Question

Find the convolution of the signals x1 (t) = e^-2t u(t) and x2 (t) = e^-3t u(t).

(a) e^-2t u(t) – e^-3t u(t)

(b) e^-2t u(t) + e^-3t u(t)

(c) e^2t u(t) – e^3t u(t)

(d) e^2t u(t) – e^-3t u(t)

The question was asked during an interview.

The origin of the question is Inverse Fourier Transform in portion Fourier Transform of Signals and Systems

Answer 1

Right option is (a) e^-2t u(t) – e^-3t u(t)

To explain I would say: Convolution property, x1 (t)*x2 (t) ↔ X1 (ω) X2 (ω)

∴  x1 (t)*x2 (t) = F^-1 [X1 (ω) X2 (ω)]

Given x1 (t) = e^-2t u(t)

∴ X1 (ω) = \(\frac{1}{jω+2}\)

Given x2 (t) = e^-3t u(t)

∴ X1 (ω) = \(\frac{1}{jω+3}\)

x1 (t)*x2 (t) = F^-1 [X1 (ω) X2 (ω)] = F^-1 \([\frac{1}{jω+2} \frac{1}{jω+3}] = F^{-1} [\frac{1}{jω+2} – \frac{1}{jω+3}] \)

∴ x1 (t)*x2 (t) = e^-2t u(t)-e^-3t u(t).