Question

For first order ODE’s replace right hand side with f(t) and determine the transfer function. By definition transfer function’s are related to a forced response and not intial conditions.

(a) 3x+4x = 15, t>0, x(0)=0 (b) 2x + x = 4, t>0, x(0)=3 (c) 2x + x = 3u(t), t>0, x(0) = 3

Answer 1

(a) 3 dx/dt + 4x = 15, x(0) = 0____(1) Apply laplace transform on both side we get 3L[dx/dt] + 4L [x] = L[15] 3[S phi (s) - x(0)] + 4 phi (s) = 15/s phi (s) [3s + 4] = 15/s phi (s) = 15/s(3s + 4) A/s + B/3s + 4 = 15 put s = -4/3 A(0) + B(-4/3) = 15 B = -45/4 & Put s = 0. A(4) = 15 A = 15/4 Therefore phi(s) = 15/4/s + -45/4/3s + 4 Apply inverse laplace transform L^-1 [phi (s)] = 15/4 L^-1[1/s) - (45/s) L^-1(1/3s + 4) = 15/4(1) - 45/4(3) L^-1 (1/s + 4/3) x(t) = 15/4 (1 - e^-4/s t). \r\n 2x + x = 4, x(0) = 3 Apply laplace transform 2L(x) + L(x) = L(4) 2(x phi (s) - x(0)) + phi (s) = 4/s 2(s phi (s) - 3) + phi (s) = 4/s phi (s)(2s + 1) - 6 = 4/s (2s + 1) phi (s) = 4/s - 6