Laplace Transform and Inverse Laplace Transform: Example Problems and Solutions - Prof. Ha, Study notes of Electrical and Electronics Engineering

Download Laplace Transform and Inverse Laplace Transform: Example Problems and Solutions - Prof. Ha and more Study notes Electrical and Electronics Engineering in PDF only on Docsity! EEL4657 - Dr. Haniph Latchman Chapter 2: Example Problems 1. Given x(t) = e−3t use the definition of the Laplace Transform to find X(s). Solution The Laplace Transform for a signal x(t), t ≥ 0− is defined by: L[x(t)] = X(s) = ∫ ∞ 0− x(t)e−stdt Thus, substituting for x(t) and solving the integral will yield X(s) 1 X(s) = L[e−3t] = ∫ ∞ 0− e−3te−stdt = ∫ ∞ 0− e−st−3tdt = ∫ ∞ 0− e(−s−3)tdt = ∫ ∞ 0− e−(s+3)tdt = −1 s + 3 [ e−(s+3)t ∣∣∣∣∞ 0− ] = −1 s + 3 [ e−(s+3)∞ − e−(s+3)0 ] = −1 s + 3 [0− 1] = −1 s + 3 [−1] X(s) = 1 s + 3 2 Solve for b, c, d 8b = 16 b = 2 b + c = 0 2 + c = 0 c = −2 4b + d = 4 4(2) + d = 4 8 + d = 4 d = −4 Substituting the above values into V (s) gives: V (s) = 2 s + −2s− 4 s2 + 4s + 8 = 2 s − 2s + 4 s2 + 4s + 8 = 2 s − 2(s + 2) (s + 2)2 + 4 = 2 s − 2 · s + 2 (s + 2)2 + 22 Apply the Inverse Laplace Transform to find v(t) L−1[V (s)] = v(t) 5 4. Given the transfer function of a system: T (s) = s + 1 s2 + 2s + 2 Find the time response of the output y(t) when the input is a unit step function. (Use Inverse Laplace Transform techniques) Solution Since the input is a unit step function, R(s) = 1s The output Y (s) is given by: Y (s) = R(s)T (s) = 1 s · s + 1 s2 + 2s + 2 = s + 1 s(s2 + 2s + 2) Use partial fraction decomposition to convert Y (s) s + 1 s(s2 + 2s + 2) = b s + cs + d s2 + 2s + 2 s + 1 = ( b s ) [ s(s2 + 2s + 2) ] + ( cs + d s2 + 2s + 2 ) [ s(s2 + 2s + 2) ] s + 1 = (b)(s2 + 2s + 2) + (cs + d)(s) s + 1 = bs2 + 2bs + 2b + cs2 + ds s + 1 = bs2 + cs2 + 2bs + ds + 2d s + 1 = (b + c)s2 + (2b + d)s + 2b 6 Equate the coefficients of the s-powers s2 : b + c = 0 s1 : 2b + d = 1 s0 : 2b = 1 Solve for b, c, d 2b = 1 b = 1 2 b + c = 0 1 2 + c = 0 c = −1 2 2b + d = 1 2 ( 1 2 ) + d = 1 1 + d = 1 d = 0 Substituting the above values into Y (s) gives: Y (s) = 1 2 s + −1 2 s + 0 s2 + 2s + 2 = 1 2 ( 1 s ) − 1 2 ( s s2 + 2s + 2 ) = 1 2 ( 1 s ) − 1 2 ( s (s + 1)2 + 1 ) Apply the Inverse Laplace transform to determine the time response of the output, y(t) L−1[Y (s)] = y(t) 7