Homework Four: Difference between revisions

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[[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/>
To begin, we know that<br/>


<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>
\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:
But recall that <math>\int_{-\infty}^{\infty}e^{j2 \pi f(t_{0}-t)}df \equiv \delta (t_{0}-t) \mbox{ or } \delta (t-t_{0})</math>


<math>
The following needs to be fixed, because the previous thing (just above this) which we just fixed wasn't an identity. Hint: <math>10^{t}</math> is related to <math>e^{t}</math>
\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>
Because of this definition, our problem has now been simplified significantly: <br/>


This now gives us a surprisingly familiar function:
<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/>


<math>
Therefore,
\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>!
<math> \mathcal{F}[10^{t}g(t)e^{j2 \pi ft_0}] = 10^{t_0}g(t_0) </math>

We can now conclude that:
<br/>

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

Looks good - Kevin

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