Solution:

The Fourier series for square wave is \(f(x)=\frac{4}{\pi} \sum_{i=1,3,5,\dots}^\infty \frac{1}{n}\sin\left(\frac{n\pi x}{L}\right)\)

The Laplace transform is \(L\left\{\sin(ax)\right\}=\frac{a}{x^2+a^2}\)

The function is \(f(x)=\frac{4}{\pi}\left[\sin\left(\frac{\pi x}{L}\right)+\frac{1}{3}\sin\left(\frac{3\pi x}{L}\right)+\frac{1}{5}\sin\left(\frac{5\pi x}{L}\right)+\dots\right]\)

Apply Laplace transform:

Conclusion:

\(L{f(x)}=\frac{4}{\pi}\left[L\left\{\sin\left(\frac{\pi x}{L}\right)\right\}+\frac{1}{3}L\left\{\sin\left(\frac{3\pi x}{L}\right)\right\}+\frac{1}{5}L\left\{\sin\left(\frac{5\pi x}{L}\right)\right\}+\dots\right]\)

\(=\frac{4}{\pi}\left[\frac{\frac{\pi}{L}}{x^2+\left(\frac{\pi}{L}\right)^2}+\frac{1}{3}\left(\frac{\frac{3\pi}{L}}{x^2+\left(\frac{3\pi}{L}\right)^2}\right)+\frac{\frac{5\pi}{L}}{x^2+\left(\frac{5\pi}{L}\right)^2}+\dots\right]\)

\(=\frac{4}{\pi}\left[\frac{\frac{\pi}{L}}{x^2+\left(\frac{\pi}{L}\right)^2}+\frac{\frac{\pi}{L}}{x^2+\left(\frac{3\pi}{L}\right)^2}+\frac{\frac{\pi}{L}}{x^2+\left(\frac{5\pi}{L}\right)^2}\dots\right]\)

\(=\frac{4}{L}\left[\frac{1}{x^2+\left(\frac{\pi}{L}\right)^2}+\frac{1}{x^2+\left(\frac{3\pi}{L}\right)^2}+\frac{1}{x^2+\left(\frac{5\pi}{L}\right)^2}+\dots\right]\)

Hence, as there are no real singularities, therefore the number of poles are 0.

The Fourier series for square wave is \(f(x)=\frac{4}{\pi} \sum_{i=1,3,5,\dots}^\infty \frac{1}{n}\sin\left(\frac{n\pi x}{L}\right)\)

The Laplace transform is \(L\left\{\sin(ax)\right\}=\frac{a}{x^2+a^2}\)

The function is \(f(x)=\frac{4}{\pi}\left[\sin\left(\frac{\pi x}{L}\right)+\frac{1}{3}\sin\left(\frac{3\pi x}{L}\right)+\frac{1}{5}\sin\left(\frac{5\pi x}{L}\right)+\dots\right]\)

Apply Laplace transform:

Conclusion:

\(L{f(x)}=\frac{4}{\pi}\left[L\left\{\sin\left(\frac{\pi x}{L}\right)\right\}+\frac{1}{3}L\left\{\sin\left(\frac{3\pi x}{L}\right)\right\}+\frac{1}{5}L\left\{\sin\left(\frac{5\pi x}{L}\right)\right\}+\dots\right]\)

\(=\frac{4}{\pi}\left[\frac{\frac{\pi}{L}}{x^2+\left(\frac{\pi}{L}\right)^2}+\frac{1}{3}\left(\frac{\frac{3\pi}{L}}{x^2+\left(\frac{3\pi}{L}\right)^2}\right)+\frac{\frac{5\pi}{L}}{x^2+\left(\frac{5\pi}{L}\right)^2}+\dots\right]\)

\(=\frac{4}{\pi}\left[\frac{\frac{\pi}{L}}{x^2+\left(\frac{\pi}{L}\right)^2}+\frac{\frac{\pi}{L}}{x^2+\left(\frac{3\pi}{L}\right)^2}+\frac{\frac{\pi}{L}}{x^2+\left(\frac{5\pi}{L}\right)^2}\dots\right]\)

\(=\frac{4}{L}\left[\frac{1}{x^2+\left(\frac{\pi}{L}\right)^2}+\frac{1}{x^2+\left(\frac{3\pi}{L}\right)^2}+\frac{1}{x^2+\left(\frac{5\pi}{L}\right)^2}+\dots\right]\)

Hence, as there are no real singularities, therefore the number of poles are 0.