Homework Three: Difference between revisions

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From this we can see that <math>\,x(t)</math> is the inverse Laplace transform of <math>\,X(f)</math>. Similarly, <math>\,X(f)</math> is the Laplace transform of <math>\,x(t)</math>
From this we can see that <math>\,x(t)</math> is the inverse Laplace transform of <math>\,X(f)</math>. Similarly, <math>\,X(f)</math> is the Laplace transform of <math>\,x(t)</math>

The next thing we did was rearranged some limits within the equations given above to give us the following:

<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>






Revision as of 18:31, 14 October 2009

October 5th, 2009, class notes (as interpreted by Nick Christman)


The topic covered in class on October 5th was about how to deal with signals that are not periodic.

Given the following Fourier series, what if the signal is not periodic?

where

To investigate this potential disaster, let's look at what happens as the period increases (i.e. not periodic). Essentially, as we can say the following:

With this, we get the following:

Given the above equivalence, we say the following:

Therefore, we have obtained an equation to relate the Fourier analysis of a function in the time-domain to the frequency-domain:


From this we can see that is the inverse Laplace transform of . Similarly, is the Laplace transform of

The next thing we did was rearranged some limits within the equations given above to give us the following:


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