Problems
AT sin(t / T )
is sampled at
t
 s /( 2 ) samples/second. Find the Fourier transform of the sampled signal. Find the output of the
Problem 1. Assume that the bandlimited signal x(t ) 
ideal low-pass filter with cut-off frequency  s / 2 . Compare the output of the ideal low-pass filter
with the original signal x(t ) (  s  2 / T ,  / T   s  2 / T ).
Solution



AT sin( t / T )  jt
2 AT sin( t / T )
AT  sin(    / T ) sin(    / T ) 
X ( )  
e dt 
cos tdt 


dt 
t
 0
t
 0 
t
t


 AT ,    / T

.
0
,



/
T

(We take into account that each of the integrals is equal to   / 2 depending on the sign of
(   / T ) ).
If  S  2 / T
x(t ) 
1
2
 /T


ATe jt d 
 /T
AT

 /T

cos td 
0
AT sin
t
t
T .
If  / T   S  2 / T
1
x(t ) 
2
 s  / T
 /2
s
1
AT
ATe d 
2  2 AT cos td 

2  s  / T

 s   / T
jt
 s  / T

0
cos td 
2

s / 2
AT
 cos td 
 s  / T

AT sin( s  )t
T  2 AT  sin  / 2t  sin     t 

s
s
t
t 
T  

Problem 2. Quantize the given sequence x =(2.3, 1.78, 1.40, 0.10, 0.75, 0.65, 0.44, 0.58,
0.53, 0.05, 0.76, 0.63, 1.89, 0.48, 0.80, 0.40, 0.71, 0.93, 0.16, 2.4) by fixed-rate uniform
scalar quantizer with rate 3, 2, and 1 bit/sample. Compute the average quantization error. Estimate
entropy of the output sequence. (The minimal value is –3, the maximal value +3). Compare with
performance of variable-rate uniform quantizer with the step size equal to 0.6.
Solution.
If R  3 then M  8 and we obtain the following sequence of thresholds:
3, 2.25, 1.5, 0.75, 0, 0.75, 1.5, 2.25, 3.
The approximating values are:
2.625, 1.875, 1.125, 0.375, 0.375, 1.125, 1.875, 2.625.
The quantized sequence looks like:
0, 1, 5, 3, 5, 4, 4, 3, 3, 4, 2, 3, 6, 4, 2, 4, 3, 2, 4, 7
or in binary form :
000001101011101100100011011100010011110100010100011010100111.
Number of bits =60.
The reconstructed sequence is
2.625, 1.875, 1.125, 0.375, 1.125, 0.375, 0.375, 0.375, 0.375, 0.375, 1.125, 0.375,
1.875, 0.375, 1.125, 0.375, 0.375, 1.125, 0.375, 2.625.
The total quantization error is equal to
  (2.3  2.625) 2  (1.78  1.875) 2  (1.4  1.125) 2  (0.1  0.375) 2  (0.75  1.125) 2 
 (0.65  0.375) 2  (0.44  0.375) 2  (0.58  0.375) 2  (0.53  0.375) 2  (0.05  0.375) 2 
 (0.76  1.125) 2  (0.63  0.375) 2  (1.89  1.875) 2  (0.48  0.375) 2  (0.80  1.125) 2 
 (0.40  0.375) 2  (0.71  0.375) 2  (0.93  1.125) 2  (0.16  0.375) 2  (2.4  2.625) 2  1.22
The average quantization error is 1.22/20=0.061. The signal energy is
E   xi2  24.95
The relative error is
 r   / E  1.22 / 24.95  0.049.
Estimates of probabilities are:
p0  1 / 20 , p1  1 / 20, p2  3 / 20, p3  5 / 20, p4  6 / 20, p5  2 / 20, p6  1 / 20, p7  1 / 20 .
The estimate of the entropy is
H   pi log 2 pi  2.6 bits.
i
Number of bits = 52.
If R  2 then M  4 and the thresholds are:
3, 1.5, 0, 1.5, 3.
The approximating values are:
2.25, 0.75, 0.75, 2.25.
The quantized sequence looks like
0,0,2,1,2,2,2,1,1,2,1,1,3,2,1,2,1,1,2,3
