Warning: Rank deficient, rank = 0... means what?

Hi,

I am using the original signal FS and nbits. When i played the sound, it does have the speech, but its very noisy. I am currently doing this:

wavwrite(sig, Fs, nbits, 'Scrambled.wav');

i = 1;
k = 1;
siz = wavread('Scrambled.wav', 'size');
tot_samples = siz(1); 
while(i<=tot_samples-15)  
    y =  wavread('Scrambled.wav',[i i+15]);
    ori_sig(:,k) = y\H;                                                                           
        if i == 1;                                          
            disp('Descrambling Speech in Progress...');         
        end                            
    i = i+16;
    k = k+1;
end
recon_sig = reshape(ori_sig,size(ori_sig,1)*size(ori_sig,2),1);
sound(recon_sig);

I have this warning: "Warning: Rank deficient, rank = 0, tol = 0.000000e+00."

What does it mean?

[Rick]

These warnings were also shown when i tried to wavwrite:

Warning: Data clipped during write to file:Scrambled.wav 
> In wavwrite>PCM_Quantize at 279
  In wavwrite>write_wavedat at 301
  In wavwrite at 138
  In Speech_Scrambler at 29

What does it mean?

[Rick]

I used:

 x(:,frame) = pinv(y\H, 0);

to solve my problem.

By doing this, I am keeping the same values for those values which are less than 0. But I am not sure whether is it the correct way. Any suggestion would be appreciated.

Answer 1

[Wayne King]

The inverse of a NxN matrix only exists only if the matrix has rank N. In other words, the linear operator that the matrix represents is surjective.

Computing the inverse in a computer has numerical subtleties. That warning is telling you that the matrix is rank deficient.

The wavwrite warning indicates that your signal values exceed +/- 1 and are being clipped when the .wav file is written. I don't think you want that clipping, that distorts the signal. You can scale the vector in MATLAB to make sure that the values do not exceed +/- 1 for writing.

[Rick]

I see, So this warning: "Warning: Rank deficient, rank = 0, tol = 0.000000e+00." can be ignored? As you said its subjective.

I have tried to use:

scram_sig = scram_sig / (max(abs(scram_sig)));

to scale the scrambled signal to within the range. But it still does clipping when i write it.

Edit:

I have just tried this:

scram_sig = scram_sig./max(abs(scram_sig(:)))*(1-(2^-(8-1)));

It does not give me the warning anymore. But I am still getting alot of noise. Is there any way where I can mask off the and make the speech clearer?

[Wayne King]

I said surjective, not subjective.

[Rick]

Oh, meaning that for every x, that might be more than 1 y?

what does rank deficient matrix affects?

[Wayne King]

surjective means that for Ax=b there is at least one solution for every b.

[Rick]

i see, than what does rank deficient matrix affects?

[Brandon Sackett]

Isn't a surjective solution valid though? I guess bijections are the "best" but is there a reason a surjective solution gets that warning? Any relation to group theory would help my brain understand the problem.

Answer 2

[Walter Roberson]

I can't tell what your "H" is there, but your "y" is all zero (or close to it) for 16 consecutive samples at some point.

The "clipped" indicates that you are attempting to write samples outside the range -1 to +1.

[Rick]

H is a hadamard matrix of order 16.

Because i scrambled the signal using frame of 16 samples using:

      y(:,k) = H*x;
      k = k + 1;

I am now trying to descrambled it using 16 samples too by doing:

ori_sig(:,k) = y\H;
k = k + 1;

I know that clipping introduces distortion. Is there anyway where i can change the range of "y" to –1.0 <= y < +1.0, because I'm using 8 bits.