6.241 Dynamic Systems and Control Lecture 9: Transfer Functions Readings: DDV, Chapters 10, 11, 12 Emilio Frazzoli Aeronautics and Astronautics Massachusetts Institute of Technology March 2, 2011 E. Frazzoli (MIT) Lecture 9: Transfer Functions Mar 2, 2011 1 / 13Asymptotic Stability (Preview) We have seen that the unforced state response (u = 0) of a LTI system is easily computed using the “A” matrix in the state-space model: x[k] = Ak x[0], or x(t) = e At x(0). A system is asymptotically stable if limt +∞ x(t) = 0, for all x0. →Assume A is diagonalizable, i.e., V −1AV = Λ, and let r = Vx be the vector of model coordinates. Then, ri [k] = λki ri [0], or ri (t) = e λit ri (0), i = 1, . . . , n. Clearly, for the system to be asymptotically stable, |λi | < 1 (DT) or Re(λi ) < 0 (CT) for all i = 1, . . . , n. It turns out that this condition extends to the general (non-diagonalizable) case. More on this later in the course. E. Frazzoli (MIT) Lecture 9: Transfer Functions Mar 2, 2011 2 / 13� � � (Time-domain) Response of LTI systems — summary Based on the discussion in previous lectures, the solution of initial value problems (i.e., the response) for LTI systems can be written in the form: k−1y[k] = CAk x[0] + C Ak−i−1Bu[i] + Du[t] i=0 or � t y(t) = C exp(At)x(0) + C exp(A(t − τ))Bu(τ ) dτ + Du(t). 0 However, the convolution integral (CT) and the sum in the DT equation are hard to interpret, and do not offer much insight. 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 (MIT) Lecture 9: Transfer Functions Mar 2, 2011 3 / 13� � The continuous-time case: elementary inputs Let us choose as elementary input u(t) = u0est , 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) = u0ejωt + u0e−jωt = 2u0 cos(ωt); in other words, if s is imaginary, then u(t) = est must be understood as a “half” of a sinusoidal signal. if s = σ + jω, then u(t) + u∗(t) = u0(e σt ejωt + u0e σt e−jωt ) = u0(e σt ejωt + e−jωt ) = 2u0e σt cos(ωt), and the input u is a “half” of a sinusoid with exponentially-changing amplitude. E. Frazzoli (MIT) Lecture 9: Transfer Functions Mar 2, 2011 4 / 13� � Output response to elementary inputs (1/2) Recall that, � t y(t) = CeAt x(0) + C e A(t−τ )Bu(τ ) dτ + Du(t). 0 Plug in u(t) = u0est : � t y(t) = CeAt x(0) + C e A(t−τ )Bu0e sτ dτ + Du0e st 0 �� t � = CeAt x(0) + C e(sI −A)τ dτ e At Bu0 + Du0e st 0 If (sI − A) is invertible (i.e., s is not an eigenvalue of A), then y(t) = CeAt x(0) + C (sI − A)−1 e(sI −A)t − I e At Bu0 + Du0e st . E. Frazzoli (MIT) Lecture 9: Transfer Functions Mar 2, 2011 5 / 13� � Output response to elementary inputs (2/2) Rearranging: y(t) = CeAt x(0) − C (sI − A)−1 e At Bu0 + C (sI − A)−1B + D u0e st . � �� �� �� � Transient response Steady−state response If the system is asymptotically stable, eAt 0, and the transient response →will converge to zero. The steady state response can be written as: yss = G (s)e st , G (s) ∈ Cny ×nu , where G (s) = C (sI − A)−1B + D is a complex matrix. The function G : s G (s) is also known as the transfer function: it → st stdescribes how the system transforms an input e into the output G (s)e . E. Frazzoli (MIT) Lecture 9: Transfer Functions Mar 2, 2011 6 / 13Laplace Transform The (one-sided) Laplace transform F : C C of a sequence f : R R is→ ≥0 →defined as � +∞ F (s) = f (t)e−st dt, 0 for all s such that the series converges (region of convergence). Given the above definition, and the previous discussion, Y (s) = G (s)U(s). U(s)e st Y (s)e st = G (s)U(s)e st ⇒ Also, G (s) is the Laplace transform of the “impulse” response. E. Frazzoli (MIT) Lecture 9: Transfer Functions Mar 2, 2011 7 / 13� � � � � The discrete-time case: elementary inputs Let us choose as elementary input u[k] = u0zk , where z ∈ C is a complex number. If z is real, then u is a simple geometric sequence. Recall k−1y[k] = CAk x[0] + C Ak−i−1Bu[i] + Du[k]. i=0 Plug in u[k] = u0zk , and substitute l = k − i − 1: k−1y[k] = CAk x[0] + C Al Bu0z k−l−1 + Du0z k l=0 k−1= CAk x[0] + Czk−1 (Az−1)i Bu0 + Du0z k . i=0 E. Frazzoli (MIT) Lecture 9: Transfer Functions Mar 2, 2011 8 / 13� � � � Matrix geometric series Recall the formula for the sum of a geometric series: k−1i 1 − mk m = . 1 − m i=0 For a matrix: k−1Mi = I + M + M2 + . . . Mk−1 . i=0 k−1 Mi (I − M) = (I + M + M2 + . . . Mk−1)(I − M) = I − Mk . i=0 i.e., k−1Mi = (I − Mk )(I − M)−1 . i=0 E. Frazzoli (MIT) Lecture 9: Transfer Functions Mar 2, 2011 9 / 13� � � � Discrete Transfer Function Using the result in the previous slide, we get y[k] = CAk x[0] + Czk−1(I − Ak z−k )(I − Az−1)−1Bu0 + Du0z k = CAk x[0] + C(z k I − Ak )(zI − A)−1Bu0 + Du0z k . Rearranging: y[k] = CAk x[0] − (zI − A)−1Bu0 + C (zI − A)−1B + D u0z k . � �� � � �� � Transient response Steady−state response If the system is asymptotically stable, the transient response will converge to zero. The steady state response can be written as: yss[k] = G (z)z k , G (z) ∈ C, where G (z) = C (zI − A)−1B + D is a complex number. The function G : z G (z) is also known as the (pulse, or discrete) transfer function: it → k kdescribes how the system transforms an input z into the output G (z)z . E. Frazzoli (MIT) Lecture 9: Transfer Functions Mar 2, 2011 10 / 13� Z-Transform The (one-sided) z-transform F …