or in binary form
0000100110101001011001011110011001011011.
Number of bits=40.
The reconstructed sequence is
2.25, 2.25, 0.75, 0.75, 0.75, 0.75, 0.75, 0.75, 0.75, 0.75, 0.75, 0.75, 2.25, 0.75, 0.75,
0.75, 0.75, 0.75, 0.75, 2.25.
The total error is:
  (2.3  2.25) 2  (1.78  2.25) 2  (1.4  0.75 2  (0.1  0.75) 2  (0.75  0.75) 2 
 (0.65  0.75) 2  (0.44  0.75) 2  (0.58  0.75) 2  (0.53  0.75) 2  (0.05  0.75) 2 
 (0.76  0.75) 2  (0.63  0.75) 2  (1.89  2.25) 2  (0.48  0.75) 2  (0.80  0.75) 2 
 (0.40  0.75) 2  (0.71  0.75) 2  (0.93  0.75) 2  (0.16  0.75) 2  (2.4  2.25) 2  2.49.
The average error is 2.49/20=0.1245.
The relative error is 2.49/24.95=0.10.
Estimates of symbol probabilities are:
p0  2 / 20, p1  8 / 20, p2  8 / 20, p3  2 / 20.
The estimate of the entropy is H  1.72 bits.
Number of bits= 35.
Compare the fixed-rate quantizer with a variable-length quantizer.
Let quantization step   0.6. The quantized sequence can be obtained as
q  round (x /  ) .
We obtain
4, 3, 2, 0, 1, 1, 1, 1, 1, 0, 1, 1, 3, 1, 1, 1, 1, 2, 0, 4.
The reconstructed sequence has the form
2.4, 1.8, 1.2, 0, 0.6, 0.6, 0.6, 0.6, 0.6, 0, 0.6, 0.6, 1.8, 0.6, 0.6, 0.6, 0.6, 1.2, 0, 2.4
Estimates of probabilities are:
p4  1 / 20, p 3  1 / 20, p2  1 / 20, p1  6 / 20, p0  3 / 20, p1  5 / 20, p2  1/ 20, p3  1 / 20,
p4  1/ 20.
The estimate of the entropy is H  2.72 bits.
Number of bits =54.
The total error is:   0.3584 , the average error 0.3584/20=0.018.
The relative error  r  0.0144.
Problem 3. Assume that X is the Gaussian variable with zero average value and variance equal to
1. For fixed-rate quantizer with M  4 and given sets of thresholds and approximating values
compute probabilities of approximating values. Estimate entropy of the discrete source formed by
the quantizer outputs. Estimate the MSE.
The sets of thresholds ti and approximating values yi are:
- t0  , t1  1.5, t2  0, t3  1.5, t4   , y0  2.25, y1  0.75, y2  0.75, y3  2.25 ,
- t0  , t1  1.0, t2  0, t3  1.0 , t4   , y0  1.5, y1  0.5, y2  0.5, y3  1.5 ,
- t0  , t1  2.0, t2  0, t3  2.0 , t4   , y0  3.0, y1  1.0, y2  1.0, y3  3.0
Solution.
1.5
p0 


0
p1 

1.5
(See qfun.m , erf ( x) 
x2
1 2
e dx  Q(1.5)  0.07 .
2

x2
x2
1 2
1 2
e dx  
e dx  0.5  1  0.5 
2
2

1.5
2
x
e
 
0
t 2
2
1.5


x2
1 2
e dx  0.43 .
2
 x 
dt , Q( x)  0.5  0.5erf 
 .)
 2
p2  0.43 ,
p3  0.07 ,
H   p0 log 2 p0  p1 log 2 p1  p2 log 2 p2  p3 log 2 p3  1.58 bits.
0
x
 1.5 1  x

1
 
( x  yi ) dx  2 
e 2 ( x  2.25) 2 dx  
e 2 ( x  0.75) 2 dx  .


i 1 t i 1
1.5 2
   2

In order to compute the quantization error first consider the following auxiliary integrals :
b  x2
b2 / 2
a2
b2
 
1
1
1   2
t
2
e xdx 
e dt 
e  e 2 .




2 a
2 a2 / 2
2 

4
ti
1
e
2
x2
2
2
2
2
b
1
e
2 a
x2
2
x 2 dx 
2

b/ 2
e
a/ 2
t 2 2
t dt 
2
1 2
( du  tet dt , u   e  t , v  t, dv  dt. )
2
b/ 2

2
2  1 t 2 b / 2 1

 e t
  e t dt .

