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 ${\hat {h}}(\lambda )={\frac {1}{\sqrt {2\pi }}}\,\left(\theta \left(\lambda \right)+\theta \left(\lambda -\lambda _{c}\right)\right)$.

For $f(t)=\theta (t)-\theta (t-t_{0})$, we have $f*h(t)={\frac {1}{\pi }}\int _{\lambda _{c}\,(t-t_{c})}^{\lambda _{c}\,t}{\frac {\sin {u}}{u}}\,\mathrm {d} u$

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

$L(f)=f*h(t)=\int _{-\infty }^{\infty }f(\tau )\,h(t-\tau )\,\mathrm {d} \tau =\int _{t_{0}}^{\infty }f(\tau )\,h(t-\tau )\,\mathrm {d} \tau$

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

$\int _{t_{0}}^{\infty }f(\tau )\cdot 0\,\mathrm {d} \tau =0$.

quod erat demonstrandum

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

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

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

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

Butterworth

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

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.

$v(t)=I\,R+{\frac {Q}{C}}$

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

${\begin{aligned}v(t)&=R\,{\frac {\mathrm {d} Q}{\mathrm {d} t}}+{\frac {Q}{C}}\\{\frac {v(t)}{R}}={\frac {\mathrm {d} Q}{\mathrm {d} t}}+{\frac {Q}{R\,C}}\end{aligned}}$

note that $R\,C$ has units of time.

Let $\alpha ={\frac {1}{R\,C}}$. 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 $t\leq 0$.

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

${\begin{aligned}{\mathcal {L}}\left\{Q=\alpha \,Q\right\}(s)&=s\,{\mathcal {L}}\left\{Q\right\}+\alpha {\mathcal {L}}\left\{Q\right\}\\&=F(s)\\{\mathcal {L}}\left\{Q\right\}&={\frac {1}{s+\alpha }}\,F(s)\\F(s)&=\int _{0}^{\infty }f(t)\,\mathrm {e} ^{-s\,t}\,\mathrm {d} t\\F(i\,\lambda )&={\frac {0}{\infty }}f(t)\,\mathrm {e} ^{-i\,\lambda \,t}\,\mathrm {d} t={\sqrt {2\pi }}\,{\mathcal {L}}\left\{f\right\}(i\,\lambda )\\{\mathcal {L}}\left\{Q\right\}(i\,\lambda )&={\frac {1}{\alpha +i\,\lambda }}\,{\sqrt {2\pi }}\,{\hat {f}}(\lambda )\\{\frac {1}{\alpha +i\,\lambda }}&={\sqrt {2\pi }}\cdot {\frac {1}{\sqrt {2\pi }}}\,{\mathcal {F}}\left[A\,\mathrm {e} ^{-\alpha \,t}\,H(t)\right]\\{\frac {\sqrt {2\pi }}{\alpha +i\,h}}&={\mathcal {F}}\left[A\,\mathrm {e} ^{-\alpha \,t}\right](i\,\lambda )\\F(i\,\lambda )={\sqrt {2\pi }}\,{\hat {f}}(\lambda )\\{\frac {1}{\sqrt {2\pi }}}\,{\mathcal {L}}[Q](i\,\lambda )&={\frac {1}{\alpha +i\,\lambda }}\,{\cancel {\sqrt {2\pi }}}\,{\hat {f}}(\lambda )\\&={\hat {Q}}(\lambda )={\frac {1}{\alpha +i\,\lambda }}\,{\hat {f}}(\lambda )\\\ldots \end{aligned}}$

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 $\left|\lambda \right|>\Omega$ for some finite $\Omega$

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

Let $f(t)$ be band-limited

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

$f(t)=\sum _{j=-\infty }^{\infty }f\left({\frac {j\,\pi }{\Omega }}\right)\,\mathrm {sinc} \left(\Omega \,t-j\,\pi \right)$

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