Parseval’s theorem

一、Parseval’s theorem介绍

帕塞瓦尔定理Parseval’s theorem表明了信号的能量在时域和频域相等。

$${\int }_{-\infty }^{\infty }|f(t){|}^2\mathrm{d}t=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }|F(\omega ){|}^2\mathrm{d}\omega ={\int }_{-\infty }^{\infty }|\hat{F}(f){|}^2\mathrm{d}f$$

对于两个不同的信号，与上式对应，一个相似的等式为

$${\int }_{-\infty }^{\infty }{f}_i(t)f_j(t)\mathrm{d}t=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }{F}_i(\omega )F_j^{\ast }(\omega )\mathrm{d}\omega$$

写成离散形式为：

$$\sum _{n=0}^{N-1}x[n]y^{\ast }[n]=\frac{1}{N}\sum _{k=0}^{N-1}X[k]Y^{\ast }[k]$$

数学本质：矢量空间信号正交变换的范数不变性。

二、 Parseval’s theorem基本形式的证明

2.1连续信号

$$\begin{array}{rl}& {\int }_{-\infty }^{\infty }|f(t){|}^{2}dt\\ =& {\int }_{-\infty }^{\infty }f(t)f^{\ast }(t)dt\\ =& {\int }_{-\infty }^{\infty }\left[\frac{1}{2\pi }{\int }_{-\infty }^{\infty }F(\omega )e^{j\omega t}d\omega \right]\left[\frac{1}{2\pi }{\int }_{-\infty }^{\infty }{F}^{\ast }\left({\omega }^{\prime }\right)e^{-j{\omega }^{\prime }t}d{\omega }^{\prime }\right]dt\\ =& \frac{1}{2\pi }{\int }_{-\infty }^{\infty }F(\omega )\frac{1}{2\pi }{\int }_{-\infty }^{\infty }{F}^{\ast }\left({\omega }^{\prime }\right)\left[{\int }_{-\infty }^{\infty }{e}^{j\left[\omega -{\omega }^{\prime }\right]t}dt\right]d{\omega }^{\prime }d\omega \\ =& \frac{1}{2\pi }{\int }_{-\infty }^{\infty }F(\omega )\frac{1}{2\pi }{\int }_{-\infty }^{\infty }{F}^{\ast }\left({\omega }^{\prime }\right)2\pi \delta \left(\omega -{\omega }^{\prime }\right)d{\omega }^{\prime }d\omega \\ =& \frac{1}{2\pi }{\int }_{-\infty }^{\infty }F(\omega )F^{\ast }(\omega )d\omega \\ =& \frac{1}{2\pi }{\int }_{-\infty }^{\infty }|F(\omega ){|}^{2}d\omega \end{array}$$

换一种写法，结论是相似的

$$\begin{array}{rl}& {\int }_{-\infty }^{\infty }|x(t){|}^{2}dt\\ =& {\int }_{-\infty }^{\infty }x(t)x^{\ast }(t)dt\\ =& {\int }_{-\infty }^{\infty }x(t)\left[{\int }_{-\infty }^{\infty }{X}^{\ast }(f)e^{-j2\pi ft}df\right]dt\\ =& {\int }_{-\infty }^{\infty }{X}^{\ast }(f)\left[{\int }_{-\infty }^{\infty }x(t)e^{-j2\pi ft}dt\right]df\\ =& {\int }_{-\infty }^{\infty }{X}^{\ast }(f)X(f)df\\ =& {\int }_{-\infty }^{\infty }|X(f){|}^{2}df\end{array}$$

2.2离散信号

$$\begin{array}{l}{\sum }_{n=0}^{N-1}x[n]y^{\ast }[n]=\frac{1}{N}{\sum }_{k=0}^{N-1}X[k]Y^{\ast }[k]\\ ={\sum }_{n=0}^{N-1}x[n]{\left(\frac{1}{N}{\sum }_{k=0}^{N-1}Y[k]W_N^{-nk}\right)}^{\ast }\\ =\frac{1}{N}{\sum }_{n=0}^{N-1}x[n]\left({\sum }_{k=0}^{N-1}{Y}^{\ast }[k]W_N^{nk}\right)\\ =\frac{1}{N}{\sum }_{k=0}^{N-1}{Y}^{\ast }[k]\left({\sum }_{n=0}^{N-1}x[n]W_N^{nk}\right)\\ =\frac{1}{N}{\sum }_{k=0}^{N-1}{Y}^{\ast }[k]X[k]\end{array}$$

三、MATLAB仿真验证

clc,clear;
T = 20;  dt = 0.001; t = dt:dt:T; N = size(t,2);  
fs = 1/dt;   ws = fs*2*pi; df = 1/T; dw = 2*pi*df;
s1 = exp(-0.2*t).*sin(2*pi*0.5*t); s2 = exp(-0.1*t).*sin(2*pi*0.6*t);
Us1 = fft(s1); Us2 = fft(s2);
Et11 = sum(s1.*s1)
Ew11 = sum(Us1.*conj(Us1))/N

Et12 = sum(s1.*s2)
Ew12 = sum(Us1.*conj(Us2))/N

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结果如图所示，可以验证公式准确性

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