Homework Six: Difference between revisions
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'''Perform the following tasks:''' |
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[[Nick Christman|<b><u>Nick Christman</u></b>]]<br/> |
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(a) Show <math> \mathcal{F}\left[ \int_{- \infty}^{t} s(\lambda ) \,d\lambda \right] = \frac{S(f)}{j2 \pi f} \mbox{ if } S(0) = 0 </math>. HINT: <math> S(0) = S(f) \vert _{_{f=0}} = \int_{- \infty}^{\infty} s(t)e^{-j2 \pi (f \rightarrow 0)t} \,dt = \int_{- \infty}^{\infty} s(t) \,dt </math> |
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(b) If <math> S(0) \neq 0 </math> can you find <math> \mathcal{F}\left[ \int_{- \infty}^{t} s(\lambda ) \,d\lambda \right] </math> in terms of <math> \displaystyle S(0) </math>? |
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(c) Do another property on the Wiki and get it reviewed (i.e. review a second property) -- [[Fourier Transform Properties]] |
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(i) '''Find <math>\mathcal{F} \left[ g(t-t_{0})e^{j2 \pi f_{0}t} \right]</math><br/>''' |
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-- Using the above definition of ''complex modulation'' and the definition from class of a ''time delay'' (a.k.a "the slacker function"), I will attempt to show a hybrid of the two... |
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By definition we know that: |
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<math> |
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\mathcal{F} \left[ g(t-t_{0})e^{j2 \pi f_{0}t} \right] = \int_{- \infty}^{\infty} \left[ g(t-t_{0})e^{j2 \pi f_{0}t} \right] e^{-j2 \pi ft} \,dt |
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</math> |
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Rearranging terms we get: |
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<math> |
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\int_{- \infty}^{\infty} \left[ g(t-t_{0})e^{j2 \pi f_{0}t} \right] e^{-j2 \pi ft} \,dt = \int_{- \infty}^{\infty} g(t-t_{0})e^{-j2 \pi (f-f_{0})t} \,dt |
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</math> |
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Now lets make the substitution <math>\lambda = t-t_{0} \rightarrow t = \lambda + t_{0}</math>. |
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This leads us to: |
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<math> |
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\int_{- \infty}^{\infty} g(t-t_{0})e^{-j2 \pi (f-f_{0})t} \,dt = \int_{- \infty}^{\infty} g(\lambda )e^{-j2 \pi (f-f_{0})(\lambda + t_{0})} \,dt |
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</math> |
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After some simplification and rearranging terms, we get: |
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<math> |
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\int_{- \infty}^{\infty} g(\lambda )e^{-j2 \pi (f-f_{0})(\lambda + t_{0})} \,dt = \int_{- \infty}^{\infty} g(\lambda )e^{-j2 \pi (f-f_{0})\lambda } e^{-j2 \pi (f-f_{0})t_{0}} \,dt |
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</math> |
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Rearranging the terms yet again, we get: |
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<math> |
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\int_{- \infty}^{\infty} g(\lambda )e^{-j2 \pi (f-f_{0})\lambda } e^{-j2 \pi (f-f_{0})t_{0}} \,dt = e^{-j2 \pi (f-f_{0})t_{0}} \left[ \int_{- \infty}^{\infty} g(\lambda )e^{-j2 \pi (f-f_{0})\lambda } \,dt \right] |
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</math> |
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We know that the exponential in terms of <math>\displaystyle t_{0}</math> is simply a constant and because of the Fourier Property of ''complex modualtion'', we finally get: |
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<math> |
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\mathcal{F} \left[ g(t)e^{j2 \pi f_{0}t} \right] = G(f-f_{0})e^{-j2 \pi (f-f_{0})t_{0}} |
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</math> |
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(ii) |
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I reviewed Max's second Fourier Transform property: <math>\mathcal{F}\bigg[\int_{-\infty}^{\infty}g(t) h^*(t) dt\bigg]</math> |
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As near as I can tell, it all looks legitimate. I made one comment about adding an additional step to make the proof/identity more complete, but that was all that I could find. |
Latest revision as of 16:31, 31 October 2009
Perform the following tasks:
(a) Show . HINT:
(b) If can you find in terms of ?
(c) Do another property on the Wiki and get it reviewed (i.e. review a second property) -- Fourier Transform Properties
(i) Find
-- Using the above definition of complex modulation and the definition from class of a time delay (a.k.a "the slacker function"), I will attempt to show a hybrid of the two...
By definition we know that:
Rearranging terms we get:
Now lets make the substitution .
This leads us to:
After some simplification and rearranging terms, we get:
Rearranging the terms yet again, we get:
We know that the exponential in terms of is simply a constant and because of the Fourier Property of complex modualtion, we finally get:
(ii) I reviewed Max's second Fourier Transform property:
As near as I can tell, it all looks legitimate. I made one comment about adding an additional step to make the proof/identity more complete, but that was all that I could find.