  2
a/ 2 2 a/ 2

In general case for M  4 :
 2  1 t 2 t / 2



1 t12 / 2
  2

Q( t1 )   2  y1 
e
 y12Q(t1 )  
  e 1

 

2
2

 2



 2  1 t 2

0

2

1
2 

(0.5  Q(t1 ))   2  y2 
et1 / 2  1  y22  0.5  Q(t1 )   .
 e t

  2

2
t1 / 2 2






Inserting our input data we obtain
 2  1 t 2  1.5 / 2



1  2.25 / 2
 e t

Q(  1.5)   2  2.25 
e
 (2.25)2 Q(1.5)  

  2

2
2





  2
 2  1 t 2


0

1
 e t
 2

(0.5  Q(1.5))   2  0.75 
e 2.25 / 2  1  (0.75) 2 0.5  Q(1.5) .

  2

 1.5 / 2 2
2






  0.1898
Let t1  1.0, t3  1.0 and y0  1.5, y1  0.5, y2  0.5, y3  1.5 .
1.0
p0 



x2
0
x2
1 2
e dx  Q(1.0)  0.16.
2
x2
1 2
1 2
p1  
e dx  
e dx  0.5  1  0.5 
1.0 2
1.0 2
H=1.90 bits,   0.1189.
1.0


x2
1 2
e dx  0.34.
2
Let t1  2.0, t3  2.0 and y0  3.0, y1  1.0, y2  1.0, y3  3.0
2.0
p0 


0
x2
1 2
e dx  Q(1.0)  0.02.
2
x2

x2
2.0
x2
1 2
1 2
1 2
p1  
e dx  
e dx  0.5  1  0.5  
e dx  0.48
2
2
2
2.0
2.0

H=1.27 bits,   0.33 .
When the quantization error reduces the distribution of approximating values becomes more
uniform (probabilities are closer to each other) and hence the entropy increases.
Problem 4. For random Gaussian variable from problem 3 find thresholds and approximating values
for the nonuniform optimal quantizer with M  4.
Since t0   , t2  0, t3  t1, t4   we should choose only t1 .
Solution
Let t1   . Set  p  0. Let the optimization accuracy be   0.0001.
The approximating values are:
1
2
y1 
1
2
t1

 xe

t1
e
x2
2
x2

2
dx

dx

t2
 1
2
e
, y2 
Q (t1 ) 2
1
2
0

 xe
x2
2
dx

t1
0
1
e
2 t1
x2

2

dx
t12
2
(e  1)
.
(0.5  Q(t1 )) 2
 2  1 t 2 t / 2



1 t12 / 2
 e 1

Q(t1 )   2  y1 
e
 ( y1 ) 2 Q(t1 )  

  2

2
2





  2


 2  1 t 2 0


2

1
 e t
 2

(0.5  Q(t1 ))   2  y2 
e t1 / 2  1  ( y2 ) 2 0.5  Q(t1 ) .

  2

t1 / 2
2
2




Compute the error by the formula
 2  1 t 2 t / 2



1 t12 / 2
 e 1
  2

Q(t1 )   2  y1 
e
 ( y1 )2 Q(t1 )   .

  2

2
2







 2  1 t 2 0


2

1
 e t
 2

(0.5  Q(t1 ))   2  y2 
e t1 / 2  1  ( y2 ) 2 0.5  Q(t1 ) .

  2

t1 / 2
2
2




y  y2
. Return to computing approximating values.
If    p   . Set  p   . Compute t1  1
2
Otherwise stop.
(See lmgauss.m)
Problem 5. For the given observed sequence x =(2.3, 1.78, 1.40, 0.10, 0.75, 0.65, 0.44, 0.58,
0.53, 0.05, 0.76, 0.63, 1.89, 0.48, 0.80, 0.40, 0.71, 0.93, 0.16, 2.4) construct the nonuniform
quantizer using the Lloyd-Max procedure.
(See lm.m and also p.37 in the book)
Problem 6. Show that for the fixed-rate uniform scalar quantizer when the quantization step   0
the MSE  
2
. Under this assumption estimate the rate-distortion function for the Gaussian
12
variable N (0, 2 ).
Solution.
Let f (x) be the probability density function (pdf) of the random variable X . Then when   0 we
can assume that pdf is a constant inside each quantization cell, that is p j  f ( y j ).
Then we obtain
tj
    ( x  y j ) f ( y j )dx  
2
j t j 1

