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}.

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One Response to “Monday Math 111”

  1. Monday Math 115 « Twisted One 151's Weblog Says:

    […] for t<0. Taking the Laplace transform of our differential equation, letting , Recall from here that we found, using the periodic function rule, that the square wave has Laplace transform , so […]

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