10/09 - Fourier Transform: Difference between revisions

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===Sifting property of the delta function===


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Revision as of 18:08, 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(λ)e−j2πfλdλ]ej2πftdf =∫−∞∞X(f)ej2πftdf =x(t)
=∫−∞∞x(λ)∫−∞∞ej2πf(t−λ)dfdλ =∫−∞∞x(λ)δ(t−λ)dλ =x(t)
=∫−∞∞[∫−∞∞x(λ)e−jωλdλ]ejωt12πdω =∫−∞∞x(λ)[12π∫−∞∞ej(t−ω)λdω]dλ =∫−∞∞x(λ)δ(t−ω)dλ =x(t)

Sifting property of the delta function

∫−∞∞ej2πfte−j2πfλdf =⟨ej2πft∣ej2πft⟩f =δ(t−λ)
∫−∞∞ej2πtfe−j2πtf0dt =⟨ej2πtf∣ej2πtf0⟩t =δ(f−f0)