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周期图法:
①先求信号频谱函数 X N ( e j w ) = ∑ n = 0 N − 1 x ( n ) e − j w n X_N(e^{jw})=\sum\limits_{n=0}^{N-1}x(n)e^{-jwn} XN(ejw)=n=0∑N−1x(n)e−jwn
②再求功率谱密度 P N ( w ) = 1 N ∣ X N ( e j w ) ∣ 2 P_N(w)=\dfrac{1}{N}|X_N(e^{jw})|^2 PN(w)=N1∣XN(ejw)∣2
自相关法:
①首先在时域求序列自相关函数 r N ( m ) = 1 N x N ( m ) ∗ x N ∗ ( − m ) = 1 N ∑ n = 0 N − 1 x N ( n ) x N ( n − m ) r_N(m)=\dfrac{1}{N}x_N(m)*x^*_N(-m)=\dfrac{1}{N}\sum\limits_{n=0}^{N-1}x_N(n)x_N(n-m) rN(m)=N1xN(m)∗xN∗(−m)=N1n=0∑N−1xN(n)xN(n−m)
②再求其傅里叶变换得到功率谱密度 P N ( w ) = ∑ m = 1 − N N − 1 r N ( m ) e − j w n P_N(w)=\sum\limits_{m=1-N}^{N-1}r_N(m)e^{-jwn} PN(w)=m=1−N∑N−1rN(m)e−jwn
周期图估计法:
x 4 ( n ) = { 1 , − 1 , 2 , − 2 } N = 4 x_4(n)=\{1,-1,2,-2\} \ \ \ \ \ N=4 x4(n)={ 1,−1,2,−2} N=4
X 4 ( e j w ) = ∑ n = 0 3 x ( n ) e − j w n = 1 − e − j w + 2 e − j 2 w − 2 e − j 3 w X_4(e^{jw})=\sum\limits_{n=0}^{3}x(n)e^{-jwn}=1-e^{-jw}+2e^{-j2w}-2e^{-j3w} X4(ejw)=n=0∑3x(n)e−jwn=1−e−jw+2e−j2w−2e−j3w
= ( 1 − c o s w + 2 c o s 2 w − 2 c o s 3 w ) + j ( s i n w − 2 s i n 2 w + 2 s i n 3 w ) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =(1-cosw+2cos2w-2cos3w)+j(sinw-2sin2w+2sin3w) =(1−cosw+2cos2w−2cos3w)+j(sinw−2sin2w+2sin3w)
P 4 ( w ) = 1 4 [ ( 1 − c o s w + 2 c o s 2 w − 2 c o s 3 w ) 2 + ( s i n w − 2 s i n 2 w + 2 s i n 3 w ) 2 ] P_4(w)=\dfrac{1}{4}[(1-cosw+2cos2w-2cos3w)^2+(sinw-2sin2w+2sin3w)^2] P4(w)=41[(1−cosw+2cos2w−2cos3w)2+(sinw−2sin2w+2sin3w)2]
通 过 积 化 和 差 : c o s α c o s β + s i n α s i n β = c o s ( α − β ) 通过积化和差:cos\alpha cos\beta+sin\alpha sin\beta=cos(\alpha-\beta) 通过积化和差:cosαcosβ+sinαsinβ=cos(α−β)
最 后 可 化 简 得 : P 4 ( w ) = 1 4 ( − 4 c o s 3 w + 8 c o s 2 w − 14 c o s w + 10 ) 最后可化简得:P_4(w)=\dfrac{1}{4}(-4cos3w+8cos2w-14cosw+10) 最后可化简得:P4(w)=41(−4cos3w+8cos2w−14cosw+10)
自相关估计法
由 于 x 4 ( n ) 为 实 序 列 , 所 以 x ∗ ( − n ) = x ( − n ) 由于x_4(n)为实序列,所以x^*(-n)=x(-n) 由于x4(n)为实序列,所以x∗(−n)=x(−n)
