Question

Find the laplace transform of f(t), where f(t) = |sin(pt)| and t>0.

(a) \(\frac{p}{s^2+p^2}×cosh⁡(\frac{s\pi}{2p})\)

(b) \(\frac{p}{s^2+p^2}×sinh⁡(\frac{s\pi}{2p})\)

(c) \(\frac{p}{s^2+p^2}×coth⁡⁡(\frac{s\pi}{2p})\)

(d) \(\frac{p}{s^2+p^2}×tanh⁡⁡(\frac{s\pi}{2p})\)

This question was posed to me during an internship interview.

Origin of the question is Laplace Transform of Periodic Function topic in portion Laplace Transform of Engineering Mathematics

Answer 1

Right choice is (c) \(\frac{p}{s^2+p^2}×coth⁡⁡(\frac{s\pi}{2p})\)

Best explanation: From this question, we know –

Period of sin(t)=2π

Period of sin⁡(pt)=\(\frac{2\pi}{p}\)

Period of |sin⁡(pt)|=\(\frac{\pi}{p}\)

\(L(f(t))=\frac{1}{1-e^{\frac{-\pi}{ps}}} \int_{0}^{\frac{\pi}{p}}e^{-st} f(t)dt\)

Since |sin⁡(pt)| is positive in all quadrants

\(L(f(t))=\frac{1}{1-e^{\frac{-\pi}{ps}}} \int_{0}^{\frac{\pi}{p}}e^{-st} sin⁡(pt)dt\)

=\(\frac{1}{1-e^{\frac{-\pi}{ps}}}\begin{bmatrix}\frac{e^{\frac{-sπ}{p}}}{s^2+p^2}×p\end{bmatrix}-\begin{bmatrix}\frac{1}{s^2+p^2}×(-p)\end{bmatrix}\)

=\(\frac{1}{1-e^{\frac{-\pi}{ps}}}×\frac{p}{s^2+p^2}×(1+e^{\frac{-π}{ps}})\)

=\(\frac{p}{s^2+p^2}×coth⁡(\frac{s\pi}{2p})\), (Multiplying and dividing by \(e^{\frac{-sπ}{2p}}\))

Thus, the answer is \(\frac{p}{s^2+p^2}×coth⁡⁡(\frac{s\pi}{2p})\).