Inverse Fourier transform_scaled inverse fourier transform-优快云博客

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Definition
$F(ω)=∫−∞∞f(t)exp(−jωt)dtF(\omega)=\int ^{\infty} _{-\infty}f(t)exp(-j\omega t)dt$
Inverse Fourier transform
$f(t)=F−1(F(ω))=12π∫−∞∞f(ω)exp(jωt)dωf(t)=\mathcal F ^{-1}( F(\omega))=\frac {1}{2\pi}\int ^{\infty} _{-\infty}f(\omega)exp(j\omega t)d\omega$
Proof
$F(ω)=∫−∞∞f(t)exp(−jωt)dtF(\omega)=\int ^{\infty} _{-\infty}f(t)exp(-j\omega t)dt$
$F(ω)=∫−∞∞12π∫−∞∞f(ω′)exp(jω′t)dω′exp(−jωt)dtF(\omega)=\int ^{\infty} _{-\infty}\frac {1}{2\pi}\int ^{\infty} _{-\infty}f(\omega ^{\prime})exp(j\omega ^{\prime} t) d\omega ^{\prime} exp(-j\omega t)dt$
$=12π∫−∞∞∫−∞∞f(ω′)exp(jω′t)dω′exp(−jωt)dt=\frac {1}{2\pi}\int ^{\infty} _{-\infty}\int ^{\infty} _{-\infty}f(\omega ^{\prime})exp(j\omega ^{\prime} t) d\omega ^{\prime} exp(-j\omega t)dt$
$=12π∫−∞∞∫−∞∞f(ω′)exp(−j(ω′−ω)t)dω′dt=\frac {1}{2\pi}\int ^{\infty} _{-\infty}\int ^{\infty} _{-\infty}f(\omega ^{\prime}) exp(-j(\omega ^{\prime}-\omega) t) d\omega ^{\prime} dt$
$=12π∫−∞∞f(ω′)[∫−∞∞exp(−j(ω′−ω)t)dt]dω′=\frac {1}{2\pi}\int ^{\infty} _{-\infty} f(\omega ^{\prime}) \left[ \int ^{\infty} _{-\infty} exp(-j(\omega ^{\prime}-\omega) t) dt \right] d\omega ^{\prime}$
Let
$δ(ω−ω′)=12π∫−∞∞exp(−j(ω′−ω)t)dt\delta (\omega - \omega ^{\prime} )= \frac {1}{2\pi} \int ^{\infty} _{-\infty} exp(-j(\omega ^{\prime}-\omega) t) dt$
$F(ω)=∫−∞∞f(ω′)δ(ω−ω′)dω′F(\omega)=\int ^{\infty} _{-\infty} f(\omega ^{\prime}) \delta (\omega - \omega ^{\prime} ) d\omega ^{\prime}$
For $δ\delta$ function
$∫−∞∞δ(x)dx=1,δ(x)=0,forx≠0\int ^{\infty} _{-\infty}\delta (x)dx=1 , \delta (x)=0, \quad for \quad x \neq 0$
$∫−∞∞f(x)δ(x−a)dx=f(a)\int ^{\infty} _{-\infty} f(x) \delta (x-a)dx=f(a)$
$12π∫−∞∞e−jωxdx=δ(ω)\frac{1}{2\pi}\int ^{\infty} _{-\infty} e^{-j\omega x}dx=\delta (\omega)$