10/09 - Fourier Transform: Difference between revisions

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|<math>=x(t)\,\!</math>
|<math>=x(t)\,\!</math>
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{| border="0" cellpadding="0" cellspacing="0"
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|<math>\int_{-\infty}^{\infty}e^{j2\pi ft}e^{-j2\pi f\lambda}df</math>
|<math>=\left\langle e^{j2\pi ft}\mid e^{j2\pi ft}\right\rangle_f</math>
|<math>=\delta(t-\lambda)\,\!</math>
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|<math>\int_{-\infty}^{\infty}e^{j2\pi tf}e^{-j2\pi tf_0}dt</math>
|<math>=\left\langle e^{j2\pi tf}\mid e^{j2\pi tf_0}\right\rangle_t</math>
|<math>=\delta(f-f_0)\,\!</math>
|}
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Revision as of 15:28, 17 November 2008

⟨ej2πnt/T∣ej2πmt/T⟩ =∫−∞∞ej2πnt/Te−j2πmt/Tdt
=∫−∞∞ej2π(n−m)t/Tdt
=∫−T/2T/2ej2π(n−m)t/Tdt Assuming the function is perodic with the period T
=Tδm,n

Fourier Transform

Remember from 10/02 - Fourier Series

  • αm=1T∫−T/2T/2x(t)e−j2πmt/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)e−j2πftdt =⟨x(t)∣ej2πft⟩t
F−1[x(t)] =x(t) =∫−∞∞X(f)ej2πftdf =⟨X(f)∣e−j2πft⟩f

Examples

F−1[F[x(t)]] =∫−∞∞[∫−∞∞x(t)e−j2πftdt]ej2πftdf
=∫−∞∞X(f)ej2πftdf
=x(t)
∫−∞∞ej2πfte−j2πfλdf =⟨ej2πft∣ej2πft⟩f =δ(t−λ)
∫−∞∞ej2πtfe−j2πtf0dt =⟨ej2πtf∣ej2πtf0⟩t =δ(f−f0)