Brandon.price at 00:47, 30 November 2009

2009-11-30T00:47:38Z

← Older revision		Revision as of 17:47, 29 November 2009
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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'''



	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
	<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>

Kevin.Starkey: 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 $\lim_{T \to \infty}\sum_{n=-\infty}^\infty (1/T\int_{-T/2...$

2009-11-03T04:35:30Z

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...

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}^{T/2} x(t^')e^{-j2\pi nt^'/T}dt^')e^{j2\pi nt/T},\!</math> where 
<math>\ 1/T\int_{-T/2}^{T/2} x(t^')e^{-j2\pi nt^'/T}dt^' = \alpha_n \!</math> 
first we need to remove the restiction x(t) = x(t + T) by following these steps. 
1/T <math>\Longrightarrow df \!</math> 
n/T <math>\Longrightarrow df \!</math> 
<math> \sum_{n=-\infty} ^\infty 1/T \Longrightarrow \int_{-\infty}^\infty ()df \!</math> 
<math> \alpha_n \Longrightarrow X(f) \!</math> 
this leads us to the equation 
<math> x(t) = \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> 
and if we replace n/T with f and take the integral with respect to f we get 
<math> x(t) = \int_{-\infty}^\infty (\int_{-\infty}^\infty x(t^')e^{-j2\pi ft^'}dt^')e^{j2\pi ft}df,\!</math> 
where <math> \int_{-\infty}^\infty x(t^')e^{-j2\pi ft^'}dt^' = X(f)\!</math> 
simplifying the equation to 
<math> x(t) = \int_{-\infty}^\infty X(f) e^{j2\pi ft}df = <X(f)|e^{j2\pi ft}> \!</math> = '''Inverse Fourier Transform''' and 
<math> X(f) = \int_{-\infty}^\infty x(t)e^{-j2 \pi ft}dt = <x(t)|e^{j2 \pi ft}> = \!</math>'''Fourier Transform''' 
Now using the x(t) equation and rearranging it gives us
<math> x(t) = \int_{-\infty}^\infty (\int_{-\infty}^\infty x(t^')e^{-j2\pi ft^'}dt^')e^{j2\pi ft}df = \int_{-\infty}^\infty x(t^') (\int_{-\infty}^\infty e^{j2\pi f(t-t^')}df)dt^' \!</math> 
where 
<math> e^{j2\pi f(t-t^')} = \delta(t - t^')\!</math> 
Similarly for X(f) 
<math> X(f) = \int_{-\infty}^\infty (\int_{-\infty}^\infty X(f^')e^{j2\pi f^'t}df^')e^{-j2\pi ft}dt = \int_{-\infty}^\infty X(f^') (\int_{-\infty}^\infty e^{j2\pi t(f^'-f)}dt)df^' \!</math> 
where 
<math> e^{j2\pi t(f^'-f)} = \delta(f^' - f) = \delta(f - f^') \!</math> 

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

3 - Non-periodic Functions - Revision history

Brandon.price at 00:47, 30 November 2009

Kevin.Starkey: 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 \lim_{T \to \infty}\sum_{n=-\infty}^\infty (1/T\int_{-T/2...

Kevin.Starkey: 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 $\lim_{T \to \infty}\sum_{n=-\infty}^\infty (1/T\int_{-T/2...$