Digital Signal Processing Reference
In-Depth Information
Table 7.3 CELP coder parameter
definition for the comparison test
Parameter
Update rate
Sampling
8kHz
LPC analysis
160 samples
LTP 1- & 3-tap
40 samples
10-bit codebook
40 samples
Weighting
γ
=
0 . 9
Table 7.4 Performance of four standard codebooks
Scheme
1-tap (dB)
3-tap (dB)
Storage (words)
Gaussian
11.11
12.52
1024
×
40
Centre-clipped Gaussian
11.20
12.53
1024
×
40
Overlapping Gaussian
11.16
12.49
( 2
×
1023 ) +
40
Overlapping centre-clipped Gaussian
11.18
12.55
( 2
×
1023 ) +
40
with the objective results in Table 7.4, the overlapping centre-clipped Gaus-
sian codebook is very attractive for its reduced memory and computational
requirements.
Vector Sum Excitation
In the normal filtering approach of CELP,
˜
s(n) is matched by exhaustively
searching a finite number of sequences
s opt (n) ,is
the sequence which gives the minimum mean square error between
ˆ
s k (n) and the best match,
ˆ
˜
s(n)
and
s k (n) . How good a match between
ˆ
s(n) and
˜
ˆ
s k (n) is determined by the
degree of freedom in
s k (n) , i.e. the size and characteristics of the codebook.
In the method previously described, the freedom in
ˆ
ˆ
s k (n) was obtained by
synthesizing many versions of
s k (n) is
limited by x(n) at the residual side of the analysis. However, it can be noted
that if the same degree of freedom can be achieved at the synthetic signal side
of the analysis whilst retaining the fact that all candidate
ˆ
s k (n) , i.e. the degree of freedom in
ˆ
s k (n) are spectrally-
shaped by the LPC and perceptual-weighting filters, then less complexity
and equal performance could be obtained. Therefore, the aim is to limit the
amount of synthesis operations and perform the vector combinations to give
the necessary freedom in
ˆ
s k (n) at the output side of the analysis. One such
method is vector sum excitation (VSE) [24].
As for the majority of speech-coding analysis, in VSE, the mean squared
error approximation is used. The formulation of VSE to derive the opti-
mum excitation x opt (n) and hence
ˆ
ˆ
s opt (n) is as follows. Let each candidate
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