MATH 414 Lecture 19

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Linear Time-Invariant Filters

Recall notation $f_a(t)=f(t-a)$, which is $f$ shifted to the right by $a$ units.

A filter is called time-invariant if and only if $L(f_a)=(L(f){)}_a$

At the end of last lecture, we discused how if $L(t)=f*h$ for fixed $h$, then $L$ is linear and time-invariant.

Theorem. $L$ is a linear time-invariant filter if and only if $L(f)=f*h$, where $h$ is the impulse response function.

Proof. We've already shown that (⇒) is true, this will only show (⇐).

Suppose $L$ is linear and time-invariant. Let's apply $L$ to ${\mathrm{e}}^{i\lambda t}$:

$L\left({\mathrm{e}}^{i\lambda t}\right)=h^{\lambda }(t)$

Applying the filter on a shifted function gives

$L\left({\mathrm{e}}^{i\lambda (t-a)}\right)=h^{\lambda }(t-a)$

Observe that ${\mathrm{e}}^{i\lambda (t-a)}={\mathrm{e}}^{-i\lambda a}\cdot {\mathrm{e}}^{i\lambda t}$

The first term is just a scalar (with respect to $\lambda$, but not with respect to $t$). Therefore

$\begin{array}{rl}{h}^{\lambda }(t-a)& =L({\mathrm{e}}^{-i\lambda a}\cdot {\mathrm{e}}^{i\lambda t})\\ & ={\mathrm{e}}^{-i\lambda a}L\left({\mathrm{e}}^{i\lambda t}\right)\\ & ={\mathrm{e}}^{-i\lambda a}{h}^{\lambda }(t)\end{array}$

holds for all $a$ and for all $t$; so we can set $a=t$.

$\begin{array}{rl}{h}^{\lambda }(t-t)& =h^{\lambda }(0)={\mathrm{e}}^{-i\lambda t}{h}^{\lambda }(t)\\ h^{\lambda }(t)& ={\mathrm{e}}^{i\lambda t}{h}^{\lambda }(0)\\ L\left({\mathrm{e}}^{i\lambda t}\right)& ={\mathrm{e}}^{i\lambda t}{h}^{\lambda }(0)\end{array}$

Let's apply $L$ to the inverse fourier transform of a function:

$\begin{array}{rl}L(f)& =L\left(\frac{1}{\sqrt{2\pi }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\hat{f}{\mathrm{e}}^{+i\lambda t}\mathrm{d}\lambda \right)\\ & =\frac{1}{\sqrt{2\pi }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\hat{f}L\left({\mathrm{e}}^{i\lambda t}\right)\mathrm{d}\lambda \\ & =\frac{1}{\sqrt{2\pi }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\hat{f}{h}^{\lambda }(0){\mathrm{e}}^{i\lambda t}\mathrm{d}\lambda \end{array}$

Let's define $\hat{h}(\lambda ):=\frac{1}{\sqrt{2\pi }}{h}^{\lambda }(0)$. Then

$\begin{array}{rlr}L(f)& =\sqrt{2\pi }\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{1}{\sqrt{2\pi }}\hat{f}(\lambda )\hat{h}(\lambda ){\mathrm{e}}^{i\lambda t}\mathrm{d}\lambda & ={{ℱ}}^{-1}(\hat{f}(\lambda )\hat{h}(\lambda ))\sqrt{2\pi }\\ & =f*h\end{array}$

quod erat demonstrandum

Impulse Response

$f_{\delta }(t)=\theta (x+\delta )\frac{1}{2\delta }-\theta (x-\delta )\frac{1}{2\delta }$

Then $\underset{\delta \to 0^{+}}{\lim }{f}_{\delta }(t)$ is a spike such that $f(t)=\{\begin{array}{ll}\mathrm{\infty }& t=0\\ 0& \text{otherwise}\end{array}$

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$\begin{array}{rl}L(f_{\delta }(t))& =\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(\tau )h(\tau -t)\mathrm{d}\tau \\ & =\dots \end{array}$

Constructing Filters

$\begin{array}{rl}L(f)& =f*h\\ (L(f))& =\sqrt{2\pi }\hat{f}(\lambda )\hat{h}(\lambda )\end{array}$

Low pass filter

Reduces amplitudes of high magnitude frequencies

High pass filter

Reverse of low pass filter

Notch filter

Removes selection of frequencies.

