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 ...