j
pj

y j  / 2

( x  y j ) dx 
2
yj  /2
y j  / 2
1
 p  (x


j
2
 2 xy j  y 2j )dx 
yj  /2
j

y j   / 2
x2 y j   / 2
 x3 y j   / 2

2
p

2
y

y
x


j
j
j
y j   / 2
 j  3 y j   / 2
2 yj  / 2

1
3
3


( y j   / 2) 2  ( y j   / 2) 2
 ( y j   / 2)  ( y j   / 2)

  pj 
 2yj
 y 2j ( y j   / 2  y j   / 2) 
 j 
3
2


2


1


 1
  p j  (( y j   / 2)2  ( y j   / 2)2  ( y 2j   2 / 4)  y j 2 y j  y 2j     p j  (3 y 2j  )  y 2j   
 j 3
4
  j
3

1

2
12
p
j
j

2
12
.
Choose   6 / 2 R since the probability that the Gaussian variable N (0, 2 ) belongs to the interval
(3 ,3 ) is rather high. It is equal to 1  2Q(3)  0.9973. Then since  2  12 we obtain
36 2
3 2
2R
  2 R  12 , 2 
. By taking logarithm of the both parts we get
2

2
1
 2 log 2 3
2 R  log 2 3 , R  log 2

 H ( )  0.7925.

2

2
2
Problem 7. For the given speech file in *.wav format :
- split speech signal into frames of the given size (N);
- for each frame find coefficients of the Yule-Walker equations of the given order using the
autocorrelation method;
- find the solution of the Yule-Walker equations using the Levinson-Durbin procedure;
- plot the amplitude function of the prediction filter;
- find the ideal excitation (prediction error) signal;
- plot the amplitude function of the inverse filter;
- scalar uniformly quantize filter coefficients and the error signal;
- compute number of bits for fixed rate encoder;
- compute compression ratio;
- reconstruct speech signal using the obtained quantized data;
- compute the relative squared error which occurs when we approximate the original speech signal
by the reconstructed speech signal.
Solution.
Let x(nTs ) be the input speech signal. Then coefficients of the Yule-Walker equations can be
computed as:
Rˆ ( i  j )  cij / N  1 / N
N 1 i  j
 x(nT ) x(nT
n 0
s
s
 i  j Ts ).
Since the coefficients are symmetric and have the Toeplitz property it is enough to compute only
m  1 values, where m is the filter order. For example, we can compute c  R( j  1) for
1j
j  1,2,..., m  1.
Then system of the Yule-Walker equations looks like
m
 a R( i  j )  R( j ),
i 1
i
j  1,..., m
We initialize the Levinson-Durbin procedure by
l  0, E0  c11, a ( 0 )  0.
For l  1 : m compute
l 1
al( l )  ( R(l )   ai( l 1) R(l  i )) / El 1
;
i 1
a(jl )  a(jl 1)  all al(l j1) , 1  j  l  1,
El  El 1 (1  (all ) 2 ).
The obtained prediction filter can be described as follows
m
e(nTs )  x(nTs )   ai x(nTs  iTs ) .
i 1
It frequency function has the form
m
