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and, as forecast, the coe
cient matrix of the conditional equations (2.187) becomes
diagonal, and for
N
=
L
their solution is easily found to be
j
=−
L
g
j
e
−
i
2π
(
m
/
M
)
t
j
j
=−
L
g
j
e
−
i
2π
(
m
/
M
)
t
j
L
L
M
2
L
G
m
=
=Δ
t
,
m
=−
L
,...,
L
,
+
1
(2.194)
just the DFT for equally spaced samples (2.148).
The mean square of the time domain representation (2.178) can be related to
the mean square in the frequency domain as represented by the DFT. Multiplying
through by g
j
and summing over the sample points, we get
j
=−
L
g
j
g
j
=
L
N
N
L
1
M
2
G
k
G
∗
m
e
i
2π((
k
−
m
)/
M
)
j
Δ
t
,
(2.195)
k
=−
N
m
=−
N
j
=−
L
for equally spaced samples. On using the orthogonality relation (2.139), we have
j
=−
L
g
j
g
j
=Δ
L
N
G
k
G
∗
k
,
Δ
t
f
(2.196)
k
=−
N
for sample interval in time
Δ
t
=
M
/(2
L
+
1) and sample interval in frequency
Δ
1/
M
. This is the form
Parseval's relation
takes for discrete, equally spaced
samples. On multiplying through by
f
=
Δ
f
, the left side becomes the power in the
time sequence,
j
=−
L
g
j
g
j
=
L
N
E
g
j
g
j
1
f
)
2
G
k
G
∗
k
,
=
(
Δ
(2.197)
2
L
+
1
k
=−
N
with
E
g
j
g
j
representing the squared magnitude of a particular realisation of the
time sequence. For stationary sequences, this expected value is independent of
whether the samples are equally spaced or unequally spaced. Under the assumption
of stationarity, Parseval's relation holds in either case.
2.3.6 Singular value decomposition
The conditional equations (2.187) for the DFT have a coe
cient matrix that is
equidiagonal and Hermitian and is thus
Toeplitz
. This suggests that the Levinson
algorithm described in Section 2.2.3 might be used in their solution. However,
experience shows that
cient matrix is
ill-conditioned, and the Levinson algorithm is inadequate for the solution of the
for
long time sequences,
the coe
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