ASN3 - Class Notes October 5: Difference between revisions

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<math> X(F)= \mathcal{F}[x(t)] \int_{-\frac{T}{2}}^{\frac{T}{2}} x(t')e^{\frac{ -j2 \pi nt'}{T}} dt' \!</math>
<math> X(F)= \mathcal{F}[x(t)] \int_{-\frac{T}{2}}^{\frac{T}{2}} x(t')e^{\frac{ -j2 \pi nt'}{T}} dt' \!</math>


Using the Fourier Transform property along with  <math> \lim_{T\to \infty} n/t </math> is <math>f </math>
Using the Fourier Transform property along with  <math> \lim_{T\to \infty} n/t = f </math> then <math >x(t)</math> is now
 
<math> x(t)=  \int_{-\infty} ^ {\infty} [\int_{-\infty} ^ {\infty} x(t')e^{ -j2 \pi ft'} dt'] e^{ j2 \pi ft}df \!</math>
<math> x(t)=  \int_{-\infty} ^ {\infty} [\int_{-\infty} ^ {\infty} x(t')e^{ -j2 \pi ft'} dt'] e^{ j2 \pi ft}df \!</math>
Reordering order of integration


<math> x(t)=  \int_{-\infty} ^ {\infty} x(t')[\int_{-\infty} ^ {\infty} e^{ j2 \pi f(t'-t)} df ]dt'  \!</math>
<math> x(t)=  \int_{-\infty} ^ {\infty} x(t')[\int_{-\infty} ^ {\infty} e^{ j2 \pi f(t'-t)} df ]dt'  \!</math>

Revision as of 22:25, 17 December 2009

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When T is very large (approaching infinity) the quanity on the left transforms to be approximately the quanity on the right.

limT


1/Tdf

n/Tf

n=1T()df


Using the relations above, can we make an unperiodic signal such as the one given below and make it periodic by taking the limit?

x(t)=limT1T(T2T2x(t)ej2πntTdt)ej2πntT

note that X(F)=[x(t)]T2T2x(t)ej2πntTdt

Using the Fourier Transform property along with limTn/t=f then x(t) is now

x(t)=[x(t)ej2πftdt]ej2πftdf

Reordering order of integration

x(t)=x(t)[ej2πf(tt)df]dt

note that the defination of the delta function is ej2πf(tt)df

x(t)=x(t)δ(tt)dt

                     THE GAME
            LTI (Linear Time Invariant System) 
Input     LTI                             Output                                  Reason

x(t)x(t)δ(tt)dt Superposition

X(f)X(f)δ(ff)df Superposition