10/09 - Fourier Transform: Difference between revisions

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==Fourier Transform==
==Fourier Transform==
Remember from [[10/02 - Fourier Series]]
*<math> \alpha_m = \frac{1}{T}\int_{-T/2}^{T/2} x(t) e^{-j2\pi mt/T}\, dt</math>
*<math>x(t) = x(t+T) = \sum_{n=-\infty}^\infty \alpha_m e^{j2\pi m/T}</math>
If we let <math> T \rightarrow \infty</math>
{| border="0" cellpadding="0" cellspacing="0"
|-
|<math>\frac{1}{T}</math>
|<math>\rightarrow df</math>
|-
|<math>\frac{n}{T}</math>
|<math>\rightarrow f</math>
|Remember <math>f=\frac{2\pi n}{T} \,\!</math>
|-
|<math>T\,\!</math>
|<math>\rightarrow \infty</math>
|-
|<math>\sum_{n=-\infty}^{\infty} \frac{1}{T}</math>
|<math>\rightarrow \int_{-\infty}^{\infty}() df</math>
|}

Revision as of 14:24, 17 November 2008

ej2πnt/Tej2πmt/T =ej2πnt/Tej2πmt/Tdt
=ej2π(nm)t/Tdt
=T/2T/2ej2π(nm)t/Tdt Assuming the function is perodic with the period T
=Tδm,n

Fourier Transform

Remember from 10/02 - Fourier Series

  • αm=1TT/2T/2x(t)ej2πmt/Tdt
  • x(t)=x(t+T)=n=αmej2πm/T

If we let T

1T df
nT f Remember f=2πnT
T
n=1T ()df