Question

The Laplace transform of the signal u(t) – u(t-2) is ______________

(a) \(\frac{e^{-2s}-1}{s}\)

(b) \(\frac{-e^{-2s}+1}{s}\)

(c) \(\frac{2}{s}\)

(d) \(\frac{-2}{s}\)

The question was asked in a national level competition.

I would like to ask this question from Common Laplace Transforms topic in division Laplace Transform and System Design of Signals and Systems

Answer 1

The correct answer is (b) \(\frac{-e^{-2s}+1}{s}\)

Easy explanation: X(s) = \(\int_0^∞ x(t)  e^{-st} \,dt\)

Here, the given signal is u (t).

Hence, u (t) = 1 for all t>0 and = 0 for all t<0.

Again, the given signal is u (t-2).

Hence, u (t-2) = 1 for all t>2 and = 0 for all t<2.

So, limit is from 0 to 2

∴ Laplace of signal= \(\int_0^2 e^{-st} \,dt\)

= \(\frac{-e^{-2s}+1}{s}\).