negative eigenvalues in sample covariance matrix

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yasser
yasser 2014-10-10
回答： Matt J 2014-10-10
clc; clear;
N=10; taps=2; snr=0; noise_var=0.05;
h1r=randn(1,taps)/sqrt(2); h1i=randn(1,taps)/sqrt(2); h1=complex(h1r,h1i); h1=h1/norm(h1);
h2r=randn(1,taps)/sqrt(2); h2i=randn(1,taps)/sqrt(2); h2=complex(h2r,h2i); h2=h2/norm(h2);
c1=[h1(1);zeros(1,N-1)']; r1=[h1 zeros(1,N-1)]; H1=toeplitz(c1,r1);
c2=[h2(1);zeros(1,N-1)']; r2=[h2 zeros(1,N-1)]; H2=toeplitz(c2,r2);
H=[H1;H2];
order=64; k=log2(order); n=(taps+N-1)*k; x = randi([0 1],n,1); hMod = comm.RectangularQAMModulator(order); hBitToInt = comm.BitToInteger(k);% Convert the bits in x into k-bit symbols. xsym = step(hBitToInt,x); D = modulate(modem.qammod(order),xsym);
X=awgn(H*D,snr,'measured');
% noise1=sqrt(noise_var/2)*(randn(1,size(H1*D,1))+i*randn(1,size(H1*D,2))); % noise2=sqrt(noise_var/2)*(randn(1,size(H2*D,1))+i*randn(1,size(H2*D,2))); % noise=[noise1.';noise2.']; % % X=H*D+noise;
R=X*X'/size(X,2);
[Q ,eig_val]=eig (R);
the problem is that matrix of eig_val has negative values and this can't happen for sample covariance matrix R any help please

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回答（1 个）

Matt J
Matt J 2014-10-10
I can't run your code, because you haven't provided all variables needed to run it. However, you can expect small magnitude negative eigenvalues due to floating point errors, if your true covariance matrix ix close to singular.

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