## Monday Math 111

The Laplace Transform
Part 12: Combining Rules

We can combine many of the previous rules we’ve found to determine the Laplace transform of some more complicated functions.

For example, what is the Laplace transform of $f(t)=t^2e^{-t}\cos{t}$? We know from here that the Laplace transform for cosine is $\mathcal{L}\left[\cos(\omega{t})\right](s)=\frac{s}{s^2+\omega^2}$, so $\mathcal{L}\left[\cos{t}\right](s)=\frac{s}{s^2+1}$ Thus, via the frequency shift formula here, $\mathcal{L}\left[e^{-t}\cos{t}\right](s)=\frac{(s+1)}{(s+1)^2+1}$.
Considering, then, our frequency differentiation formula,
$\begin{array}{rcl}\mathcal{L}\left[t^2e^{-t}\cos{t}\right](s)&=&\frac{d^2}{ds^2}\left(\frac{(s+1)}{(s+1)^2+1}\right)\\\mathcal{L}\left[t^2e^{-t}\cos{t}\right](s)&=&\frac{2(s+1)\left((s+1)^2-3\right)}{\left((s+1)^2+1\right)^3}\\&=&\frac{2(s+1)\left(s^2+2s-2\right)}{\left(s^2+2s+2\right)^3}\\&=&\frac{2\left(s^3+3s^2-2\right)}{\left(s^2+2s+2\right)^3}\end{array}$.

Now consider the “sine cardinal” or “sinc” function $\mathrm{sinc}(t)=\frac{\sin{t}}t$. What is the Laplace transform of sinc(kt)?
We recall from here that the Laplace transform for the sine is
$\mathcal{L}\left[\sin(\omega{t})\right](s)=\frac{\omega}{s^2+\omega^2}$. We apply to this our frequency integration formula to get the transform of sinc(kt)
$\begin{array}{rcl}\mathcal{L}\left[\mathrm{sinc}(kt)\right](s)&=&\mathcal{L}\left[\frac{\sin(kt)}{kt}\right](s)\\&=&\frac{1}{k}\mathcal{L}\left[\frac{\sin(kt)}{t}\right](s)\\&=&\frac{1}{k}\int_s^{\infty}\mathcal{L}\left[\sin(kt)\right](\sigma)\,d\sigma\\&=&\frac{1}{k}\int_s^{\infty}\frac{k}{\sigma^2+k^2}\,d\sigma\\&=&\int_s^{\infty}\frac{d\sigma}{\sigma^2+k^2}\\&=&\left.\frac{1}{k}\tan^{-1}\left(\frac{\sigma}k\right)\right|_s^{\infty}\\&=&\frac{1}{k}\left(\frac{\pi}2-\tan^{-1}\left(\frac{s}k\right)\right)\\\mathcal{L}\left[\mathrm{sinc}(kt)\right](s)&=&\frac{1}{k}\tan^{-1}\left(\frac{k}s\right)\end{array}$,
$\Re(s)>0$.

Lastly, let’s consider the square wave with period 2π given by $f(t)=\mathrm{sgn}(\sin{t})=\left\{\begin{array}{rcl}1&\,\text{if}&\sin{t}>{0}\\-1&\,\text{if}&\;\sin{t}<{0}\end{array}\right.$. As this is a periodic function, we use our periodic function Laplace transform
$\mathcal{L}\left[f(t)\right](s)=\frac{1}{1-e^{-sT}}\int_0^{T}f(u)e^{-su}\,du$,
where T is the period.
For our current function, this gives us:
$\begin{array}{rcl}\mathcal{L}\left[f(t)\right](s)&=&\frac{1}{1-e^{-sT}}\int_0^{T}f(u)e^{-su}\,du\\&=&\frac{1}{1-e^{-s(2\pi)}}\int_0^{2\pi}f(u)e^{-su}\,du\\&=&\frac{1}{1-e^{-2\pi{s}}}\left(\int_0^{\pi}e^{-su}\,du-\int_{\pi}^{2\pi}e^{-su}\,du\right)\\&=&\frac{1}{1-e^{-2\pi{s}}}\left(\left[-\frac{e^{-su}}s\right]_{u=0}^{\pi}-\left[-\frac{e^{-su}}s\right]_{u=\pi}^{2\pi}\right)\\&=&\frac{1}{1-e^{-2\pi{s}}}\left(\frac1s-\frac{e^{-\pi{s}}}s+\frac{e^{-2\pi{s}}}s-\frac{e^{-\pi{s}}}s\right)\\&=&\frac{1}{1-e^{-2\pi{s}}}\frac1s\left(1-2e^{-\pi{s}}+e^{-2\pi{s}}\right)\\&=&\frac1s\frac{\left(1-e^{-\pi{s}}\right)^2}{1-e^{-2\pi{s}}}\\&=&\frac1s\frac{1-e^{-\pi{s}}}{1+e^{-\pi{s}}}\\&=&\frac1s\frac{e^{\pi{s}/2}-e^{-\pi{s}/2}}{e^{\pi{s}/2}+e^{-\pi{s}/2}}\\\mathcal{L}\left[f(t)\right](s)&=&\frac1s\tanh\left(\frac{\pi}2s\right)\end{array}$.