Homework Four: Difference between revisions
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[[Nick Christman|<b><u>Nick Christman</u></b>]]<br><br> |
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'''Find <math>\mathcal{F}[ |
1. '''Find <math>\mathcal{F} \left[ g(t)e^{j2 \pi f_{0}t} \right] </math><br/>''' |
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This is a fairly straightforward property and is known as ''complex modulation''<br/> |
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To begin, we know that<br/> |
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<math> |
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</math> |
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Combining terms, we get: |
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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> |
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<math> |
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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> |
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\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 |
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</math> |
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<br/> |
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Now let's make the following substitution <math> \displaystyle \theta = f-f_{0}</math> |
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Because of this definition, our problem has now been simplified significantly: <br/> |
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This now gives us a surprisingly familiar function: |
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<math> |
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Therefore, |
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</math> |
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<br/> |
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This looks just like <math> \displaystyle G(\theta )</math>! |
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<math> \mathcal{F}[10^{t}g(t)e^{j2 \pi ft_0}] = 10^{t_0}g(t_0) </math> |
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We can now conclude that: |
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<br/> |
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<math> |
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\mathcal{F} \left[ g(t)e^{j2 \pi f_{0}t} \right] = G(\theta ) = G(f-f_{0}) |
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</math> |
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<br> |
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Looks good - Kevin |
Latest revision as of 08:36, 8 November 2009
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