MATH 414 Lecture 18

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Convolution Theorem

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

$\left(f*g\right)(t)=\int _{-\infty }^{\infty }f(\tau )\,g(t-\tau )\,\mathrm {d} \tau$

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

Proof. (omitted)

quod erat demonstrandum

Plancheral (Parseval) Theorem

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

$\left\langle f,g\right\rangle _{L^{2}}=\int _{-\infty }^{\infty }f(t)\,g(t)\,\mathrm {d} t$

Show that $\left\langle f,g\right\rangle _{L^{2}}=\left\langle {\hat {f}},{\hat {g}}\right\rangle _{L^{2}}$

What does this mean?

let $f=g$

${\begin{aligned}{\left\|f\right\|_{L^{2}}}^{2}&=\int _{-\infty }^{\infty }\left|f(t)\right|^{2}\,\mathrm {d} t\\{\left\|{\hat {f}}\right\|_{L^{2}}}^{2}&=\int _{-\infty }^{\infty }\left|{\hat {f}}(t)\right|^{2}\,\mathrm {d} t\\{\hat {f}}(\lambda )&={\frac {1}{\sqrt {2\pi }}}\,\int _{-\infty }^{\infty }f(t)\,\mathrm {e} ^{-i\,\lambda \,t}\,\mathrm {d} t\\f(t)&={\frac {1}{\sqrt {2\pi }}}\,\int _{-\infty }^{\infty }{\hat {f}}(\lambda )\,\mathrm {e} ^{+i\,\lambda \,t}\,\mathrm {d} \lambda \\\end{aligned}}$

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

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

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

$f(t)=\sum _{n=-\infty }^{\infty }c_{n}\,\mathrm {e} ^{i\,n\,t}$

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

Theorem. [Plancheral / Parseval].

Proof.

${\begin{aligned}\int _{-\infty }^{\infty }f(t)\,{\overline {g(t)}}\,\mathrm {d} t&=\int _{-\infty }^{\infty }f(t)\,{\overline {\left({\frac {1}{\sqrt {2\pi }}}\,\int _{-\infty }^{\infty }{\hat {g}}(\lambda )\,\mathrm {e} ^{+i\,\lambda \,t}\,\mathrm {d} \lambda \right)}}\,\mathrm {d} t\\&=\int _{-\infty }^{\infty }f(t)\,\left({\frac {1}{\sqrt {2\pi }}}\,\int _{-\infty }^{\infty }{\overline {{\hat {g}}(\lambda )}}\,\mathrm {e} ^{-i\,\lambda \,t}\,\mathrm {d} \lambda \right)\,\mathrm {d} t\\&{\underset {=}{fubini}}\int _{-\infty }^{\infty }{\overline {{\hat {g}}(\lambda )}}\,\left({\frac {1}{\sqrt {2\pi }}}\,\int _{-\infty }^{\infty }f(t)\,\mathrm {e} ^{-i\,\lambda \,t}\,\mathrm {d} t\right)\,\mathrm {d} \lambda \\&=\int _{-\infty }^{\infty }{\hat {f}}(\lambda )\,{\overline {{\hat {g}}(\lambda )}}\,\mathrm {d} \lambda \\&=\left\langle {\hat {f}},{\hat {g}}\right\rangle _{L^{2}}\end{aligned}}$

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 $L\left[\alpha \,f+\beta \,g\right]=\alpha \,L[f]+\beta \,L[g]$, 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

$L[f](t)={\frac {1}{d}}\,\int _{t-d}^{t}f(\tau )\,\mathrm {d} \tau$

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