MATH 414 Lecture 16

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Convolution Theorem

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \mathcal{F} \left[ f * g \right] &= \sqrt{2\pi} \hat{f}(\lambda) \, \hat{g}(\lambda) \\ (f * g)(t) &= \frac{1}{\sqrt{2\pi}} \, \mathcal{F}^{-1} \left[ \hat{f}(\lambda) \, \hat{g}(\lambda) \right] \end{align}}

Example

Let $f(t)={\begin{cases}1&x\in \left[-\pi ,\pi \right]\\0&{\mbox{otherwise}}\end{cases}}$

${\begin{aligned}(f*f)(t)&=\int _{-\infty }^{\infty }f(\tau )\,f(t-\tau )\,\mathrm {d} \tau \\&=\int _{t-\pi }^{t+\pi }f(\tau )\,\mathrm {d} \tau \\\end{aligned}}$

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

Observe that $G'(\tau )=f(\tau )$, so

$(f*f)(t)=\int _{t-\pi }^{t+\pi }G'(\tau )\,\mathrm {d} \tau =G(t+\pi )-G(t-\pi )$ by the fundamental theorem of calculus

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