MATH 414 Lecture 20

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Last Time

"Structure Theorem": $L$ is a linear time-invariant filter if and only if $L(f)=f*h$, where $h$ is the impulse response function, and $\hat{h}(\lambda )$ is the system function.

We also discussed an acausal filter $\hat{h}(\lambda )=\frac{1}{\sqrt{2\pi }}(\theta \left(\lambda \right)+\theta (\lambda -{\lambda }_c))$.

For $f(t)=\theta (t)-\theta (t-t_0)$, we have $f*h(t)=\frac{1}{\pi }\underset{{\lambda }_c(t-t_c)}{\overset{{\lambda }_{c}t}{\int }}\frac{\sin u}{u}\mathrm{d}u$

Plotting this filtered function gives a function that seems to give an output before the signal actually arrives (hence acausal), violating the laws of physics!

Causal Filters

Don't violate physical laws

A linear time-invariant filter $L$ is said to be causal if and only if the output is $0$ until the input function arrives.

Let our input be $f(t)=0$ for all $t<t_0$

Then $L(f)=f*h(t)=0$ for all $t<t_0$ as well.

The Butterworth filter is causal

Theorem. $L$ is causal if and only if $h(t)=0$ for all $t<0$.

Partial Proof. (⇐) Assume $f(t)=0$ for all $t<t_0$. We write

$L(f)=f*h(t)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(\tau )h(t-\tau )\mathrm{d}\tau =\underset{t_0}{\overset{\mathrm{\infty }}{\int }}f(\tau )h(t-\tau )\mathrm{d}\tau$

Suppose $t<t_0$. Then for $t_0\le \tau <\mathrm{\infty }$, $t-\tau <t_0-\tau <0$, so therefore $h(t-\tau )=0$, so

$\underset{t_0}{\overset{\mathrm{\infty }}{\int }}f(\tau )\cdot 0\mathrm{d}\tau =0$.

quod erat demonstrandum

Corollary. $L$ is causal if and only if $\hat{h}(\lambda )=\frac{1}{\sqrt{2\pi }}{ℒ}\left\{h\right\}(i\lambda )$

Proof. If $h$ is causal, then ${ℒ}\left\{h\right\}=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}h(t){\mathrm{e}}^{-st}\mathrm{d}t=\underset{0}{\overset{\mathrm{\infty }}{\int }}h(t){\mathrm{e}}^{-st}$. If we let $s=i\lambda$, we get ${ℒ}\left\{h\right\}(i\lambda )=\underset{0}{\overset{\mathrm{\infty }}{\int }}h(t){\mathrm{e}}^{-i\lambda t}\mathrm{d}t=\sqrt{2\pi }\hat{h}(\lambda )$

Suppose that $\hat{h}(\lambda )$ is the system function $\hat{h}(\lambda )=\frac{1}{\sqrt{2\pi }}\underset{0}{\overset{\mathrm{\infty }}{\int }}H(t){\mathrm{e}}^{i\lambda t}\mathrm{d}t$, hence $\hat{h}(\lambda )=\frac{1}{\sqrt{2\pi }}{ℒ}\left\{H\right\}(s=i\lambda )$

Define $h(t):=\{\begin{array}{ll}H(t)& 0\le t\\ 0& t<0\end{array}$. Then $\hat{h}(\lambda )=\frac{1}{\sqrt{2\pi }}{ℒ}\left\{H\right\}(i\lambda )=\frac{1}{\sqrt{2\pi }}\underset{0}{\overset{\mathrm{\infty }}{\int }}H(t){\mathrm{e}}^{-i\lambda t}\mathrm{d}t={ℱ}[h]$

Butterworth

$\hat{h}(\lambda )=\frac{A}{\sqrt{2\pi }(\alpha +i\lambda )}=\frac{A}{\sqrt{2\pi }}\frac{1}{\alpha +s}$, where $\frac{1}{\alpha +s}=\underset{0}{\overset{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-\alpha t}\cdot {\mathrm{e}}^{-st}\mathrm{d}t={ℒ}\left\{{\mathrm{e}}^{-\alpha t}\right\}$

Look up Papoulis in regard to signal processing

Example

Consider the simple circuit containing an input voltage $v$, a resistor of resistance $R$, and a capacitor with capacitance $R$ connected in series with a current of $I$ running through it.

