Homework Six: Difference between revisions

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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 S(0)?
(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)
(c) Do another property on the Wiki and get it reviewed (i.e. review a second property) -- [[Fourier Transform Properties]]
 
'''Find <math>\mathcal{F} \left[ g(t-t_{0})e^{j2 \pi f_{0}t} \right]</math><br/>'''
 
-- 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:
 
<math>
\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
</math>
 
Rearranging terms we get:
 
<math>
\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
</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:
 
<math>
\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
</math>
 
After some simplification and rearranging terms, we get:
 
<math>
\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
</math>
 
Rearranging the terms yet again, we get:
 
<math>
\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]
</math>
 
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:
 
<math>
\mathcal{F} \left[ g(t)e^{j2 \pi f_{0}t} \right] = G(f-f_{0})e^{-j2 \pi (f-f_{0})t_{0}}
</math>
 
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Revision as of 15:03, 31 October 2009

Perform the following tasks:


Nick Christman

(a) Show ℱ[∫−∞ts(λ)dλ]=S(F)j2πf if S(0)=0


(b)If S(0)≠0 can you find ℱ[∫−∞ts(λ)dλ] in terms of S(0)?


(c) Do another property on the Wiki and get it reviewed (i.e. review a second property) -- Fourier Transform Properties

Find ℱ[g(t−t0)ej2πf0t]

-- 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:

ℱ[g(t−t0)ej2πf0t]=∫−∞∞[g(t−t0)ej2πf0t]e−j2πftdt

Rearranging terms we get:

∫−∞∞[g(t−t0)ej2πf0t]e−j2πftdt=∫−∞∞g(t−t0)e−j2π(f−f0)tdt

Now lets make the substitution λ=t−t0→t=λ+t0.
This leads us to:

∫−∞∞g(t−t0)e−j2π(f−f0)tdt=∫−∞∞g(λ)e−j2π(f−f0)(λ+t0)dt

After some simplification and rearranging terms, we get:

∫−∞∞g(λ)e−j2π(f−f0)(λ+t0)dt=∫−∞∞g(λ)e−j2π(f−f0)λe−j2π(f−f0)t0dt

Rearranging the terms yet again, we get:

∫−∞∞g(λ)e−j2π(f−f0)λe−j2π(f−f0)t0dt=e−j2π(f−f0)t0[∫−∞∞g(λ)e−j2π(f−f0)λdt]

We know that the exponential in terms of t0 is simply a constant and because of the Fourier Property of complex modualtion, we finally get:

ℱ[g(t)ej2πf0t]=G(f−f0)e−j2π(f−f0)t0