Laplace transform of a square wave function

Question

What is the right way to find the Laplace transform of this function:

The thing I noticed was:

$$f(t)=\text{A}\space\space\space\space\space\space\space\space\space\space 0\le t<\frac{\text{T}}{2}\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space f(t)=-\text{A}\space\space\space\space\space\space\space\space\space\space \frac{\text{T}}{2}<t<\text{T}$$

But how do find the complete Laplace transform of this function?

Answer 1

We can write the square wave function as $$ f(t)=A\sum_{k=0}^{\infty} \left[u\left(t-kT\right)-2u\left(t-\frac{2k+1}{2}T\right)+u\left(t-(k+1)T\right)\right] $$ where $u(t)$ is the Heaviside's function. So the Laplace transform of $f(t)$ is \begin{align} F(s)&=A\sum_{k=0}^{\infty} \frac{1}{s}\left[\mathrm e^{-kTs}-2\mathrm e^{-\frac{2k+1}{2}Ts}+\mathrm e^{-(k+1)Ts}\right]\\ &=A\sum_{k=0}^{\infty} \frac{\mathrm e^{-kTs}}{s}\left[1-\mathrm e^{-\frac{Ts}{2}}\right]^2\\ &=\frac{A}{s}\left(1-\mathrm e^{-\frac{Ts}{2}}\right)^2\sum_{k=0}^{\infty} \mathrm e^{-kTs}\\ &=\frac{A}{s}\left(1-\mathrm e^{-\frac{Ts}{2}}\right)^2\frac{1}{1-\mathrm e^{-sT}}\\ &=\frac{A}{s}\frac{1-\mathrm e^{-\frac{Ts}{2}}}{1+\mathrm e^{-\frac{Ts}{2}}}\\ &=\frac{A}{s}\tanh\left(\frac{sT}{4}\right) \end{align}