Assignment

Summery of the class notes from Oct. 5:

What if a periodic signal had an infinite period? We would no longer be able to tell the difference between it and a non periodic signal. We can use this property to look at signals that do not have a period (an observable one at least).

Begining with a Fourier Series:

(1) $x(t)=x(t+T)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{\alpha }_n{e}^{\frac{j2\pi nt}{T}}$

where

${\alpha }_n=\frac{1}{T}\underset{-\frac{T}{2}}{\overset{\frac{T}{2}}{\int }}x(t^{\text{'}})e^{\frac{-j2\pi n{t}^{\text{'}}}{T}}d{t}^{\text{'}}$

We then take the limit of a Fourier series as its period T approaches infinity:

(2)$\underset{T\to \mathrm{\infty }}{\lim }\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}(\frac{1}{T}\underset{-\frac{T}{2}}{\overset{\frac{T}{2}}{\int }}x(t^{\text{'}})e^{\frac{-j2\pi n{t}^{\text{'}}}{T}}d{t}^{\text{'}})e^{\frac{j2\pi nt}{T}}$

In order to evaluate this limit we need the following relationships:

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

We can now write out the following:

(3)$x(t)=\underset{T\to \mathrm{\infty }}{\lim }\left[\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}(\frac{1}{T}\underset{-\frac{T}{2}}{\overset{\frac{T}{2}}{\int }}x(t^{\prime })e^{-\frac{j2\pi n{t}^{\prime }}{T}}d{t}^{\prime })e^{\frac{j2\pi nt}{T}}\right]$

which can also be written as:

(4)$x(t)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}(\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t^{\prime })e^{-j2\pi f{t}^{\prime }}d{t}^{\prime })e^{j2\pi tf}df$

using,

${\alpha }_n\to X(f)$

we now have

(5)$X(f)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t^{\prime })e^{-j2\pi f{t}^{\prime }}d{t}^{\prime }$

We can now relate a signal in the time domain to a signal in the frequency domain. Using vector notation we can show this relationship as such:

$X(f)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t)e^{-j2\pi ft}dt$	$\equiv$	$⟨x(t)|e^{j2\pi tf}⟩$	$x(t)\text{ projected onto }{e}^{j2\pi tf}$
$x(t)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}X(f)e^{-j2\pi ft}df$	$\equiv$	$⟨X(f)|e^{j2\pi tf}⟩$	$x(f)\text{ projected onto }{e}^{j2\pi tf}$

Where,

$<x(t)|e^{j2\pi ft}>$ Is the Fourier transform, or ${ℱ}[x(t)]$

$<X(f)|e^{j2\pi ft}>$ Is the inverse Fourier transform, or ${{ℱ}}^{-1}[X(f)]$

Now we can use our new tool, the Fourier transform on equation (2) to give us the following:

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}(\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t^{\prime })e^{-j2\pi f{t}^{\prime }}d{t}^{\prime })e^{j2\pi tf}df$

$\equiv$

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t^{\prime })\left(\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{j2\pi f(t-t^{\prime })}df\right)d{t}^{\prime }$

Notice

$e^{j2\pi f(t-t^{\prime })}\equiv \delta (t-t^{\prime })$

Similarly,

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}(\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}X(f^{\prime })e^{j2\pi f^{\prime }t}d{t}^{\prime })e^{-j2\pi tf}df$

$\equiv$

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}X(f^{\prime })\left(\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{j2\pi t(f^{\prime }-f)}df\right)d{t}^{\prime }$

Again, notice

$e^{j2\pi f(f^{\prime }-f)}\equiv \delta (f^{\prime }-f)=\delta (f-f^{\prime })$

Both the time-domain and frequency domain have non-zero integrals when $t=t^{\prime }\text{ and }f=f^{\prime }$ respectively.