Homework Six: Difference between revisions

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[[Nick Christman]]
'''Perform the following tasks:'''
 
<br/>
 
[[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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<br/>
(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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<br/>
(c) Do another property on the Wiki and get it reviewed (i.e. review a second property) -- [[Fourier Transform Properties]]
(i) '''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...
<br/>
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>
<br/>
Now lets make the substitution <math>\lambda = t-t_{0} \rightarrow t = \lambda + t_{0}</math>.
<br/>
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>
<br/>
(ii)
I reviewed Max's second Fourier Transform property: <math>\mathcal{F}\bigg[\int_{-\infty}^{\infty}g(t) h^*(t) dt\bigg]</math>
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 17:31, 31 October 2009

Perform the following tasks:


Nick Christman



(a) Show [ts(λ)dλ]=S(f)j2πf if S(0)=0. HINT: S(0)=S(f)|f=0=s(t)ej2π(f0)tdt=s(t)dt




(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

(i) 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


(ii) I reviewed Max's second Fourier Transform property: [g(t)h*(t)dt]

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.