Convolution theorem. Using the Convolution Theorem find the inverse Laplace tran
ID: 2985392 • Letter: C
Question
Convolution theorem.
Using the Convolution Theorem find the inverse Laplace transform of the function: F(s) =1/S2-1 First rewrite F(s) as a product of two functions F(s) = G(s)H(s); If G(s) = 1/s-1 then H(s) = Answer is a Expression Then find: g(t) = (G(s)) = h(t) = -1(H(s)) = Answer is a Expression Finally, using the Convolution Theorem you can calculate f(t) = -1(F(s)) = g(t) * h(t) = g(tau)h(t - tau)d tau = Answer is a Expression The convolution Theorem h(t) = (f*g) (t) = f(tau) g(t-rau)d tau then (h) = (f) (g) Laplace Transform tableExplanation / Answer
F(s) = 1/(s-1) * 1/(s+1) G(s) = 1/(s-1) H(s) = 1/(s+1) g(t) = e^t h(t) = e^(-t) f(t) = g * h = integral from 0 to t [ e^T * e^(-t+T) ]dT = integral from 0 to t [ e^(-t) * e^(2T) ]dT =e^(-t) * integral from 0 to t [ e^(2T) ]dT = e^(-t) * [ e^(2T)]/2 apply limits = e^(-t) * [ e^(2t) - e^(0)]/2 = [e^t - e^-t]/2 = sinh(t)