Two functions of f(t)f(t)f(t) and g(t)g(t)g(t)
The Fourier transform of f(t)f(t)f(t) is
F(f(t))=F(ω)=∫−∞∞f(t)exp(−jωt)dt\mathcal{F}(f(t))=F(\omega)=\int ^{\infty} _{-\infty}f(t)exp(-j\omega t)dtF(f(t))=F(ω)=∫−∞∞f(t)exp(−jωt)dt
The Fourier transform of g(t)g(t)g(t) is
F(g(t))=G(ω)=∫−∞∞g(t)exp(−jωt)dt\mathcal{F}(g(t))=G(\omega)=\int ^{\infty} _{-\infty}g(t)exp(-j\omega t)dtF(g(t))=G(ω)=∫−∞∞g(t)exp(−jωt)dt
The convolution f(t)f(t)f(t) and g(t)g(t)g(t) is
h(t)=f(t)∗g(t)=∫−∞∞f(τ)g(t−τ)dτ→F(ω)G(ω)h(t)=f(t)*g(t)=\int ^{\infty} _{-\infty}f(\tau ) g(t-\tau)d\tau \rightarrow F(\omega)G(\omega)h(t)=f(t)∗g(t)=∫−∞∞f(τ)g(t−τ)dτ→F(ω)G(ω)
Proof:
H(t)=F[f(t)∗g(t)]=F(ω)G(ω)H(t)=\mathcal{F}\left [f(t)*g(t)\right]=F(\omega) G(\omega)H(t)=F[f(t)∗g(t)]=F(ω)G(ω)
F[f(t)∗g(t)]=F(ω)=∫−∞∞H(t)exp(−jωt)dt\mathcal{F}\left [f(t)*g(t)\right]=F(\omega)=\int ^{\infty} _{-\infty}H(t)exp(-j\omega t)dtF[f(t)∗g(t)]=F(ω)=∫−∞∞H(t)exp(−jωt)dt
=∫−∞∞∫−∞∞f(τ)g(t−τ)dτexp(−jωt)dt=\int ^{\infty} _{-\infty}\int ^{\infty} _{-\infty}f(\tau ) g(t-\tau)d\tau exp(-j\omega t)dt=∫−∞∞∫−∞∞f(τ)g(t−τ)dτexp(−jωt)dt
=∫−∞∞∫−∞∞f(τ)e−jωτdτg(t−τ)e−jω(t−τ)d(t−τ)=\int ^{\infty} _{-\infty}\int ^{\infty} _{-\infty}f(\tau ) e^{-j\omega \tau} d\tau g(t-\tau) e^{-j\omega (t-\tau)}d(t-\tau)=∫−∞∞∫−∞∞f(τ)e−jωτdτg(t−τ)e−jω(t−τ)d(t−τ)
=∫−∞∞f(τ)e−jωτdτ∫−∞∞g(t−τ)e−jω(t−τ)d(t−τ)=\int ^{\infty} _{-\infty}f(\tau ) e^{-j\omega \tau} d\tau \int ^{\infty} _{-\infty}g(t-\tau) e^{-j\omega (t-\tau)}d(t-\tau)=∫−∞∞f(τ)e−jωτdτ∫−∞∞g(t−τ)e−jω(t−τ)d(t−τ)
let t‾=t−τ\overline t=t-\taut=t−τ
∫−∞∞f(τ)e−jωτdτ∫−∞∞g(t−τ)e−jω(t−τ)d(t−τ)\int ^{\infty} _{-\infty}f(\tau ) e^{-j\omega \tau} d\tau \int ^{\infty} _{-\infty}g(t-\tau) e^{-j\omega (t-\tau)}d(t-\tau)∫−∞∞f(τ)e−jωτdτ∫−∞∞g(t−τ)e−jω(t−τ)d(t−τ)
=∫−∞∞f(τ)e−jωτdτ∫−∞∞g(t‾)e−jωt‾dt‾=\int ^{\infty} _{-\infty}f(\tau ) e^{-j\omega \tau} d\tau \int ^{\infty} _{-\infty}g(\overline t) e^{-j\omega \overline t}d \overline t=∫−∞∞f(τ)e−jωτdτ∫−∞∞g(t)e−jωtdt
=F(ω)G(ω)=F(\omega) G(\omega)=F(ω)G(ω)
本文详细介绍了傅里叶变换的基本概念,并通过数学公式解释了如何从时域信号转换到频域信号。同时,文章还探讨了两个函数的卷积过程及其与傅里叶变换之间的关系。
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