Homework Six: Difference between revisions

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Back to [[Nick Christman|<b><u>Nick Christman</u></b>]]<br/>
'''Perform the following tasks:'''
'''Perform the following tasks:'''


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[[Nick Christman|<b><u>Nick Christman</u></b>]]<br/>
 


(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>
(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>

Revision as of 16:05, 31 October 2009

Back to Nick Christman

Perform the following tasks:



(a) Show [ts(λ)dλ]=S(F)j2πf if S(0)=0


(b)If S(0)0 can you find [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 [g(tt0)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:

[g(tt0)ej2πf0t]=[g(tt0)ej2πf0t]ej2πftdt

Rearranging terms we get:

[g(tt0)ej2πf0t]ej2πftdt=g(tt0)ej2π(ff0)tdt

Now lets make the substitution λ=tt0t=λ+t0.
This leads us to:

g(tt0)ej2π(ff0)tdt=g(λ)ej2π(ff0)(λ+t0)dt

After some simplification and rearranging terms, we get:

g(λ)ej2π(ff0)(λ+t0)dt=g(λ)ej2π(ff0)λej2π(ff0)t0dt

Rearranging the terms yet again, we get:

g(λ)ej2π(ff0)λej2π(ff0)t0dt=ej2π(ff0)t0[g(λ)ej2π(ff0)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:

[g(t)ej2πf0t]=G(ff0)ej2π(ff0)t0