"Ideal" Low Pass Filter

Take $\hat{h}(\lambda )=\{\begin{array}{ll}\frac{1}{\sqrt{2\pi }}& \left|\lambda \right|\le {\lambda }_0\\ 0& \text{otherwise}\end{array}$

Taking the convolution $L(f)=f*h$, where

$\begin{array}{rl}h& ={{ℱ}}^{-1}(\hat{h}(\lambda ))\\ & =\frac{1}{\sqrt{2\pi }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\hat{h}(\lambda ){\mathrm{e}}^{+i\lambda t}\mathrm{d}\lambda \\ & =\frac{1}{2\pi }\underset{-{\lambda }_0}{\overset{{\lambda }_0}{\int }}{\mathrm{e}}^{i\lambda t}\mathrm{d}\lambda & =\frac{\sin ({\lambda }_{0}t)}{\pi t}\end{array}$

$L(f)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(\tau )\frac{\sin ({\lambda }_0(t-\tau ))}{\pi (t-\tau )}\mathrm{d}\tau$

Put in $f(t)=\theta (t)-\theta (t-t_0)$. Then

$L(f)=\underset{0}{\overset{b}{\int }}\frac{\sin ({\lambda }_{0}t)}{\pi t}$

Perform $u$-substitution with $u={\lambda }_0(t-\tau )$

$\begin{array}{rl}L(f)& =\underset{{\lambda }_{0}t}{\overset{{\lambda }_0(t-t_0)}{\int }}\frac{\sin u}{\pi \frac{u}{{\lambda }_0}}\frac{-\mathrm{d}u}{{\lambda }_0}\\ & =\underset{{\lambda }_0(t-t_0)}{\overset{{\lambda }_{0}t}{\int }}\frac{\sin u}{\pi u}\mathrm{d}u\\ & =\frac{1}{\pi }(\mathrm{S}\mathrm{i}({\lambda }_{0}t)-\mathrm{S}\mathrm{i}({\lambda }_0(t-t_0)))\end{array}$

See Definition of $\mathrm{S}\mathrm{i}(x)$ below.

Butterworth Filter

$h(t)=\{\begin{array}{ll}a{\mathrm{e}}^{-\alpha t}& t\ge 0\\ 0& t<0\end{array}$

$\begin{array}{rl}\hat{h}(\lambda )& =\frac{1}{\sqrt{2\pi }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}h(t){\mathrm{e}}^{-i\lambda t}\mathrm{d}t\\ & =\frac{A}{\sqrt{2\pi }}\underset{0}{\overset{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-\alpha t-i\lambda t}\mathrm{d}t\\ & =\frac{A}{\sqrt{2\pi }(\alpha +i\lambda )}\end{array}$

$\hat{L}(f)=\cancel{\frac{\sqrt{2\pi }}{\sqrt{2\pi }}}\frac{A\hat{f}(\lambda )}{\alpha +i\lambda }=\frac{A}{a+i\lambda }\hat{f}(\lambda )$

Now ${|\hat{L}(f)|}^2=\frac{A^2}{{\alpha }^2+{\lambda }^2}{|\hat{f}(\lambda )|}^2$. As $\lambda$ gets big, then the frequencies returned by $\hat{f}$ are reduced. Therefore this is essentially a low-pass filter

The Sine Integral

$\mathrm{S}\mathrm{i}(x):=\underset{0}{\overset{x}{\int }}\frac{\sin u}{u}\mathrm{d}u$