MATH 414 Lecture 16

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Convolution Theorem

$\begin{array}{rl}{ℱ}[f*g]& =\sqrt{2\pi }\hat{f}(\lambda )\hat{g}(\lambda )\\ (f*g)(t)& =\frac{1}{\sqrt{2\pi }}{{ℱ}}^{-1}[\hat{f}(\lambda )\hat{g}(\lambda )]\end{array}$

Example

Let $f(t)=\{\begin{array}{ll}1& x\in [-\pi ,\pi ]\\ 0& \text{otherwise}\end{array}$

$\begin{array}{rl}(f*f)(t)& =\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(\tau )f(t-\tau )\mathrm{d}\tau \\ & =\underset{t-\pi }{\overset{t+\pi }{\int }}f(\tau )\mathrm{d}\tau \\ \end{array}$

Let $G(\tau )=\tau \theta (\tau +\pi )-\tau \theta (\tau -\pi )$

Observe that $G^{\prime }(\tau )=f(\tau )$, so

$(f*f)(t)=\underset{t-\pi }{\overset{t+\pi }{\int }}{G}^{\prime }(\tau )\mathrm{d}\tau =G(t+\pi )-G(t-\pi )$ by the fundamental theorem of calculus

In this case, $(f*f)(t)=(t+\pi )\theta (t+2\pi )-(t+\pi )\theta (t)-((t-\pi )\theta (t)-(t-\pi )\theta (t-2\pi ))=(t+\pi )\theta (t+2\pi )+(t-\pi )\theta (t-2\pi )$