Class lecture notes October 5 - HW3

Max Woesner

Back to my Home Page

Homework #3 - Class lecture notes October 5

The following notes are my interpretation of the material covered in class on October 5, 2009

Nonperiodic Signals

In real life, most systems have a finite time period and can be fairly easily evaluated as periodic. However, we still want to be able to work with nonperiodic signals.
A nonperiodic signal can be thought of as periodic signal with an infinite period. To deal with such signals we can take the limit of the Fourier series as the period $T$ goes to infinity, or
$\underset{T\to \mathrm{\infty }}{\lim }\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}(\frac{1}{T}\underset{-\frac{T}{2}}{\overset{\frac{T}{2}}{\int }}x(t^{\text{'}})e^{\frac{-j2\pi n{t}^{\text{'}}}{T}}d{t}^{\text{'}})e^{\frac{j2\pi nt}{T}}$, where $\frac{1}{T}\underset{-\frac{T}{2}}{\overset{\frac{T}{2}}{\int }}x(t^{\text{'}})e^{\frac{-j2\pi n{t}^{\text{'}}}{T}}d{t}^{\text{'}}$ is the ${\alpha }_n$ term.
We want to remove the restriction $x(t)=x(t+T)$, which we can do as follows.

$\frac{1}{T}⟶df$

$\frac{n}{T}⟶f$

$\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{T}⟶\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}()df$

${\alpha }_n⟶X(f)$

So $x(t)=\underset{T\to \mathrm{\infty }}{\lim }\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{T}(\underset{-\frac{T}{2}}{\overset{\frac{T}{2}}{\int }}x(t^{\text{'}})e^{\frac{-j2\pi n{t}^{\text{'}}}{T}}d{t}^{\text{'}})e^{\frac{j2\pi nt}{T}}$
Using $\frac{n}{T}=f$ and the information above, we can rewrite the equation.
$x(t)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}(\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t^{\text{'}})e^{-j2\pi f{t}^{\text{'}}}d{t}^{\text{'}})e^{j2\pi ft}df$, where $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t^{\text{'}})e^{-j2\pi f{t}^{\text{'}}}d{t}^{\text{'}}=X(f)$
So $$\begin{gathered} x(t)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}X(f)e^{+j2\pi ft}df= \\ <X(f)|e^{j2\pi ft}> \end{gathered}$$ This is the inverse Fourier transform, or ${{ℱ}}^{-1}[X(f)]$
Also, $$\begin{gathered} X(f)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t)e^{-j2\pi ft}dt= \\ <x(t)|e^{-j2\pi ft}> \end{gathered}$$ This is the Fourier transform, or ${ℱ}[x(t)]$
So $x(t)={{ℱ}}^{-1}[X(f)]$ and $X(f)={ℱ}[x(t)]$
From above, $x(t)=\underset{-{\mathrm{\infty }}_f}{\overset{\mathrm{\infty }}{\int }}(\underset{-{\mathrm{\infty }}_{t^{\text{'}}}}{\overset{\mathrm{\infty }}{\int }}x(t^{\text{'}})e^{-j2\pi f{t}^{\text{'}}}d{t}^{\text{'}})e^{j2\pi ft}df$, so
$x(t)=\underset{-{\mathrm{\infty }}_{t^{\text{'}}}}{\overset{\mathrm{\infty }}{\int }}x(t^{\text{'}})(\underset{-{\mathrm{\infty }}_{\delta (t-t^{\text{'}})}}{\overset{\mathrm{\infty }}{\int }}{e}^{j2\pi f(t-t^{\text{'}})}df)d{t}^{\text{'}}$, where $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{j2\pi f(t-t^{\text{'}})}df=\delta (t-t^{\text{'}})$, which is the projection $<e^{j2\pi ft}|e^{j2\pi f{t}^{\text{'}}}>$ with respect to $f$.
This identity for $x(t)$ is the defining property of the impulse function.
Similarly, $X(f)=\underset{-{\mathrm{\infty }}_t}{\overset{\mathrm{\infty }}{\int }}(\underset{-{\mathrm{\infty }}_{f^{\text{'}}}}{\overset{\mathrm{\infty }}{\int }}X(f^{\text{'}})e^{+j2\pi f{t}^{\text{'}}}d{f}^{\text{'}})e^{-j2\pi ft}dt$, where $\underset{-{\mathrm{\infty }}_{f^{\text{'}}}}{\overset{\mathrm{\infty }}{\int }}X(f^{\text{'}})e^{+j2\pi f{t}^{\text{'}}}d{f}^{\text{'}}=x(t)$,so
$X(f)=\underset{-{\mathrm{\infty }}_{f^{\text{'}}}}{\overset{\mathrm{\infty }}{\int }}X(f^{\text{'}})(\underset{-{\mathrm{\infty }}_t}{\overset{\mathrm{\infty }}{\int }}{e}^{j2\pi t(f^{\text{'}}-f)}dt)d{f}^{\text{'}}$, where $\underset{-{\mathrm{\infty }}_t}{\overset{\mathrm{\infty }}{\int }}{e}^{j2\pi t(f^{\text{'}}-f)}dt=\delta (f^{\text{'}}-f)=\delta (f-f^{\text{'}})$, which is the projection $<e^{j2\pi ft}|e^{j2\pi f^{\text{'}}t}>$ with respect to $t$.

