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:

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

where

$\alpha _{n}\!={\frac {1}{T}}\int _{-{\frac {T}{2}}}^{\frac {T}{2}}x(t^{'})e^{\frac {-j2\pi nt^{'}}{T}}dt^{'}\!$

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

$\lim _{T\to \infty }\sum _{n=-\infty }^{\infty }({\frac {1}{T}}\int _{-{\frac {T}{2}}}^{\frac {T}{2}}x(t^{'})e^{\frac {-j2\pi nt^{'}}{T}}dt^{'})e^{\frac {j2\pi nt}{T}}\!$

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

${\frac {1}{T}}$	$\rightarrow$	$\,df$
${\frac {n}{T}}$	$\rightarrow$	$\,f$
$\sum _{n=-\infty }^{\infty }{\frac {1}{T}}$	$\rightarrow$	$\int _{-\infty }^{\infty }({\mbox{ }})\,df$

We can now write out the following:

$x(t)=\lim _{T\to \infty }\left[\sum _{n=-\infty }^{\infty }\left({\frac {1}{T}}\int _{-{\frac {T}{2}}}^{\frac {T}{2}}x(t')e^{-{\frac {j2\pi nt'}{T}}}\,dt'\right)e^{\frac {j2\pi nt}{T}}\right]$

which can also be written as:

$x(t)=\int _{-\infty }^{\infty }\left(\int _{-\infty }^{\infty }\,x(t')e^{-j2\pi ft'}\,dt'\right)e^{j2\pi tf}\,df$

using,

$\alpha _{n}\rightarrow X(f)\!$

we now have

$\,X(f)=\int _{-\infty }^{\infty }\,x(t')e^{-j2\pi ft'}\,dt'$