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(New page: <b><u>Nick Christman</u></b><br><br> '''Find <math>\mathcal{F}[10^{t}g(t)e^{j2 \pi ft_{0}}]</math><br/>''' To begin, we know that<br/> <math> \mathcal{F}[10^{t}g(t)e^{...)
 
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[[Fourier Transform Properties|<b>Fourier Transform Properties</b>]]

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[[Nick Christman|<b><u>Nick Christman</u></b>]]<br><br>
[[Nick Christman|<b><u>Nick Christman</u></b>]]<br><br>
'''Find <math>\mathcal{F}[10^{t}g(t)e^{j2 \pi ft_{0}}]</math><br/>'''
1. '''Find <math>\mathcal{F} \left[ g(t)e^{j2 \pi f_{0}t} \right] </math><br/>'''

This is a fairly straightforward property and is known as ''complex modulation''<br/>

<math>
\mathcal{F} \left[ g(t)e^{j2 \pi f_{0}t} \right] = \int_{- \infty}^{\infty} \left[ g(t)e^{j2 \pi f_{0}t} \right] e^{-j2 \pi ft} \,dt
</math>

Combining terms, we get:


<math>
To begin, we know that<br/>
\int_{- \infty}^{\infty} \left[ g(t)e^{j2 \pi f_{0}t} \right] e^{-j2 \pi ft} \,dt = \int_{- \infty}^{\infty} g(t)e^{-j2 \pi (f-f_{0})t} \,dt
</math>
<br/>


Now let's make the following substitution <math> \displaystyle \theta = f-f_{0}</math>
<math> \mathcal{F}[10^{t}g(t)e^{j2 \pi ft_0}] = \int_{-\infty}^{\infty}10^{t}g(t)e^{j2 \pi ft_0}e^{-j2 \pi ft}\,dt = \int_{-\infty}^{\infty}10^{t}g(t)e^{j2 \pi f(t_{0}-t)}\,dt</math>


This now gives us a surprisingly familiar function:
But recall that <math>e^{j2 \pi f(t_{0}-t)} \equiv \delta (t_{0}-t) \mbox{ or } \delta (t-t_{0})</math>


<math>
\int_{- \infty}^{\infty} g(t)e^{-j2 \pi (f-f_{0})t} \,dt = \int_{- \infty}^{\infty} g(t)e^{-j2 \pi \theta t} \,dt
</math>
<br/>


This looks just like <math> \displaystyle G(\theta )</math>!
Because of this definition, our problem has now been simplified significantly: <br/>


We can now conclude that:
<math> \mathcal{F}[10^{t}g(t)e^{j2 \pi ft_0}] = \int_{-\infty}^{\infty}10^{t}g(t) \delta (t-t_{0})\,dt = 10^{t_0}g(t_0)</math> <br/>
<br/>


<math>
Therefore,
\mathcal{F} \left[ g(t)e^{j2 \pi f_{0}t} \right] = G(\theta ) = G(f-f_{0})
</math>
<br>


Looks good - Kevin
<math> \mathcal{F}[10^{t}g(t)e^{j2 \pi ft_0}] = 10^{t_0}g(t_0) </math>

Latest revision as of 08:36, 8 November 2009

Fourier Transform Properties


Nick Christman

1. Find

This is a fairly straightforward property and is known as complex modulation

Combining terms, we get:


Now let's make the following substitution

This now gives us a surprisingly familiar function:


This looks just like !

We can now conclude that:


Looks good - Kevin