有关傅里叶变换有一个重要的定理,Parseval定理:
∫−∞∞∣Ff(s)∣2ds=∫−∞∞∣f(t)∣2dt(1)\huge\int_{-\infty}^{\infty} |\mathscr{F}f(s)|^2ds = \int_{-\infty}^{\infty} |f(t)|^2dt \tag 1∫−∞∞∣Ff(s)∣2ds=∫−∞∞∣f(t)∣2dt(1)
Parseval定理是能量守恒的一种体现,即能量在时域和频域内总是相同的
证明如下:
设 fff,ggg 为给定的两个可积函数,则
g(x)=F−1(Fg(s))=∫−∞∞e2πisxFg(s)dsg(x) = \mathscr{F}^{-1}(\mathscr{F}g(s)) = \int_{-\infty}^{\infty} e^{2\pi isx} \mathscr{F}g(s) dsg(x)=F−1(Fg(s))=∫−∞∞e2πisxFg(s)ds
那么
g(x)‾=∫−∞∞e−2πisxFg(s)‾ds\overline{g(x)} = \int_{-\infty}^{\infty} e^{-2\pi isx} \overline{\mathscr{F}g(s)} dsg(x)=∫−∞∞e−2πisxFg(s)ds
于是
∫−∞∞f(x)g(x)‾dx=∫−∞∞f(x)(∫−∞∞e−2πisxFg(s)‾ds)dx=∫−∞∞(∫−∞∞e−2πisxf(x)dx)Fg(s)‾ds=∫−∞∞Ff(s)Fg(s)‾ds\begin{aligned} \int_{-\infty}^{\infty} f(x) \overline{g(x)}dx &= \int_{-\infty}^{\infty}f(x) \left(\int_{-\infty}^{\infty} e^{-2\pi isx} \overline{\mathscr{F}g(s)} ds \right)dx\\ &=\int_{-\infty}^{\infty} \left( \int_{-\infty}^{\infty}e^{-2\pi isx}f(x)dx \right)\overline{\mathscr{F}g(s)} ds\\ &=\int_{-\infty}^{\infty} \mathscr{F}f(s) \overline{\mathscr{F}g(s)} ds \end{aligned}∫−∞∞f(x)g(x)dx=∫−∞∞f(x)(∫−∞∞e−2πisxFg(s)ds)dx=∫−∞∞(∫−∞∞e−2πisxf(x)dx)Fg(s)ds=∫−∞∞Ff(s)Fg(s)ds
即:
∫−∞∞f(x)g(x)‾dx=∫−∞∞Ff(s)Fg(s)‾ds(2)\huge \int_{-\infty}^{\infty} f(x) \overline{g(x)}dx = \int_{-\infty}^{\infty} \mathscr{F}f(s) \overline{\mathscr{F}g(s)} ds\tag 2∫−∞∞f(x)g(x)dx=∫−∞∞Ff(s)Fg(s)ds(2)
令(2)(2)(2)式中的g=fg=fg=f,即为(1)(1)(1)式的另一种写法。得证。