Digital Signal Processing Reference
In-Depth Information
whichingeneralcanbewrittenas,
(m
i
)
φ(m
i
m
i
)
g
i
=
(7.50)
where
(m
i
)
is the cross-correlation between the perceptually-weighted
target speech and the combined LP and perceptual-weighting filter impulse
response;
φ(m
i
m
i
)
is the autocorrelation of the combined LPC and perceptual-
weighting filter impulse response at positions
m
i
;and0
1. Sub-
stituting equation (7.50) into (7.47) gives an expression for the perceptually-
weighted squared error in terms of the pulse locations,
≤
i
≤
M
−
L
−
1
2
(m
i
)
φ(m
i
m
i
)
s
2
(n)
=
˜
−
E
(7.51)
n
=
0
Tominimize the error in equation (7.51), it can be seen that the best position for
a single pulse is that value of
m
i
which maximizes the term,
2
(m
i
)/φ(m
i
m
i
)
.
Once the search for the optimum pulse has finalized, the effect of this newly-
found pulse is removed from the perceptually-weighted input speech to give
a new reference signal to be used in determining the next pulse location.
Hence the updated reference speech is,
˜
s
i
+
1
(n)
= ˜
s
i
(n)
−
g
i
h(n
−
m
i
)
(7.52)
The steps carried out from equations (7.49)-(7.52) are repeated to find
the remaining pulse locations and amplitudes. Figure 7.18 shows a typi-
cal example of speech signal and the excitation signal produced in an AbS
manner, as discussed above. It can clearly be seen from Figure 7.18 that the
multi-pulse structure is very effective in producing a flexible excitation signal
in modelling the glottal characteristics, especially the pitch pulses.
Optimum Amplitude Excitation MPLPC
The sequential AbS method described is simple and fast but it has several
shortcomings. Successive optimization of individual pulses becomes inaccu-
rate when the number of pulses per frame increases. In order to improve the
performance one needs to consider the interactions amongst all the pulses
during optimization. To consider the interaction between the pulses, let the
weighted mean squared error after having placed
M
pulses at positions
m
i
,
be given by,
m
i
)
2
L
−
1
M
−
1
E
=
˜
s(n)
−
g
i
h(n
−
(7.53)
n
=
0
i
=
0
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