10/09 - Fourier Transform: Difference between revisions

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|<math>=\int_{-\infty}^{\infty}x(\lambda) \int_{-\infty}^{\infty}e^{j2\pi f(t-\lambda)} df d\lambda</math>
|<math>=\int_{-\infty}^{\infty}x(\lambda) \int_{-\infty}^{\infty}e^{j2\pi f(t-\lambda)} df d\lambda</math>
|<math>=\int_{-\infty}^{\infty}x(\lambda) \delta(t-\lambda) d\lambda</math>
|<math>=\int_{-\infty}^{\infty}x(\lambda) \delta(t-\lambda) d\lambda</math>
|<math>=x(t)\,\!</math>
|-
|
|<math>=\int_{-\infty}^{\infty}\left [ \int_{-\infty}^{\infty} x(\lambda) e^{-j\omega\lambda}d\lambda\right ]e^{j\omega t} \frac{1}{2\pi}d\omega </math>
|<math>=\int_{-\infty}^{\infty}x(\lambda) \left [ \frac{1}{2\pi} \int_{-\infty}^{\infty}  e^{j(t-\omega) \lambda}d\omega\right ] d\lambda</math>
|<math>=\int_{-\infty}^{\infty}x(\lambda) \delta(t-\omega) d\lambda</math>
|<math>=x(t)\,\!</math>
|<math>=x(t)\,\!</math>
|}
|}

Revision as of 17:53, 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

Definitions

F[x(t)] =X(f) =x(t)ej2πftdt =x(t)ej2πftt
F1[x(t)] =x(t) =X(f)ej2πftdf =X(f)ej2πftf

Examples

F1[F[x(t)]] =[x(λ)ej2πfλdλ]ej2πftdf =X(f)ej2πftdf =x(t)
=x(λ)ej2πf(tλ)dfdλ =x(λ)δ(tλ)dλ =x(t)
=[x(λ)ejωλdλ]ejωt12πdω =x(λ)[12πej(tω)λdω]dλ =x(λ)δ(tω)dλ =x(t)
ej2πftej2πfλdf =ej2πftej2πftf =δ(tλ)
ej2πtfej2πtf0dt =ej2πtfej2πtf0t =δ(ff0)