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:

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

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:

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

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:

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

which can also be written as:

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

using,

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

we now have

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