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(New page: 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 ...)
 
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Begining with a Fourier Series:
Begining with a Fourier Series:


<math>x(t) = x(t+T) = \sum_{n=- \infty}^{\infty} \alpha _n e^{\frac{j2 \pi nt}{T}}</math>
(1) <math>x(t) = x(t+T) = \sum_{n=- \infty}^{\infty} \alpha _n e^{\frac{j2 \pi nt}{T}}</math>


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


<math>\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}}\!</math>
(2)<math>\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}}\!</math>


In order to evaluate this limit we need the following relationships:
In order to evaluate this limit we need the following relationships:
<table>
<tr>
<td width=100>
<math>\frac{1}{T}</math>
</td>
<td width=50>
<math>\rightarrow</math>
</td>
<td widt=100>
<math>\,df</math>
</td>
</tr>
<tr>
<td>
<math>\frac{n}{T}</math>
</td>
<td>
<math>\rightarrow</math>
</td>
<td>
<math>\,f</math>
</td>
</tr>
<tr>
<td>
<math>\sum_{n=- \infty}^{\infty} \frac{1}{T}</math>
</td>
<td>
<math>\rightarrow</math>
</td>
<td>
<math>\int_{-\infty}^{\infty}(\mbox{ })\,df</math>
</td>
</tr>
</table>

We can now write out the following:

(3)<math>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] </math>

which can also be written as:

(4)<math>x(t) =\int_{- \infty}^{\infty} \left( \int_{-\infty}^{\infty} \,x(t') e^{-j2 \pi ft'} \,dt' \right) e^{j2 \pi tf} \,df</math>

using,

<math>\alpha_n\rightarrow X(f) \!</math>

we now have

(5)<math>\,X(f) = \int_{-\infty}^{\infty} \,x(t') e^{-j2 \pi ft'} \,dt'

</math>

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:


<table>
<tr>
<td width=100><math>\,X(f) = \int_{-\infty}^{\infty} \,x(t) e^{-j2 \pi ft} \,dt</math></td>
<td width=100 align="center"><math>\equiv</math></td>
<td width=100><math>\langle x(t) | e^{j2 \pi tf} \rangle </math></td>
<td width=200><math>\,x(t) \mbox{ projected onto } e^{j2 \pi tf}</math></td>
</tr>

<tr>
<td><math>\,x(t) = \int_{-\infty}^{\infty} \,X(f) e^{-j2 \pi ft} \,df</math></td>
<td align="center"><math>\equiv</math></td>
<td><math>\langle X(f) | e^{j2 \pi tf} \rangle </math></td>
<td><math>\,x(f) \mbox{ projected onto } e^{j2 \pi tf}</math></td>
</tr>

</table>

Where,

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



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

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


<table>
<tr>
<td width=100>
<math>\int_{- \infty}^{\infty} \left( \int_{-\infty}^{\infty} \,x(t') e^{-j2 \pi ft'} \,dt' \right) e^{j2 \pi tf} \,df</math>
</td>
<td width=100 align="center">
<math>\equiv</math>
</td>
<td width=100>
<math>\int_{- \infty}^{\infty} \,x(t') \left( \int_{-\infty}^{\infty} e^{j2 \pi f(t-t')} \,df \right) \,dt'</math>
</td>
</tr>
</table>

Notice

<math>e^{j2 \pi f(t-t')} \equiv \delta (t-t')</math>

Similarly,

<table>
<tr>
<td width=100>
<math>\int_{- \infty}^{\infty} \left( \int_{-\infty}^{\infty} \,X(f') e^{j2 \pi f't} \,dt' \right) e^{-j2 \pi tf} \,df</math>
</td>
<td width=100 align="center">
<math>\equiv</math>
</td>
<td width=100>
<math>\int_{- \infty}^{\infty} \,X(f') \left( \int_{-\infty}^{\infty} e^{j2 \pi t(f'-f)} \,df \right) \,dt'</math>
</td>
</tr>
</table>

Again, notice

<math>e^{j2 \pi f(f'-f)} \equiv \delta (f'-f) = \delta (f-f')</math>

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

Latest revision as of 14:58, 15 October 2009

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)

where

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

(2)

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

We can now write out the following:

(3)

which can also be written as:

(4)

using,

we now have

(5)

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:


Where,

Is the Fourier transform, or


Is the inverse Fourier transform, or

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


Notice

Similarly,

Again, notice

Both the time-domain and frequency domain have non-zero integrals when respectively.