Control Systems I. Lecture 5: Transfer Functions. Readings: Emilio Frazzoli. Institute for Dynamic Systems and Control DMAVT ETH Zürich


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1 Control Systems I Lecture 5: Transfer Functions Readings: Emilio Frazzoli Institute for Dynamic Systems and Control DMAVT ETH Zürich October 20, 2017 E. Frazzoli (ETH) Lecture 5: Control Systems I 20/10/ / 24
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 5: Control Systems I 20/10/ / 24
3 Vibrations of a string E. Frazzoli (ETH) Lecture 5: Control Systems I 20/10/ / 24
4 Today s learning objectives Response to exponential inputs Transfer functions Conversions from StateSpace models to/from Transfer Functions Laplace transform and inverse Laplace transform E. Frazzoli (ETH) Lecture 5: Control Systems I 20/10/ / 24
5 Motivation Real System: Modeling: Linearization/Normalization: Transfer Function(Today!) Easy Easy Hard Control System Design E. Frazzoli (ETH) Lecture 5: Control Systems I 20/10/ / 24
6 Statespace formulation of LTI response Recall: y(t) = Ce At x(0) + C t 0 e A(t τ) Bu(τ) dτ + Du(t). This formula gives a complete characterization of the output response of a LTI system. We have fully described the homogeneous solution (Ce At x(0)), but the convolution integral is hard to interpret. In order to gain a better understanding, we will study the response to elementary inputs of a form that is particularly easy to analyze: the output has the same form as the input. very rich and descriptive: most signals/sequences can be written as linear combinations of such inputs. Then, using the superposition principle, we will recover the response to general inputs, written as linear combinations of the easy inputs. E. Frazzoli (ETH) Lecture 5: Control Systems I 20/10/ / 24
7 Exponential inputs Let us choose as elementary input u(t) = e st, where s C is a complex number. If s is real, then u is a simple exponential. If s = jω is imaginary, then the elementary input must always be accompanied by the conjugate, i.e., u(t) + u (t) = e jωt + e jωt = 2 cos(ωt); in other words, if s is imaginary, then u(t) = e st must be understood as a half of a sinusoidal signal. if s = σ + jω, then u(t) + u (t) = e σt e jωt + e σt e jωt = e σt ( e jωt + e jωt) = 2e σt cos(ωt), and the input u is a half of a sinusoid with exponentiallychanging amplitude. E. Frazzoli (ETH) Lecture 5: Control Systems I 20/10/ / 24
8 Output response to elementary inputs (1/2) Recall y(t) = Ce At x(0) + C t 0 e A(t τ) Bu(τ) dτ + Du(t). Plug in u(t) = e st : y(t) = Ce At x(0) + C t 0 e A(t τ) Be sτ dτ + De st t = Ce At x(0) + Ce At e (si A)τ B dτ + De st 0 If (si A) is invertible (i.e., s is not an eigenvalue of A), then [ ] y(t) = Ce At x(0) + Ce At (si A) 1 e (si A)t I B + De st. E. Frazzoli (ETH) Lecture 5: Control Systems I 20/10/ / 24
9 Output response to elementary inputs (2/2) Rearranging: y(t) = Ce At [ x(0) (si A) 1 B ] + [ C(sI A) 1 B + D ] e st. }{{}}{{} Transient response Steady state response If the system is asymptotically stable, the transient response will converge to zero. The steady state response to an input u(t) = e st can be written as: y ss = G(s)e st, G(s) C, where G(s) = C(sI A) 1 B + D is a complex number. The function G : s G(s) is also known as the transfer function: it describes how the system transforms an input e st into the output G(s)e st. E. Frazzoli (ETH) Lecture 5: Control Systems I 20/10/ / 24
10 Frequency response Consider the case in which the input is a sinusoidal signal (s = ±jω): u(t) = e jωt + e jωt = 2 cos(ωt). The output is u(t) = G(jω)e jωt + G( jω)e jωt = Me φ e jωt + Me φ e jωt = 2M cos(ωt + φ). In other words, the output is another sinusoid of frequency ω, the magnitude of which is M = G(jω) times the magnitude of the input, and with phase leading the input by φ = G(jω). E. Frazzoli (ETH) Lecture 5: Control Systems I 20/10/ / 24
11 Linearity Review Using linearity and the solution for u(t) = e st we can find the solution to more complex inputs. Remember, if y 1(t) is the output for u 1(t), y 2(t) is the output for u 2(t), then αy 1(t) + βy 2(t) is the output for αu 1(t) + βu 2(t). E. Frazzoli (ETH) Lecture 5: Control Systems I 20/10/ / 24
12 Building more general inputs from e st Suppose we have two inputs u 1(t) = U 1e s 2t and u 2(t) = U 2e s 2t. The corresponding outputs are Then if y 1(t) = g(s 1)U 1e s 1t and y 2(t) = g(s 2)U 2e s 2t, u(t) = U 1e s 1t + U 2e s 2t, we get the output y(t) = g(s 1)U 1e s 1t + g(s 2)U 2e s 2t. More generally, u(t) = i U i e s i t y(t) = i g(s i )U i e s i t (1) E. Frazzoli (ETH) Lecture 5: Control Systems I 20/10/ / 24
