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

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:

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

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:

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

which can also be written as:

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

using,

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

we now have

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

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:

$\,X(f)=\int _{-\infty }^{\infty }\,x(t)e^{-j2\pi ft}\,dt$	$\equiv$	$\langle x(t)|e^{j2\pi tf}\rangle$	$\,x(t){\mbox{ projected onto }}e^{j2\pi tf}$
$\,x(t)=\int _{-\infty }^{\infty }\,X(f)e^{-j2\pi ft}\,df$	$\equiv$	$\langle X(f)|e^{j2\pi tf}\rangle$	$\,x(f){\mbox{ projected onto }}e^{j2\pi tf}$

Where,

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

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

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

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

$\equiv$

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

Notice

$e^{j2\pi f(t-t')}\equiv \delta (t-t')$

Similarly,

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

$\equiv$

$\int _{-\infty }^{\infty }\,X(f')\left(\int _{-\infty }^{\infty }e^{j2\pi t(f'-f)}\,df\right)\,dt'$

Again, notice

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

Both the time-domain and frequency domain have non-zero integrals when $\,t=\,t'{\mbox{ and }}\,f=\,f'$ respectively.