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 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 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, 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 (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)}