Question

The inverse Laplace transform of F(s) = \(\frac{2}{s+c}e^{-bs}\) is _____________

(a) 2e^-k (t+b) u (t+b)

(b) 2e^-k (t-b) u (t-b)

(c) 2e^k (t-b) u (t-b)

(d) 2e^k (t-b) u (t+b)

The question was asked in semester exam.

Question is from Exponential Fourier Series and Fourier Transforms topic in portion Fourier Series of Signals and Systems

Answer 1

The correct answer is (b) 2e^-k (t-b) u (t-b)

Explanation: Let G(s) = \(\frac{2}{s+c}\)

Or, G (t) = L^-1{G(s)} = 2e^-ct

∴ F (t) = L^-1{G(s) e^-bs}

= 2e^-k (t-b) u (t-b).