Question

Find the Laplace transform for f(t) = \(\frac{1}{t}\) [e^-2t – e^-3t]u(t).

(a) ln\(\left(\frac{s-2}{s-3}\right)\)

(b) ln\(\left(\frac{s+2}{s+3}\right)\)

(c) ln\(\left(\frac{s-2}{s+3}\right)\)

(d) ln\(\left(\frac{s+2}{s-3}\right)\)

I got this question in a job interview.

The query is from Properties of the Laplace Transform topic in portion Laplace Transform and System Design of Signals and Systems

Answer 1

Right choice is (b) ln\(\left(\frac{s+2}{s+3}\right)\)

The best I can explain: Given f(t) = \(\frac{1}{t}\) [e^-2t – e^-3t]u(t)

We know that L{e^-2t) u(t)} = \(\frac{1}{s+2}\); L{^-3t u(t)} = \(\frac{1}{s+3}\)

Integration in s-domain property states that \(\int_s^∞ F(s)ds \leftrightarrow \frac{f(t)}{t}\)

L{\(\frac{1}{t}\) [e^-2t – e^-3)]u(t)} = \(\int_s^∞ \left(\frac{1}{s+2} – \frac{1}{s+3}\right)ds = [ln⁡(s+2) – ln⁡(s+3)]|_s^∞ = ln\left(\frac{s+2}{s+3}\right)\).