Databases Reference
In-Depth Information
160 samples, while for the lower rate option the data are divided into 30 millisecond blocks
containing 240 samples each. For the higher bit rate each block is encoded using 304 bits per
block while for the lower bit rate each block is encoded using 400 bits. Each block is divided
into subblocks of 40 samples each.
For the computation of the autocorrelation coefficients we use a portion of the samples
from the previous block along with the samples from the current block. For the 15.2 kbit
encoding we use 80 samples from the previous block, while for the 13.33 kbit encoding we use
60 samples from the previous block. The use of samples from the previous blocks helps in the
continuity of the reconstruction. Because the computed LPC coefficients are transmitted to the
decoder the use of the previous block in generating the LPC coefficients does not detract from
the independent block coding feature of the iLBC algorithm. The samples from the current
block along with the samples from the previous block are windowed before the autocorrelation
coefficients are computed. For the high-rate case a single set of LPC coefficients are computed
while in the low-rate case two sets of LPC coefficients are computed. For the high-rate case
a single window of length 240 samples centered on the third subblock is used, while for the
low-rate case two windows centered on the second and fifth subblocks are used.
To reduce problems with numerical precision the autocorrelation coefficients are smoothed
using a window given by
1
.
0001
k
=
0
2
120
π
k
f
s
2
w
ac
[
k
]=
(35)
1
e
−
k
=
1
,
2
,...,
10
The multiplication of
w
ac
[
0
]
by 1.0001 corresponds to the addition of a white noise floor
40 dB below the signal power.
In each case the encoder computes LPC coefficients using the Durbin-Levinson algorithm
that are then represented using the line spectral frequency (LSF) representation, described
below. Let
A
k
(
z
)
be the tenth-order linear predictive filter
10
z
−
i
A
k
(
z
)
=
1
+
a
k
(
i
)
i
=
1
Define the two functions
f
1
(
and
f
2
(
z
)
z
)
as
f
1
(
z
−
11
A
k
(
z
−
1
z
)
=
A
k
(
z
)
+
)
f
2
(
z
−
11
A
k
(
z
−
1
z
)
=
A
k
(
z
)
−
)
These polynomials have roots only on the unit circle. Clearly, the polynomial
f
1
(
z
)
has a root
1 while the polynomial
f
2
(
at
z
=−
z
)
has a root at
z
=
1. Removing these roots we obtain
the polynomials
f
1
(
z
)
f
1
(
z
)
=
z
−
1
1
+
f
2
(
)
z
f
2
(
z
)
=
z
−
1
1
−