The Linear Time Invariant System Game

Recall the Linear Time Invariant System Game that can be used to help us understand the impulse response of a linear time invariant system.

Input$-----⟶$	Linear Time Invariant System	$⟶$Output	Reason
$\delta (t)$	$-------⟶$	$h(t)$	Given
$\delta (t-t_0)$	$-------⟶$	$h(t-t_0)$	Time Invariance
$x(t_0)\delta (t-t_0)$	$-------⟶$	$x(t_0)h(t-t_0)$	Proportionality
$$\begin{gathered} \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t_0)\delta \\ (t-t_0)d{t}_0 \end{gathered}$$	$-------⟶$	$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t_0)h(t-t_0)d{t}_0$	Superposition

where $$\begin{gathered} \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t_0)\delta \\ (t-t_0)d{t}_0=x(t) \end{gathered}$$ for any $x(t)$ and $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t_0)h(t-t_0)d{t}_0$ is the convolution integral.

We can expand the game further.

Input$-----⟶$	Linear Time Invariant System	$⟶$Output	Reason
$\delta (t)$	$-------⟶$	$h(t)$	Given
$\delta (t-t_0)$	$-------⟶$	$h(t-t_0)$	Time Invariance
$$\begin{gathered} x(t_0)\delta \\ (t-t_0) \end{gathered}$$	$-------⟶$	$x(t_0)h(t-t_0)$	Proportionality
$x(t)$	$-------⟶$	$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t_0)h(t-t_0)d{t}_0$	Superposition
$e^{j2\pi ft}$	$-------⟶$	$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{j2\pi f{t}_0}h(t-t_0)d{t}_0$	Superposition

Let $$\begin{gathered} \lambda \\ =t-t_0 \end{gathered}$$, so $t_0=t-\lambda$ and $d{t}_0=-d\lambda$
Therefore $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{j2\pi f{t}_0}h(t-t_0)d{t}_0=\underset{+\mathrm{\infty }}{\overset{-\mathrm{\infty }}{\int }}h(\lambda )e^{j2\pi f(t-\lambda )}(-d\lambda )=e^{j2\pi ft}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}h(\lambda )e^{-j2\pi f\lambda }d\lambda$
This tells us that $e^{j2\pi ft}$ is the eigenfunction and $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}h(\lambda )e^{-j2\pi f\lambda }d\lambda$ is the eigenvalue of all linear time invariant systems.
Also, the eigenvalue $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}h(\lambda )e^{-j2\pi f\lambda }d\lambda$ is $H(f)$, which equals the Fourier transform of $h(t)$, or $H(f)={ℱ}[h(t)]$
We can expand the game even further.

Input$-----⟶$	Linear Time Invariant System	$⟶$Output	Reason
$\delta (t)$	$-------⟶$	$h(t)$	Given
$\delta (t-t_0)$	$-------⟶$	$h(t-t_0)$	Time Invariance
$$\begin{gathered} x(t_0)\delta \\ (t-t_0) \end{gathered}$$	$-------⟶$	$x(t_0)h(t-t_0)$	Proportionality
$x(t)$	$-------⟶$	$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t_0)h(t-t_0)d{t}_0$	Superposition
$e^{j2\pi ft}$	$-------⟶$	$H(f)e^{j2\pi ft}$	Superposition
$X(f)e^{j2\pi ft}$	$-------⟶$	$X(f)H(f)e^{j2\pi ft}$	Proportionality
$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}X(f)e^{j2\pi ft}df$	$-------⟶$	$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}X(f)H(f)e^{j2\pi ft}df$	Superposition

where $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}X(f)e^{j2\pi ft}df=x(t)$ and $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}X(f)H(f)e^{j2\pi ft}df={{ℱ}}^{-1}[H(f)X(f)]$
This is helpful because in frequency space, when we go through a linear time invariant system, it multiplies by the transfer function, compared to time space, which convolves the impulse response, and we would all prefer to do multiplication rather than convolution.