Homework Four: Difference between revisions

From Class Wiki
Jump to navigation Jump to search
No edit summary
No edit summary
 
(4 intermediate revisions by 3 users not shown)
Line 1: Line 1:
[[Fourier Transform Properties|Fourier Transform Properties]]
[[Fourier Transform Properties|<b>Fourier Transform Properties</b>]]


----
'''Find <math>\mathcal{F}[10^{t}g(t)e^{j2 \pi ft_{0}}]</math><br/>'''


[[Nick Christman|<b><u>Nick Christman</u></b>]]<br><br>
To begin, we know that<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}[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>


<math>
But recall that <math>e^{j2 \pi f(t_{0}-t)} \equiv \delta (t_{0}-t) \mbox{ or } \delta (t-t_{0})</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>
Because of this definition, our problem has now been simplified significantly: <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) \delta (t-t_{0})\,dt = 10^{t_0}g(t_0)</math> <br/>


This now gives us a surprisingly familiar function:
Therefore,


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

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 09:36, 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