Thank you for your company , knowledge has no limit .

<>1, Let the observed values of ergodic stationary random signals be x ( n ) = x(n)= x(n)={1,-1,2,-2;n=0~3}; The period method and correlation method are used to estimate the power spectrum
P N ( w ) P_N(w)PN​(w), Verify its consistency .

Periodogram method :
① First find the signal spectrum function 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−1​x(n)e−jwn
② Further calculation of power spectral density P N ( w ) = 1 N ∣ X N ( e j w ) ∣ 2
P_N(w)=\dfrac{1}{N}|X_N(e^{jw})|^2PN​(w)=N1​∣XN​(ejw)∣2

Autocorrelation method :
① First, the sequence autocorrelation function is obtained in time domain 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)=N1​xN​(m)∗xN∗​(−m)=N1​n=0∑N−1​xN​(n)xN​(n−m)
② The power spectral density is obtained by Fourier transform 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−1​rN​(m)e−jwn

Periodogram estimation method :

x 4 ( n ) = { 1 , − 1 , 2 , − 2 }       N = 4 x_4(n)=\{1,-1,2,-2\} \ \ \ \ \
N=4x4​(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​(e
jw)=n=0∑3​x(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−2cos
3w)+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]

through too product turn and difference : c o s α c o s β + s i n α s i n β = c o s ( α − β )
By integrating sum and difference :cos\alpha cos\beta+sin\alpha sin\beta=cos(\alpha-\beta) By integrating sum and difference :cosαcosβ+si
nαsinβ=cos(α−β)

most after can turn simple have to : P 4 ( w ) = 1 4 ( − 4 c o s 3 w + 8 c o s 2 w − 14 c o s w + 10
) Finally, it can be reduced to :P_4(w)=\dfrac{1}{4}(-4cos3w+8cos2w-14cosw+10) Finally, it can be reduced to :P4​(w)=41​(−4cos3w+
8cos2w−14cosw+10)

Autocorrelation estimation

from to x 4 ( n ) by real order column , place with x ∗ ( − n ) = x ( − n )
because x_4(n) Is a real sequence , therefore x^*(-n)=x(-n) because x4​(n) Is a real sequence , therefore 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)=41​n=0∑3​x4​(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,...2,3r4​(
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∑3​r4​(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 Draw and verify the power spectrum image of them :

Matlab code :
clc; clear; close all; xn=[1,-1,2,-2]; N=length(xn); rm=xcorr(xn); % Computational autocorrelation
figure('name',' Power spectrum estimation '); 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(' Periodogram estimation '); 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
(' Autocorrelation estimation '); 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); % Periodogram estimation results P2=0.25*(-4*cos(3*w)+8*cos(2*w)-14*cos(w)+10); % Autocorrelation estimation results
subplot(2,2,3);plot(w/pi,P1,'r'); xlabel('\omega/\pi');ylabel('P_4(\omega)');
grid on;title(' Periodogram estimation '); subplot(2,2,4);plot(w/pi,P2,'b'); xlabel('\omega/\pi');
ylabel('P_4(\omega)');grid on;title(' Autocorrelation estimation ');

Technology
©2019-2020 Toolsou All rights reserved,
Hikvision - Embedded software written test questions C Language application 0 The length of array in memory and structure is 0 In depth analysis data structure --- The preorder of binary tree , Middle order , Subsequent traversal How to do it ipad Transfer of medium and super large files to computer elementui Shuttle box el-transfer Display list content text too long 2019 The 10th Blue Bridge Cup C/C++ A Summary after the National Games ( Beijing Tourism summary )unity Shooting games , Implementation of first person camera python of numpy Module detailed explanation and application case Study notes 【STM32】 Digital steering gear Horizontal and vertical linkage pan tilt Vue Used in Element Open for the first time el-dialog Solution for not getting element