Question

Find the laplace transform of y(t)=e^|t-1| u(t).

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

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

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

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

This question was posed to me in a national level competition.

My doubt stems from Laplace Transform by Properties topic in chapter Laplace Transform of Engineering Mathematics

Answer 1

Correct answer is (d) \(\frac{2s}{1-s^2} e^{-s}\)

Explanation: y(t)=\(e^{|t-1|}\)

Laplace transform of e^|t| =\(\int_{-\infty}^\infty e^{|t|} e^{-st} dt\)

=\(\int_0^∞ e^t e^{-st} dt-\int_{-∞}^0 e^{-t} e^{-st} dt\)

=\(\int_0^∞ e^{-(s-1)t} dt-∫_{-∞}^0 e^{(-s-1)t} dt\)

Now,\(\int_0^∞ e^{-(s-1)t} dt=\left [-\frac{1}{s-1} [e^{-(s-1)t}]\right ]_∞^0\)

=\(\left [-\frac{1}{s-1} [e^{-(s-1)t} ]\right ]_∞^0=\frac{-1}{s-1}\)

Now, \(∫_{-∞}^0 e^{(-s-1)t} dt=\left [\frac{1}{-(s+1)} [e^{(-s-1)t}]\right ]_0^{-∞}\)

=\(\left [-\frac{1}{s+1} [e^{(-s-1)t} ]\right ]_0^{-∞}=-\frac{1}{(s+1)}\)

Laplace transform of |t| e^|t| =\(\int_{-\infty}^\infty e^{|t|} e^{-st} dt=-\left [\frac{1}{s-1}+\frac{1}{s+1}\right ]=-\left [\frac{2s}{s^2-1}\right ]\)

Laplace transform of |t| e^|t| = \(\int_{-\infty}^\infty e^{|t-1|} e^{-st} dt=\frac{2s}{1-s^2} e^{-s}\)