MATH 414 Lecture 18

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Convolution Theorem

By definition, the convolution of two functions $f(t)$ and $g(t)$ is

$(f*g)(t)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(\tau )g(t-\tau )\mathrm{d}\tau$

Theorem. ${ℱ}[f*g](\lambda )=\sqrt{2\pi }\hat{f}(\lambda )\hat{g}(\lambda )$

Proof. (omitted)

quod erat demonstrandum

Plancheral (Parseval) Theorem

Given $f$, $\hat{f}$, $g$, and $\hat{g}$ in $L^2$ ,

${⟨f,g⟩}_{L^2}=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(t)g(t)\mathrm{d}t$

Show that ${⟨f,g⟩}_{L^2}={⟨\hat{f},\hat{g}⟩}_{L^2}$

What does this mean?

let $f=g$

$\begin{array}{rl}{{\Vert f\Vert }_{L^2}}^2& =\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{|f(t)|}^2\mathrm{d}t\\ {{\Vert \hat{f}\Vert }_{L^2}}^2& =\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{|\hat{f}(t)|}^2\mathrm{d}t\\ \hat{f}(\lambda )& =\frac{1}{\sqrt{2\pi }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(t){\mathrm{e}}^{-i\lambda t}\mathrm{d}t\\ f(t)& =\frac{1}{\sqrt{2\pi }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\hat{f}(\lambda ){\mathrm{e}}^{+i\lambda t}\mathrm{d}\lambda \\ \end{array}$

Note that ${\mathrm{e}}^{i\lambda t}$ is a "pure tone frequency" for fixed $\lambda$.

Then $\hat{f}(\lambda )$ represents the amplitude and phase.

When we compute the synthesis

$f(t)=\frac{1}{\sqrt{2\pi }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\hat{f}(\lambda ){\mathrm{e}}^{i\lambda t}\mathrm{d}\lambda$

We are expressing $f(t)$ as a "linear combination" of "pure" tones. In Fourier Series, we do the same thing (synthesis of "pure" tones)

$f(t)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{c}_n{\mathrm{e}}^{int}$

Parseval States that $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{|f(t)|}^2\mathrm{d}t$ is the total energy in $f$.

Theorem. [Plancheral / Parseval].

Proof.

$\begin{array}{rl}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(t)\overline{g(t)}\mathrm{d}t& =\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(t)\stackrel{‾}{(\frac{1}{\sqrt{2\pi }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\hat{g}(\lambda ){\mathrm{e}}^{+i\lambda t}\mathrm{d}\lambda )}\mathrm{d}t\\ & =\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(t)\left(\frac{1}{\sqrt{2\pi }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\overline{\hat{g}(\lambda )}{\mathrm{e}}^{-i\lambda t}\mathrm{d}\lambda \right)\mathrm{d}t\\ & \underset{=}{fubini}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\overline{\hat{g}(\lambda )}(\frac{1}{\sqrt{2\pi }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(t){\mathrm{e}}^{-i\lambda t}\mathrm{d}t)\mathrm{d}\lambda \\ & =\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\hat{f}(\lambda )\overline{\hat{g}(\lambda )}\mathrm{d}\lambda \\ & ={⟨\hat{f},\hat{g}⟩}_{L^2}\end{array}$

quod erat demonstrandum

Linear Filters

space of signals: $L^2$ (work with piecewise continuous)

A Linear Filter $l$ is a linear transformation from the space of signals into the space of signals ($L^2$). This means $L[\alpha f+\beta g]=\alpha L[f]+\beta L[g]$, where $\alpha$ and $\beta$ are constants, and $f$ and $g$ are signals.

We shall use the notation $L[f](t)=\tilde{f}(t)$

Time invariance

$f(t)$, start at 12 noon, but we perform a shift (time translation) by $a$ and apply again, we'll get the same signal back.

Define $f_a(t):=f(t-a)$

When $a>0$, then the signal shifts to the right; and when $a<0$, the signal shifts to the left.

A linear filter is said to be time-invariant if and only if $L[f_a]=(L[f]{)}_a=\tilde{f}(t-a)$ Intuitively, when we shift a function and filter, we should get the same thing as if we would have applied the filter and then shifted.

All time-invariant filters look like convolutions

Example: Running Filter

Take some time $d>0$ and a function $f$, we define

$L[f](t)=\frac{1}{d}\underset{t-d}{\overset{t}{\int }}f(\tau )\mathrm{d}\tau$

This is the average of $f$ over the interval from $t-d$ to $d$. This has the effect of smoothing a "rough" function.