3 - Non-periodic Functions

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Revision as of 21:35, 2 November 2009 by Kevin.Starkey (talk | contribs) (New page: Lets look at what happens if our signals are not periodic. We can achieve this by setting our period T to infinity such that <math>\lim_{T \to \infty}\sum_{n=-\infty}^\infty (1/T\int_{-T/2...)
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Lets look at what happens if our signals are not periodic. We can achieve this by setting our period T to infinity such that limTn=(1/TT/2T/2x(t')ej2πnt'/Tdt')ej2πnt/T,
where
1/TT/2T/2x(t')ej2πnt'/Tdt'=αn
first we need to remove the restiction x(t) = x(t + T) by following these steps.
1/T df
n/T df
n=1/T()df
αnX(f)
this leads us to the equation
x(t)=limTn=(1/TT/2T/2x(t')ej2πnt'/Tdt')ej2πnt/T,
and if we replace n/T with f and take the integral with respect to f we get
x(t)=(x(t')ej2πft'dt')ej2πftdf,
where x(t')ej2πft'dt'=X(f)
simplifying the equation to
x(t)=X(f)ej2πftdf=<X(f)|ej2πft> = Inverse Fourier Transform
and
X(f)=x(t)ej2πftdt=<x(t)|ej2πft>=Fourier Transform
Now using the x(t) equation and rearranging it gives us x(t)=(x(t')ej2πft'dt')ej2πftdf=x(t')(ej2πf(tt')df)dt'
where
ej2πf(tt')=δ(tt')
Similarly for X(f)
X(f)=(X(f')ej2πf'tdf')ej2πftdt=X(f')(ej2πt(f'f)dt)df'
where
ej2πt(f'f)=δ(f'f)=δ(ff')

This works out nicely for us in both the time and frequency domain because this give us the inpulse function for both where they are non-zero only when t = t' or f = f' depending on which equation you use