-phase zeros depends on the choice of the output to make your life easier, choose another output and/or move the sensors!) E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/ / 33

32 Minimum-phase vs. non-minimum-phase zeros Impulse Response no zero z = -1 z= Amplitude Time (seconds) E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/ / 33

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/ / 33

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