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Solution Via Matrix Exponential EE263 Autumn 2015 S. Boyd and S. Lall Solution via matrix exponential I matrix exponential I solving x_ = Ax via matrix exponential I state transition matrix I qualitative behavior and stability 1 Matrix exponential define matrix exponential as M 2 eM = I + M + + ··· 2! n×n I converges for all M 2 R I looks like ordinary power series (ta)2 eat = 1 + ta + + ··· 2! with square matrices instead of scalars . 2 Matrix exponential solution of autonomous LDS n×n solution of x_ = Ax, with A 2 R and constant, is x(t) = etAx(0) the matrix etA is called the state transition matrix, usually written Φ(t) generalizes scalar case: solution of x_ = ax, with a 2 R and constant, is x(t) = etax(0) 3 Properties of matrix exponential I matrix exponential is meant to look like scalar exponential I some things you'd guess hold for the matrix exponential (by analogy with the scalar exponential) do in fact hold I but many things you'd guess are wrong example: you might guess that eA+B = eAeB , but it's false (in general) 0 1 0 1 A = ;B = −1 0 0 0 0:54 0:84 1 1 eA = ; eB = −0:84 0:54 0 1 0:16 1:40 0:54 1:38 eA+B = 6= eAeB = −0:70 0:16 −0:84 −0:30 4 Properties of matrix exponential eA+B = eAeB if AB = BA i.e., product rule holds when A and B commute (tA+sA) tA sA thus for t; s 2 R, e = e e with s = −t we get etAe−tA = etA−tA = e0 = I so etA is nonsingular, with inverse −1 etA = e−tA 5 Example: matrix exponential 0 1 let's find etA, where A = 0 0 the power series gives t2A2 t3A3 etA = I + tA + + + ··· 2 3! = I + tA since A2 = 0 1 t = 0 1 1 t we have x(t) = x(0) 0 1 6 Example: Double integrator 0 1 x_ = x 0 0 2 1.5 1 0.5 0 −0.5 −1 −1.5 −2 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 7 Example: Harmonic oscillator 0 1 A = , so state transition matrix is −1 0 t2A2 t3A3 t4A4 etA = I + tA + + + ··· 2 3! 4! t2 t4 t3 t5 = 1 − + − · · · I + t − + − · · · A 2 4! 3! 5! = (cos t)I + (sin t)A cos t sin t = − sin t cos t a rotation matrix (−t radians) cos t sin t so we have x(t) = x(0) − sin t cos t 8 Example 1: Harmonic oscillator 0 1 x_ = x −1 0 2 1.5 1 0.5 0 −0.5 −1 −1.5 −2 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 9 Time transfer property for x_ = Ax we know x(t) = Φ(t)x(0) = etAx(0) interpretation: the matrix etA propagates initial condition into state at time t more generally we have, for any t and τ, x(τ + t) = etAx(τ) (to see this, apply result above to z(t) = x(t + τ)) interpretation: the matrix etA propagates state t seconds forward in time (backward if t < 0) 10 Time transfer property I recall first order (forward Euler) approximate state update, for small t: x(τ + t) ≈ x(τ) + tx_(τ) = (I + tA)x(τ) I exact solution is x(τ + t) = etAx(τ) = (I + tA + (tA)2=2! + ··· )x(τ) I forward Euler is just first two terms in series 11 Sampling a continuous-time system suppose x_ = Ax sample x at times t1 ≤ t2 ≤ · · · : define z(k) = x(tk) then z(k + 1) = e(tk+1−tk)Az(k) for uniform sampling tk+1 − tk = h, so z(k + 1) = ehAz(k); a discrete-time LDS (called discretized version of continuous-time system) 12 Piecewise constant system consider time-varying LDS x_ = A(t)x, with 8 A0 0 ≤ t < t1 <> A(t) = A1 t1 ≤ t < t2 > . : . where 0 < t1 < t2 < ··· (sometimes called jump linear system) for t 2 [ti; ti+1] we have x(t) = e(t−ti)Ai ··· e(t3−t2)A2 e(t2−t1)A1 et1A0 x(0) (matrix on righthand side is called state transition matrix for system, and denoted Φ(t)) 13 Stability we say system x_ = Ax is stable if etA ! 0 as t ! 1 meaning: I state x(t) converges to 0, as t ! 1, no matter what x(0) is I all trajectories of x_ = Ax converge to 0 as t ! 1 fact: x_ = Ax is stable if and only if all eigenvalues of A have negative real part: <λi < 0; i = 1; : : : ; n 14 Stability the `if' part is clear since lim p(t)eλt = 0 t!1 for any polynomial, if <λ < 0 we'll see the `only if' part next lecture more generally, maxi <λi determines the maximum asymptotic logarithmic growth rate of x(t) (or decay, if < 0) 15.
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