H (e jTs )  1   ai e  jiTs
i 1
or using the normalized frequency    / s
m
H (e j 2 )  1   ai e  j 2i .
i 1
The amplitude function has the form
m
m
i 1
i 1
A( )  (1   ai cos(2i)) 2  ( ai sin( 2i)) 2 .
The amplitude function of the inverse filter has the form
1
AI ( ) 
m
(1   ai cos( 2i ))  ( ai sin( 2i ))
i 1
.
m
2
2
i 1
Let m  4 and the coefficients c1 j  R( j  1) are equal to
3.79  107 , 2.85  107 , 6.62  106 ,  1.46  107 ,  2.48  107 .
The normalized by R (0) coefficients are:
1 , 0.75, 0.17 ,  0.38 , 0.65.
The Levinson-Durbin procedure:
Step 1: a11  c12 / c11  0.75
E  E (1  (a11 )2 )  3.79  107 (1  (0.75)2 )  1.645  107.
Step 2: a22  (c13  a11c12 ) / E  (6.62  106  0.75  2.85  107 ) /(1.65  107 )  0.902
a12  a11 (1  a22 )  0.75  (1  0.902)  1.43
E  E (1  (a22 )2 )  1.645  107 (1  (0.902)2 )  3.054  106
Step 3:
a33  (c14  a12c13  a22c12 ) / E  (1.46  107  1.43  6.62  106  0.902  2.85  107 ) /(3.054  106 )  0.55
a13  a12  a33a22  1.43  0.55  (0.902)  1.928
a23  a22  a33a12  0.902  0.55  1.43  1.69
E  E (1  (a33 ) 2 )  3.054 106 (1  (0.55) 2 )  2.13 106
Step 4:
a44  (c15  a13c14  a23c13  a33c12 ) / E  (2.48  107  1.928  1.46  107  1.69  6.62  106  0.55  2.85  107 ) /
(2.13  106 )  0.539
a14  a13  a44 a33  1.928  0.539  0.55  2.224
a24  a23  a44a23  1.69  (0.539)  (1.69)  2.6
a34  a33  a44 a13  0.55  (0.539)  1.98  1.589
E  E (1  (a44 )2 )  2.13  106 (1  (0.539)2 )  1.51 106
The normalized error is 0.0399.
Let quantization step be equal to 0.02. Then quantized coefficients are:
111, 130, 79, 27
The number of bits for storing coefficients are :
bits _ c  m  log 2 (max_ coeff  min_ coeff  1)  4  log 2 (111  130  1)  32 .
Let the quantization step for error signal be equal to step then the number of bits for storing the
quantized error signal is
bits _ err  N  log 2(max_ err  min_ err  1) ,
where max_ err and min_ err are the maximum and the minimum values of error samples.
The compression ratio is
N  16
compr 
.
bits _ c  bits _ err
(See also autocorr.m)
Problem 8. For the given analysis filter bank (h0 (n), h1 (n)) find the frequency response of the
lowpass and highpass filters. For the given input sequence check can the filter bank ( g0 (n), g1 (n))
be the inverse filter bank with respect to (h0 (n), h1 (n)) .
Let h0 (n)  (1,1,8,8,1,1) /(16 / 2 ) , h1 (n)  (1,1) / 2
g 0 (n)  (1,1) / 2 , g1 (n)  (1,1,8,8,1,1) /(16 / 2 )
The input sequence x(n)  (1,1,2,3,1,5,2)
Solution
The frequency response of the lowpass filter is

