LPC spectra estimate 
LPC spectra estimate 
Feb 6 2009, 03:26
Post
#1


Group: Members Posts: 99 Joined: 2August 07 From: Shanghai,China Member No.: 45817 
LPC spectra should correspond with the envelope of DFT spectra.
I plot LPC spectra with matlab. The LPC spectra keep the same shape with the envelope of DFT spectra. But it is larger with a offset than the DFT envelope. Why? Where it go wrong? Here is the matlab script which illustrate it. N = 2560; [x,fs] = wavread('lpc.wav',N); % any speech signal can be used p=12; a = lpc(x,p); X =fft(x); X = X(1:N/2+1)'; X = 10*log10(abs(X).^2); Z = fft(a,N); Z = 1./Z(1:N/2+1); Z = 10*log10(abs(Z).^2); figure; plot(X'); hold on; plot(Z,'r'); grid; offset = 10; figure; plot(X+offset); hold on; plot(Z,'r'); grid; 


Feb 7 2009, 15:51
Post
#2


Group: Members Posts: 99 Joined: 2August 07 From: Shanghai,China Member No.: 45817 
LPC spectra should correspond with the envelope of DFT spectra. I plot LPC spectra with matlab. The LPC spectra keep the same shape with the envelope of DFT spectra. But it is larger with a offset than the DFT envelope. Why? Where it go wrong? What you call "LPC spectra" (by which you probably mean the LPC synthesis filter response) is independent of the input signal's level. If I remember correctly Matlab's LPC function can return more than the set of filter coefficients  one of those return values gives you a hint about the scaling of the input. Cheers! SG Thank you. Great job. It must be here that lpc coefficients were scaled. I would confirm it soon. LPC function in matlab return LPC coefficients and prediction error variance sigma^2. And LPC coefficients lost important signal energy information. But I still do not get the exact scale factor from returned value of lpc function. Cheers 


Feb 7 2009, 16:54
Post
#3


Group: Developer Posts: 1317 Joined: 20March 04 From: Göttingen (DE) Member No.: 12875 
What you call "LPC spectra" (by which you probably mean the LPC synthesis filter response) is independent of the input signal's level. If I remember correctly Matlab's LPC function can return more than the set of filter coefficients  one of those return values gives you a hint about the scaling of the input. LPC function in matlab return LPC coefficients and prediction error variance sigma^2. Bingo! Guess what happens to this sigma^2 if you scale your input by factor two... Cheers! SG 


Feb 8 2009, 02:58
Post
#4


Group: Members Posts: 99 Joined: 2August 07 From: Shanghai,China Member No.: 45817 
What you call "LPC spectra" (by which you probably mean the LPC synthesis filter response) is independent of the input signal's level. If I remember correctly Matlab's LPC function can return more than the set of filter coefficients  one of those return values gives you a hint about the scaling of the input. LPC function in matlab return LPC coefficients and prediction error variance sigma^2. Bingo! Guess what happens to this sigma^2 if you scale your input by factor two... Cheers! SG Thank you. It cause sigma^2 to rise 4 times by scaling the input by factor two preceding lpc computation,. but I still can not get the corresponding input signal power from prediction error variance sigma^2. N = 25600; [x,fs] = wavread('origsign.wav',N); % any speech signal can be used y = 2*x; p=20; [a,ae] = lpc(x,p); [b,be] = lpc(y,p); be/ae This post has been edited by hyeewang: Feb 9 2009, 06:53 


Feb 10 2009, 09:44
Post
#5


Group: Developer Posts: 1317 Joined: 20March 04 From: Göttingen (DE) Member No.: 12875 
but I still can not get the corresponding input signal power from prediction error variance sigma^2. If you still havn't figured out the details try Z = fft(a,N); Z = sqrt(e*N) ./ Z(1:N/2+1); % < changed Z = 10*log10(abs(Z).^2); instead  where 'e' is this sigma^2. Cheers! SG 


Feb 11 2009, 04:17
Post
#6


Group: Members Posts: 99 Joined: 2August 07 From: Shanghai,China Member No.: 45817 
but I still can not get the corresponding input signal power from prediction error variance sigma^2. If you still havn't figured out the details try Z = fft(a,N); Z = sqrt(e*N) ./ Z(1:N/2+1); % < changed Z = 10*log10(abs(Z).^2); instead  where 'e' is this sigma^2. Cheers! SG Thank you. I follow u and it work well. But I did not get the principle,. why the multiplication factor is energy,not variance? According to LPC theory,synthesis filter H = G./[1sum(ai.*Z.(i))], E(n) = s(n)  sp(n) = G*e(n); here,E(n) refer to as lpc pediction error, and e(n) refer to as excitation. s(n) is the input speech,sp(n) is the estimated signal. Assume the variance of e(n) to be 1. sum( E(n)*E(n)) = G.^2*sum( e(n)*e(n)) = G.^2* N; then G^2 = [sum( E(n)*E(n) ) ] ./ N. It should be variance(power),not energy. Why? Thank for your teaching. This post has been edited by hyeewang: Feb 11 2009, 14:11 


Feb 11 2009, 10:30
Post
#7


Group: Developer Posts: 1317 Joined: 20March 04 From: Göttingen (DE) Member No.: 12875 
Z = fft(a,N); Z = sqrt(e*N) ./ Z(1:N/2+1); % < changed Z = 10*log10(abs(Z).^2); instead  where 'e' is this sigma^2. I follow u and it work well. But I did not get the principle 'a' and 'e' don't depend on the length of your signal (assuming its characteristics don't change over time). But you wanted both curves to match which is why I included 'N' in the scaling. Also, are you confusing power with energy? Cheers! SG This post has been edited by SebastianG: Feb 11 2009, 10:31 


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