3 - Non-periodic Functions: Difference between revisions
(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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'''Look carefully at the signs in your exponential for the Fourier transform (<math>e^{-j2\pi ft}</math>) and its inverse (<math>e^{j2\pi ft}</math>). It is correct in the integral form, but not in the bra-ket notation that follows... -Brandon''' |
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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 |
Lets look at what happens if our signals are not periodic. We can achieve this by setting our period T to infinity such that |
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<math>\lim_{T \to \infty}\sum_{n=-\infty}^\infty (1/T\int_{-T/2}^{T/2} x(t^')e^{-j2\pi nt^'/T}dt^')e^{j2\pi nt/T},\!</math> <br> where <br> |
<math>\lim_{T \to \infty}\sum_{n=-\infty}^\infty (1/T\int_{-T/2}^{T/2} x(t^')e^{-j2\pi nt^'/T}dt^')e^{j2\pi nt/T},\!</math> <br> where <br> |
Latest revision as of 17:47, 29 November 2009
Look carefully at the signs in your exponential for the Fourier transform () and its inverse (). It is correct in the integral form, but not in the bra-ket notation that follows... -Brandon
Lets look at what happens if our signals are not periodic. We can achieve this by setting our period T to infinity such that
where
first we need to remove the restiction x(t) = x(t + T) by following these steps.
1/T
n/T
this leads us to the equation
and if we replace n/T with f and take the integral with respect to f we get
where
simplifying the equation to
= Inverse Fourier Transform
and
Fourier Transform
Now using the x(t) equation and rearranging it gives us
where
Similarly for X(f)
where
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