r 4 ( m ) = x ( m ) ∗ x ( − m ) = 1 4 ∑ n = 0 3 x 4 ( n ) x 4 ( n − m ) r_4(m)=x(m)*x(-m)=\dfrac{1}{4}\sum\limits_{n=0}^{3}x_4(n)x_4(n-m) r4(m)=x(m)∗x(−m)=41n=0∑3x4(n)x4(n−m)
r 4 ( m ) = 1 4 { − 2 , 4 , − 7 , 10 , − 7 , 4 , − 2 } m = − 3 , − 2 , . . . , 2 , 3 r_4(m)=\dfrac{1}{4}\{-2,4,-7,10,-7,4,-2\}\ \ \ \ m=-3,-2,...,2,3 r4(m)=41{ −2,4,−7,10,−7,4,−2} m=−3,−2,...,2,3
P 4 ( w ) = ∑ m = − 3 3 r 4 ( m ) e − j w m P_4(w)=\sum\limits_{m=-3}^{3}r_4(m)e^{-jwm} P4(w)=m=−3∑3r4(m)e−jwm
= 1 4 ⋅ ( − 2 e j 3 w + 4 e j 2 w − 7 e j w + 10 − 7 e − j w + 4 e − j 2 w − 2 e − j 3 w ) \ \ \ \ \ \ \ \ \ \ \ \ =\dfrac{1}{4}\cdot(-2e^{j3w}+4e^{j2w}-7e^{jw}+10-7e^{-jw}+4e^{-j2w}-2e^{-j3w}) =41⋅(−2ej3w+4ej2w−7ejw+10−7e−jw+4e−j2w−2e−j3w)
= 1 4 [ − 2 ( e j 3 w + e − j 3 w ) + 4 ( e j 2 w + e − j 2 w ) − 7 ( e j w + e − j w ) + 10 ] \ \ \ \ \ \ \ \ \ \ \ \ =\dfrac{1}{4}[-2(e^{j3w}+e^{-j3w})+4(e^{j2w}+e^{-j2w})-7(e^{jw}+e^{-jw})+10] =41[−2(ej3w+e−j3w)+4(ej2w+e−j2w)−7(ejw+e−jw)+10]
= 1 4 ( − 4 c o s 3 w + 8 c o s 2 w − 14 c o s w + 10 ) \ \ \ \ \ \ \ \ \ \ \ \ =\dfrac{1}{4}(-4cos3w+8cos2w-14cosw+10) =41(−4cos3w+8cos2w−14cosw+10)
Matlab绘图验证二者功率谱图像:
Matlab代码:
clc;
clear;
close all;
xn=[1,-1,2,-2];
N=length(xn);
rm=xcorr(xn); %计算自相关
figure('name','功率谱估计');
B=xn;
A=1;
[H1,w]=freqz(B,A,'whole');
subplot(2,2,1);
plot(w/pi,1/N*(abs(H1)).^2,'r');
xlabel('\omega/\pi');ylabel('P_4(\omega)');grid on;title('周期图估计');
B=rm/N;
A=1;
[H2,w]=freqz(B,A,'whole');
subplot(2,2,2);
plot(w/pi,abs(H2),'b');
xlabel('\omega/\pi');ylabel('P_4(\omega)');grid on;title('自相关估计');
P1=0.25*((1-cos(w)+2*cos(2*w)-2*cos(3*w)).^2+(-sin(w)+2*sin(2*w)-2*sin(3*w)).^2); %周期图估计结果
P2=0.25*(-4*cos(3*w)+8*cos(2*w)-14*cos(w)+10); %自相关估计结果
subplot(2,2,3);plot(w/pi,P1,'r');
xlabel('\omega/\pi');ylabel('P_4(\omega)');grid on;title('周期图估计');
subplot(2,2,4);plot(w/pi,P2,'b');
xlabel('\omega/\pi');ylabel('P_4(\omega)');grid on;title('自相关估计');