Digital Signal Processing Reference
In-Depth Information
This leads to the assertion
¼
ln{
s
(
n
)}
þ
1
i¼
1
s
(
n
)
A
(
z
,
n
)
c
i
(
n
)
z
i
ln
:
(7
:
15)
If the
N
pre
predictor coefficients
a
i
(
n
) [or
a
i
(
n
)] are known we can derive a simple
recursive computation of the linear predictive cepstral coefficients
c
i
(
n
) by differentiat-
ing both sides of (7.15) with respect to
z
and equating the coefficients of alike
powers of
z
"
(
)
#
"
#
N
pre
ln 1
X
dz
1
i¼
1
d
dz
d
a
i
(
n
)
z
i
c
i
(
n
)
z
i
¼
i¼
1
(7
:
16)
"
#
1
X
N
pre
1
þ
X
N
pre
¼
1
i¼
1
ia
i
(
n
)
z
i
1
a
i
(
n
)
z
i
ic
i
(
n
)
z
i
1
:
i¼
1
i¼
1
Multiplying both sides of (7.16) with [
1
þ
P
N
pre
i¼
1
a
i
(
n
)
z
i
] leads to
X
N
pre
¼
1
i¼
1
1
k¼
1
X
N
pre
ia
i
(
n
)
z
i
1
ic
i
(
n
)
z
i
1
kc
k
(
n
)
a
i
(
n
)
z
ki
1
:
(7
:
17)
i¼
1
i¼
1
If we now consider the equation above for equal powers of
z
, we find that starting from
the left, the first two terms only contribute a single term each up to
z
N
pre
1
. We will
label the order with
i
. The last term in contrast produces an amount of terms that
depends on
i
. Setting all terms that belong to the same power of
z
equal results in
ia
i
(
n
)
¼ ic
i
(
n
)
X
i
1
kc
k
(
n
)
a
ik
(
n
),
for
i
[
{1,
...
,
N
pre
}
:
(7
:
18)
k¼
1
Solving this equation for
c
i
(
n
) results in
i
X
i
1
1
c
i
(
n
)
¼ a
i
(
n
)
þ
kc
k
(
n
)
a
ik
(
n
),
for
i
[
{1,
...
,
N
pre
}
:
(7
:
19)
k¼
1
For
i
.
N
pre
the first term on the right side still needs to be considered in (7.17)
whereas the term on the left side does not contribute any more to powers of
z
larger
than
N
pre
1 and can therefore be omitted. Therefore, we can solve (7.17)
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