13 Example (1/4) ẋ = x + u, y = x. A = [ 1] B = [1] C = [1] D = [0] g(s) = C(sI A) 1 B + D = 1 (s ( 1)) = s+1 The transfer function of an LTI system will always be a ratio of two polynomials in s. What can you say about the eigenvalues of A and the values of s where g(s)? Let s find the steady state output y(t) for the input u(t) = cos(ωt) E. Frazzoli (ETH) Lecture 5: Control Systems I 20/10/ / 24
14 Example (2/4) We can write u(t) as u(t) = cos(ωt) Notice that U 1,2 = 1 2 and s 1,2 = ±jω Then = 1 2 ejωt }{{} u 1(t) e jωt }{{} u 2(t) y 1 (t) = g(s 1 )U 1 e s1t = g(jω) 1 2 ejωt = e jωt 2(1+jω) y 2 (t) = g(s 2 )U 1 e s2t = g( jω) 1 2 e jωt = e jωt 2(1 jω) E. Frazzoli (ETH) Lecture 5: Control Systems I 20/10/ / 24
15 Example (3/4) y(t) = y 1 (t) + y 2 (t) = e jωt 2(1 jω) + ejωt 2(1+jω) = ejωt (1 jω)+e jωt (1+jω) 2(1+ω 2 ) = e jωt +e jωt +jω(e jωt e jωt ) 2(1+ω 2 ) = 1 1+ω 2 (cos(ωt) + ω sin(ωt)) Sinusoid in  sinusoid out The system changes the amplitude and phase E. Frazzoli (ETH) Lecture 5: Control Systems I 20/10/ / 24
16 Example (4/4) E. Frazzoli (ETH) Lecture 5: Control Systems I 20/10/ / 24
17 From Statespace to Transfer Function Recall, the transfer function is defined as g(s) = C(sI A) 1 B + D If A is diagonal, with eigenvalues λ 1, λ 2,..., λ n, this simply becomes g(s) = c 1b 1 + c 2b c nb n. s λ 1 s λ 2 s λ n In general, the transfer function is a rational function of the form g(s) = b n 1s n 1 + b n 2 s n b 0 s n + a n 1 s n a 0 + d. Note: the denominator of g(s) is the characteristic polynomial of the matrix A, i.e., det(si A). E. Frazzoli (ETH) Lecture 5: Control Systems I 20/10/ / 24
18 From Transfer Function to Statespace (1/2) Given a transfer function g(s), there are many statespace models (A, B, C, D) such that g(s) = C(sI A) 1 B + D. We are interested in minimal realizations, i.e., (A, B, C, D) matrices that are controllable, observable, and generate the given transfer function. If the transfer function is written as a partial fraction expansion of the form g(s) = p 1 + p p n + d, s λ 1 s λ 2 s λ n then a realization is A = λ 1... λ n, B = p1. pn C = [ p1... pn ], D = d. E. Frazzoli (ETH) Lecture 5: Control Systems I 20/10/ / 24
19 From Transfer Function to Statespace (2/2) In the general case g(s) = b n 1s n 1 + b n 2 s n b 0 s n + a n 1 s n a 0 + d, you can verify that the following is a minimal realization of g(s): A =.... 1, B = 0. a 0 a 1... a n 1 1 C = [ b 0 b 1... b n 1 ], D = [d]; This particular realization is called the controllable canonical form. E. Frazzoli (ETH) Lecture 5: Control Systems I 20/10/ / 24
20 The Laplace Transform (1/3) Can all inputs u(t) be expressed as a (infinite) sum of complex exponentials? Yes! The tool for this is the Laplace transform and the inverse Laplace transform. To denote the Laplace transform of a signal u, we write L [u] = U, (2) The Laplace transform of u is defined (don t worry you will never have to calculate these integrals), U(s) := 0 u(t)e st dt (3) E. Frazzoli (ETH) Lecture 5: Control Systems I 20/10/ / 24
21 The Laplace Transform (2/3) The inverse Laplace transform is denoted L 1 [U] = u, (4) and is defined, u(t) = 1 2πj lim ω σ+jω σ jω U(s)e st ds (5) Notice that the inverse Laplace transform tells us how to write u(t) as a linear combination of e st weighted by U(s). E. Frazzoli (ETH) Lecture 5: Control Systems I 20/10/ / 24
22 The Laplace Transform (3/3) Remember, u(t) = i U i e s i t y(t) = i g(s i )U i e s i t Similarly we have (notice that the sum is replaced by an integral), u(t) = 1 2πj lim ω σ+jω σ jω U(s)e st ds y(t) = 1 2πj lim ω σ+jω σ jω g(s)u(s)e st ds. We can easily obtain the output for (almost) any input by using the Laplace transform. E. Frazzoli (ETH) Lecture 5: Control Systems I 20/10/ / 24
23 Systems represented in the frequency domain We can write the output as y(t) = 1 2πj lim ω σ+jω σ jω g(s)u(s)e st ds. Notice that this is y(t) = L 1 [gu] (t) Taking the Laplace transform we find L[y](s) = g(s)u(s) Y (s) = g(s)u(s) Once we are comfortable with looking at the signals in the frequency domain, we obtain the (steadystate) output by simply multiplying the input by the transfer function! E. Frazzoli (ETH) Lecture 5: Control Systems I 20/10/ / 24
24 Today s learning objectives Response to exponential inputs Transfer functions Conversions from StateSpace models to/from Transfer Functions Laplace transform and inverse Laplace transform E. Frazzoli (ETH) Lecture 5: Control Systems I 20/10/ / 24
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