Solutions to EE422G Homework #3: Laplace Transform and Differential Equations, Assignments of Signals and Systems

Download Solutions to EE422G Homework #3: Laplace Transform and Differential Equations and more Assignments Signals and Systems in PDF only on Docsity! EE422G Homework #3 Solution 1. (2 points) (a) By definition, we have L [∫ t −∞ x(τ)dτ ] = ∫ ∞ 0 ∫ t −∞ x(τ)dτe−stdt = ∫ ∞ 0 ∫ t −∞ x(τ)dτd (−e−st s ) Again, we will use integration by parts: ∫ a b u(t)dv(t) = u(t)v(t)|t=bt=a − ∫ b a v(t)du(t) Make the following substitutions: −e−st s ⇒ v(t) ∫ t −∞ x(τ)dτ ⇒ u(t) Then, L [∫ t −∞ x(τ)dτ ] = ∫ ∞ 0 ∫ t −∞ x(τ)dτd (−e−st s ) = (∫ t −∞ x(τ)dτ ) −e−st s ∣∣∣∣ t=∞ t=0 − ∫ ∞ 0 −e−st s d (∫ t −∞ x(τ)dτ ) = lim t→∞ −e−st s ∫ t ∞ x(τ)dτ + e−s0 s ∫ 0 −∞ x(τ)dτ − ∫ ∞ 0 e−st s x(t)dt = 0 + y(0) s − X(s) s (b) For simplicity, we assume that all the initial conditions are zero. If you assume otherwise, your answer may be different. L−1 [ s2Y (s) + 2sY (s) ] = L−1 [ s3X(s) + X(s) s − e −s s ] d2y dt2 + 2 dy dt = d3x dt3 + ∫ t −∞ x(τ)dτ − u(t− 1) 2. (2 points) The strategy is to first apply Laplace Transform to both differential equa- tions turning them into algebraic equations involving X(s) and Y (s), then solve for X(s) and Y (s) and finally convert them back to time domain by applying inverse Laplace Transform. 1 sX(x)− x(0−) + 3X(s) + 2Y (s) = 1s with x(0) = 0, we get (s + 3)X(s) + 2Y (s) = 1 s sY (s)− y(0−)−X(s) = 0 with y(0) = 0, we get sY (s)−X(s) = 0 Solving for X(s) and Y (s), we get X(s) = 1 s2 + 3s + 2 = 1 (s + 1)(s + 2) Y (s) = 1 s3 + 3s2 + 2s = 1 s(s + 1)(s + 2) To apply the inverse Laplace Transform, we expand X(s) and Y (s) into partial frac- tions: X(s) = a s + 1 + b s + 2 Y (s) = c s + d s + 1 + e s + 2 Heaviside’s theorem allows us to compute a, b, c, d, and e: a = (X(s)(s + 1)) |s=−1 = 1 b = (X(s)(s + 2)) |s=1 = −1 c = (Y (s)s) |s=0 = 1/2 d = (Y (s)(s + 1)) |s=−1 = −1 e = (Y (s)(s + 2)) |s=−2 = 1/2 Thus, we have X(s) = 1 s + 1 − 1 s + 2 Y (s) = 1 2s − 1 s + 1 + 1/2 s + 2 Taking inverse Laplace Transform will result in x(t) = e−t − e−2t y(t) = 1 2 u(t)− e−t + 1 2 e−2t I will not take off any points if you use MATLAB to perform the inverse Laplace transform. 2