5

H 0 (e jTs )   h0 (n)e jnTs  16 / 2  1  e jTs  8e j 2Ts  8e j 3Ts  e j 4Ts  e j 5Ts .
i 0
1


H1 (e jTs )   h1 (n)e jnTs  1 / 2  1  e jTs .
i 0
The output of the lowpass filter is:
y0 (n)  (1,0,7,15,27,36,36,51,54,20,3,2)
The output of the highpass filter is:
y1 (n)  (1,0,1,1,2,4,3,2)
The decimated output of the lowpass filter:
v0 (n)  (1,7, 27,36,54, 3)
The decimated output of the highpass filter:
v1 (n)  (1,1,2,3)
The upsampled output of the lowpass filter is:
~
y0 (n)  (1,0,7,0,27,0,36,0,54,0,3, )
The upsampled output of the highpass filter is:
~
y1 (n)  (1,0,1,0,2,0,3)
The lowpass filtered ~
y0 (n) is:
w0 (n)  (1,1,7,7,27,27,36,36,54,54,3,3)
The highpass filtered ~
y1 (n) is:
w1 (n)  (1,1,7,9,11,5,12,20,26,22,3,3)
Summing up w0 (n) and w1 (n) , and dividing by 16 we obtain
xˆ (n)  (0,0,0,1,1,2,3,1,5,2,0,0)  x(n  3).
Problem 9. Consider orthogonal analysis and synthesis wavelet filters :
h0 (n)  [2 6 3 1] / 50 ,
h1 (n)  [1 3 - 6 2] / 50 ,
g0 (n)  [1 3 6 2] / 50 ,
g1 (n)  [2 - 6 3 1] / 50 .
For the given input sequence of length N show that the use of cyclic extension allows to reconstruct
the input sequence from the reference and detail sequences of length N / 2.
Solution.
Let input sequence be equal to:
1,2,3,4,5,6,7,8 .
To avoid increasing of the transform length we perform cyclic extension of the input sequence that
is we place L  1 first symbols to the end of the sequence, where L is the length of impulse
response. In our case L  4 .
Cyclically extended input sequence is:
1,2,3,4,5,6,7,8,1,2,3
Lowpass filtered sequence is:
20,10,21,31,41,51,61,71,65,27,13,23,7,3
Lowpass filtered sequence multiplied by window of length N delayed by L  1 symbols is:
31,41,51,61,71,65,27,13
Highpass filtered sequence is:
1,5,3,3,3,3,3,3,5,29,19,1,14,6
Highpass filtered sequence multiplied by window is:
3,3,3,3,3,3,5,29,19
Reference sequence is:
31,51,71,27
Detail sequence is:
3,3,3,29
Upsampled reference sequence is:
31,0,51,0,71,0,27,0
Upsampled detail sequence is:
3,0,3,0,3,0,29,0
Cyclically extended reference sequence is:
0,27,0,31,0,51,0,71,0,27,0
Cyclically extended detail sequence is:
0,29,0,3,0,3,0,3,0,29,0
Reconstructed low-frequency part is:
0,27,81,131,147,135,215,235,315,399,223,162,54,0
Reconstructed high-frequency part is:
0,58,174,81,47,15,15,15,15,49,177,87,29,0
Sum of low-frequency and high-frequency parts is:
0,85,255,50,100,150,200,250,300,350,400,75,25,0
After normalization by 1/ 50 and multiplying by window we obtain:
1,2,3,4,5,6,7,8 .
(See also cyclext.m)
Problem 10. Let matrix of the quantized DCT coefficients have the form:
 115 3 1  1 0 0 0 0 


  12 2 1  1 0 0 0 0 
 0
1 1 0 0 0 0 0


1 0 0 0 0 0 0
 3
 1 0 0 0 0 0 0 0


 1 0 0 0 0 0 0 0


0 0 0 0 0 0 0
 0
 0
0 0 0 0 0 0 0 

Construct codewords for run-length coding for AC coefficients. Construct codeword for DC
coefficient if DC coefficient of the previous block is equal to 113.
Solution
The 1-D array of DCT coefficients obtained by scanning in zigzag order is:
115,3,12,0,2,1,1,1,1,3,1,1,1,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0
The corresponding codewords for run-length coding are:
((0,2),11) ((0,4),0011) ((1,2),10) ((0,1),1) (0,1),0) ((0,1),1) ((0,1),1) ((0,2),11) ((0,1),0) ((0,1),0)
((0,1),0) ((0,1),0) ((6,1),0) EOB
Difference of DC coefficients is equal to 115113=2. It is coded as (2,10).
(See also batdctcod.m)
Problem 11. For the given frame of speech samples (bold font)
461
90 102 144
494 792 606 147 136 75
107 244 387 672 478 8 296 496 368 104
31 32 268 371 218 114 329 208 16 163
160
6 236 133
7 55 16 252 411 226
32 158
223 24 282 324 92 90 53 262
465 302 159
722 835 527 177 22 137 378
381 247 16 18 108 485 584 372 43 280
280 237 218 395 671 743 445 183 1 66
10 15
74
251 275 144 177 390 176 325
699 755
535
233 123 244 368 335 83 495
508 235 209
369 149 211 284 9 191 105
find pitch period (in the range 16-45).
Solution
To estimate the pitch period it is necessary to find the maximum over j of the match function
2
 59

  s (i ) s (i  j ) 
 , 16  j  45 .
m( j )   i059
 s(i  j )2
i 0
We obtain
m  106 ( 2.6070 1.0125 0.1362 0.1775 1.2626
1.5385 0.0011 1.2221 1.7722 0.3718
0.1492 0.1103 1.2155 1.2199 0.0629
0.3767 0.1611 2.8998 4.2192 1.4495
0.6896 0.2670 0.0811 0.3296 0.0123
The maximum value 4.2192  106 corresponds to j  38
(See also pitch_p.m)
2.6932
0.0674
0.3893
0.0092
1.1721).
that is pitch period estimate is equal to 38.