MATH 414 Lecture 20

« previous | Monday, March 3, 2014 | next »

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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{h}(\lambda) = \frac{A}{\sqrt{2\pi} \, (\alpha + i\,\lambda)} = \frac{A}{\sqrt{2\pi}} \, \frac{1}{\alpha + s}} , where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v} , a resistor of resistance Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R} , and a capacitor with capacitance Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R} connected in series with a current of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I} running through it.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v(t) = I \, R + \frac{Q}{C}}

We also have that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I = \frac{\mathrm{d}Q}{\mathrm{d}t}} , hence

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} 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{align}}

note that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R\,C} has units of time.

Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha = \frac{1}{R \, C}} . then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{v(t)}{R} = \frac{\mathrm{d}Q}{\mathrm{d}t} + \alpha \, Q = f(t)} . Suppose that at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t=0} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Q(0) = 0} , and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(t)=0} for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t \le 0} .

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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \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{align}}

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

Sampling Theorem

We say that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(t)} is band-limited if and only if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{f}(\lambda) = 0} for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left| \lambda \right| > \Omega} for some finite Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Omega}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Omega} is a circular or angular frequency, and the frequency we would work with is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \nu_{sig} = \frac{\Omega}{2\pi}}

Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(t)} be band-limited

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{f}(\lambda) = 0} outside of the band Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[ -\Omega, \Omega \right]} . Then

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)} to reconstruct the entire signal