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

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

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

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

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

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

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

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

quod erat demonstrandum

Impulse Response

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

Then $\lim _{\delta \to 0^{+}}f_{\delta }(t)$ is a spike such that $f(t)={\begin{cases}\infty &t=0\\0&{\mbox{otherwise}}\end{cases}}$

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${\begin{aligned}L(f_{\delta }(t))&=\int _{-\infty }^{\infty }f(\tau )\,h(\tau -t)\,\mathrm {d} \tau \\&=\ldots \end{aligned}}$

Constructing Filters

${\begin{aligned}L(f)&=f*h\\\left(L(f)\right)&={\sqrt {2\pi }}{\hat {f}}(\lambda )\,{\hat {h}}(\lambda )\end{aligned}}$

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 ${\hat {h}}(\lambda )={\begin{cases}{\frac {1}{\sqrt {2\pi }}}&\left|\lambda \right|\leq \lambda _{0}\\0&{\mbox{otherwise}}\end{cases}}$

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

${\begin{aligned}h&={\mathcal {F}}^{-1}\left({\hat {h}}(\lambda )\right)\\&={\frac {1}{\sqrt {2\pi }}}\,\int _{-\infty }^{\infty }{\hat {h}}(\lambda )\,\mathrm {e} ^{+i\,\lambda \,t}\,\mathrm {d} \lambda \\&={\frac {1}{2\pi }}\int _{-\lambda _{0}}^{\lambda _{0}}\mathrm {e} ^{i\,\lambda t}\,\mathrm {d} \lambda &={\frac {\sin {(\lambda _{0}\,t)}}{\pi \,t}}\end{aligned}}$

$L(f)=\int _{-\infty }^{\infty }f(\tau )\,{\frac {\sin {(\lambda _{0}\,(t-\tau ))}}{\pi \,(t-\tau )}}\,\mathrm {d} \tau$

Put in $f(t)=\theta (t)-\theta (t-t_{0})$. Then

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

Perform $u$-substitution with $u=\lambda _{0}\,(t-\tau )$

${\begin{aligned}L(f)&=\int _{\lambda _{0}\,t}^{\lambda _{0}\,(t-t_{0})}{\frac {\sin {u}}{\pi {\frac {u}{\lambda _{0}}}}}\,{\frac {-\mathrm {d} u}{\lambda _{0}}}\\&=\int _{\lambda _{0}\,(t-t_{0})}^{\lambda _{0}\,t}{\frac {\sin {u}}{\pi \,u}}\,\mathrm {d} u\\&={\frac {1}{\pi }}\left(\mathrm {Si} (\lambda _{0}\,t)-\mathrm {Si} (\lambda _{0}\,(t-t_{0}))\right)\end{aligned}}$

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

Butterworth Filter

$h(t)={\begin{cases}a\,\mathrm {e} ^{-\alpha \,t}&t\geq 0\\0&t<0\end{cases}}$

${\begin{aligned}{\hat {h}}(\lambda )&={\frac {1}{\sqrt {2\pi }}}\,\int _{-\infty }^{\infty }h(t)\,\mathrm {e} ^{-i\,\lambda \,t}\,\mathrm {d} t\\&={\frac {A}{\sqrt {2\pi }}}\,\int _{0}^{\infty }\mathrm {e} ^{-\alpha \,t-i\,\lambda \,t}\,\mathrm {d} t\\&={\frac {A}{{\sqrt {2\pi }}\,\left(\alpha +i\,\lambda \right)}}\end{aligned}}$

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

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

The Sine Integral

$\mathrm {Si} (x):=\int _{0}^{x}{\frac {\sin {u}}{u}}\,\mathrm {d} u$