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24 Pages

### lec9

Course: EE 6362, Spring 2008
School: Dallas
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Word Count: 927

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Spectrum H LPC (e ) = j G 1 - k e - j k k =1 p x = s .* hamming(301); X = fft( x , 1000 ) [ A , G , r ] = autolpc( x , 10 ) H = G ./ fft(A,1000); 50 Comparison of LPC Solutions Covariance Method Autocorrelation Method 51 Prediction Error as a Function of p p En Rn [k ] Vn = = 1 - k Rn [0] Rn [0] k =1 Model order is usually determined by the following rule of thumb: Fs/1000 poles for vocal tract 2-4...

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Spectrum H LPC (e ) = j G 1 - k e - j k k =1 p x = s .* hamming(301); X = fft( x , 1000 ) [ A , G , r ] = autolpc( x , 10 ) H = G ./ fft(A,1000); 50 Comparison of LPC Solutions Covariance Method Autocorrelation Method 51 Prediction Error as a Function of p p En Rn [k ] Vn = = 1 - k Rn [0] Rn [0] k =1 Model order is usually determined by the following rule of thumb: Fs/1000 poles for vocal tract 2-4 poles for radiation 2 poles for glottal pulse 52 Durbin's Algorithm (Levinson Recursion) E 0 = R[0] for i = 1, 2,... , p ( i -1) ( i -1) ki = R[i ] - j R[i - j ] / E j =1 i(i ) = ki i -1 The ki are called PARCOR coefficients (i ) ( i -1) if i > 1, then for j = 1, 2,... , i - 1 i (ji ) = (ji -1) - ki i(--j1) A ( z) = A - ki z A -i ( z) -1 ( i -1) (z ) end E end (i ) = (1 - k ) E 2 i ( i -1) The E ( i ) are the prediction error for an i th -order predictor 53 j = (j p ) j = 1, 2,... , p Fallout from the Durbin Algorithm each iteration computes an i-th order predictor and the associatedi error, i.e., A(i ) ( z ) = 1 - (ji ) z - j j =1 i (i ) E = R[0] - (ji ) R[ j ] j =1 it can be seen from the Durbin algorithm that E ( p ) = R[0] (1 - ki2 ) > 0 i =1 p the quantities ki are called the PARCOR (partial correlation) coefficients. The above result implies -1 < ki < 1 54 PARCORs to Prediction Coefficients assume that ki , i = 1, 2,... , p are given. Then we can skip the computation of ki in the Levinson recursion. for i = 1, 2,... , p i(i ) = ki if i > 1, then for j = 1, 2,... , i - 1 end end (i ) j = ( i -1) j - ki ( i -1) i- j j = (j p ) j = 1, 2,... , p 55 Prediction Coefficients to PARCORs assume that j , j = 1, 2,... , p are given. Then we can work backwards through the Levinson Recursion. (j p ) = j k p = (pp ) for j = 1, 2,... , p for i = p, p - 1,... , 2 for j = 1, 2,... , i - 1 (ji -1) = end i (ji ) + ki i(-)j 1 - ki2 i- ki -1 = i(-11) end 56 12th-Order Example G H ( z) = = A( z ) G 1 - i z -i i =1 p = G (1 - zi z -1 ) i =1 p == Gz p (z - z ) i =1 i 57 p Minimum-Phase Property of A(z) A( z ) has all its zeros inside the unit circle Proof: Assume that zi ( zi > 1) is a zero of A( z ) 2 A( z ) = (1 - zi z -1 ) A( z ) The minimum mean-squared error is En = 1 2 1 - zi e - 2 - j 2 2 j j A(e ) S (e ) d > 0 n 1 - zi e - j 2 = zi 1 - (1/ z )e 2 i - j 2 Thus, A( z ) could not be the optimum filter because we could replace zi by (1/ zi ) and decrease the error. 58 Pole-Zero Plot for Model Which poles correspond to formant frequencies? 59 Pole Locations F1 F2 F3 F4 zi 0.9308 0.9308 0.9317 0.9317 0.7837 0.7837 0.9109 0.9109 0.5579 0.5579 0.9571 0.9571 zi (in Hz) 288 -288 719 -719 976 -976 2294 -2294 2540 -2540 2897 -2897 Note that the angles of the poles can be expressed in Hz by using the relation Fs F= = 2T 2 60 Estimating Formant Frequencies compute A(z) and factor it. find roots that are close to the unit circle. compute equivalent analog frequencies from the angles of the roots. plot formant frequencies as a function of time. 61 Narrowband Spectrogram 62 Root Angles of Prediction Filter 63 Root Angles of Prediction Filter 64 LP Spectrogram 65 LP Spectrogram the spectrogram is an image plot of the shorttime Fourier transform X r [k ] = m =0 x[rR + m]w[m]e - j (2 / N ) km N -1 tr = rRT kFs Fk = N the LP spectrogram is a plot of H r [k ] = Ar (e j (2 / N ) k ) Gr 66 Root Angles for Prediction Filter 67 Root Angles for Prediction Filter 68 Frequency Domain Interpretation of LPC i from the Parseval theorem we get: En = N + p -1 m =0 ( e ( m) ) n p 2 1 = 2 - S (e n j ) A(e ) d 2 j 2 i the optimum prediction error filter frequency response is: A(e j ) = 1 - k z - k k =1 En (e j ) G = = j H (e ) S n (e j ) 2 - i the minimum mean-squared prediction error becomes: En = N + p -1 m =0 ( e ( m) ) n 2 G = 2 S n (e ) H (e j ) 2 j 2 d 69 Frequency Domain Interpretation of LPC i the minimum mean-squared prediction error becomes: En = N + p -1 m =0 ( e ( m) ) n 2 G = 2 2 - S n (e ) H (e j ) j 2 2 d i as p increases, i.e., : p N R(k ) = Rn (k ), k N - 1 H (e ) S n (e ) j 2 j 2 En G 2 2 j ...

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