Question

The convolution of two continuous time signals x(t) = e^-t u(t) and h(t) = e^-2t u(t) is ________________

(a) (e^t – e^2t) u (t)

(b) (e^-2t – e^-t) u (t)

(c) (e^-t – e^-2t) u (t)

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

This question was addressed to me in examination.

The question is from Trigonometric Fourier Series topic in chapter Fourier Series of Signals and Systems

Answer 1

Right choice is (c) (e^-t – e^-2t) u (t)

Easiest explanation: Given x (t) = e^-t u (t) and h (t) = e^-2t u (t).

Now, y (t) = \(∫_{-∞}^∞ x(τ)h(t-τ)dτ\)

= e^-2t (e^t -1); t≥0

Or, y (t) = e^-t – e^-2t; t≥0

Since, y (t) = 0 for t<0

Therefore, y (t) = (e^-t – e^-2t) u (t).