$v(t)=IR+\frac{Q}{C}$

We also have that $I=\frac{\mathrm{d}Q}{\mathrm{d}t}$, hence

$\begin{array}{rl}v(t)& =R\frac{\mathrm{d}Q}{\mathrm{d}t}+\frac{Q}{C}\\ \frac{v(t)}{R}=\frac{\mathrm{d}Q}{\mathrm{d}t}+\frac{Q}{RC}\end{array}$

note that $RC$ has units of time.

Let $\alpha =\frac{1}{RC}$. then $\frac{v(t)}{R}=\frac{\mathrm{d}Q}{\mathrm{d}t}+\alpha Q=f(t)$. Suppose that at $t=0$, $Q(0)=0$, and $f(t)=0$ for all $t\le 0$.

Nothing can stop us from taking the Laplace transform of the function:

$\begin{array}{rl}{ℒ}\{Q=\alpha Q\}(s)& =s{ℒ}\left\{Q\right\}+\alpha {ℒ}\left\{Q\right\}\\ & =F(s)\\ {ℒ}\left\{Q\right\}& =\frac{1}{s+\alpha }F(s)\\ F(s)& =\underset{0}{\overset{\mathrm{\infty }}{\int }}f(t){\mathrm{e}}^{-st}\mathrm{d}t\\ F(i\lambda )& =\frac{0}{\mathrm{\infty }}f(t){\mathrm{e}}^{-i\lambda t}\mathrm{d}t=\sqrt{2\pi }{ℒ}\left\{f\right\}(i\lambda )\\ {ℒ}\left\{Q\right\}(i\lambda )& =\frac{1}{\alpha +i\lambda }\sqrt{2\pi }\hat{f}(\lambda )\\ \frac{1}{\alpha +i\lambda }& =\sqrt{2\pi }\cdot \frac{1}{\sqrt{2\pi }}{ℱ}[A{\mathrm{e}}^{-\alpha t}H(t)]\\ \frac{\sqrt{2\pi }}{\alpha +ih}& ={ℱ}\left[A{\mathrm{e}}^{-\alpha t}\right](i\lambda )\\ F(i\lambda )=\sqrt{2\pi }\hat{f}(\lambda )\\ \frac{1}{\sqrt{2\pi }}{ℒ}[Q](i\lambda )& =\frac{1}{\alpha +i\lambda }\cancel{\sqrt{2\pi }}\hat{f}(\lambda )\\ & =\hat{Q}(\lambda )=\frac{1}{\alpha +i\lambda }\hat{f}(\lambda )\\ \dots \end{array}$

Signals can be filtered by running them through a bunch of circuits

Sampling Theorem

We say that $f(t)$ is band-limited if and only if $\hat{f}(\lambda )=0$ for $\left|\lambda \right|>\mathrm{\Omega }$ for some finite $\mathrm{\Omega }$

$\mathrm{\Omega }$ is a circular or angular frequency, and the frequency we would work with is ${\nu }_{sig}=\frac{\mathrm{\Omega }}{2\pi }$

Let $f(t)$ be band-limited

$\hat{f}(\lambda )=0$ outside of the band $[-\mathrm{\Omega },\mathrm{\Omega }]$. Then

$f(t)=\sum _{j=-\mathrm{\infty }}^{\mathrm{\infty }}f\left(\frac{j\pi }{\mathrm{\Omega }}\right)\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}(\mathrm{\Omega }t-j\pi )$

Hence, we only need a finite number of discrete samples of the signal $f(x)$ to reconstruct the entire signal