why PSD is different?

Hello all, I am new to digital signal processing. I understand that there are 2 ways to get the PSD of a signal: 1- by fft the auto-correlation of the signal 2- by multiplying the fft of the signal by its conjugate. when I use these 2 methods in matlab on a maximum-length PRBS signal I always get different results! can you explain it for me ? here is my code:

x=idinput(2^14-1,'PRBS')       % prbs 

[rxx lags]=xcorr(x)    

X=fft(x,length(x))

rxx_temp=rxx(1:(length(x)))
rxx_temp=fliplr(rxx_temp')
rxx=rxx_temp'

Rxx1=fft(rxx)

subplot(211)
plot(abs(Rxx1))
title('S(f)=Rxx(f)=fft(rxx)')   

Rxx2= X.*conj(X)
subplot(212)
plot(abs(Rxx2))
title('S(f)=Rxx(f)=X X*')

[Honglei Chen]

Converted to code format

Answer 1

[Youssef Khmou]

hi,

They are the same with your approach here is an example :

x=sin(2*pi*40*(0:1/90:10));
[rxx lags]=xcorr(x)
X=fft(x,length(x))
rxx_temp=rxx(1:(length(x)))
rxx_temp=fliplr(rxx_temp')
rxx=rxx_temp'
Rxx1=fft(rxx)
subplot(211)
plot(abs(Rxx1))
title('S(f)=Rxx(f)=fft(rxx)')
Rxx2= X.*conj(X)
subplot(212)
plot(abs(Rxx2))
title('S(f)=Rxx(f)=X X*')

[Ahmed]

Thanks Youssef

The 2 methods give identical results if the input signal is sinusoidal, I just dont understand why the output is different when the input signal is a MLS prbs signal, although mathematically they must be the same !

Answer 2

[Gaurav]

clc;

close all;

clear all;

% User Inputs

bit_size=input('Enter the bit size '); % Bit size is taken as input

nf=7; % Flip-flops in LFSR

% Feedback tap positions

taps=[6 7]; % taps in LFSR

ri=randi([0 1],1,nf); % LFSR with random seed inputlr=zeros(1,nf); % LFSR initialization

l=128; % PRBS length

ls=zeros(l,nf); % sequence initialization in LFSR

for i=1:l

% LFSR shift operation

y= sum(ri(taps)); % xor operation output

x= mod(y,2);

lr(2:end) = ri(1:end-1) ; % Shifting in LFSR

lr(1)=x;

% Feedback in LFSR

ri=lr;

% updating value of ri

ls(i,:)=lr;

% generation of LFSR Sequence

end