Digital Signal Processing Reference
In-Depth Information
1
R
k
z
k+1
H
k+1
(c
n
)
H
k
(c
n
)
d
k+1
=
n−→∞ (2.206)
k
R
k
=
(z
k+1
−z
k
′
).
(2.207)
k
′
=1
This expression for R
k
is deduced by simplifying the original formula R
k
=
V
k+1
/V
k
derived in Ref. [5] where V
k
is the Vandermond determinant
=
1 1 1 1
z
1
z
2
z
3
c
k
k
k
z
1
z
2
z
3
z
k
V
k
=
(z
s
−z
s
′
)
= 0 .
(2.208)
.
.
.
.
.
.
.
s=1
s
′
=1,s
′
=s
z
k−1
1
z
k−1
2
z
k−1
3
z
k−1
k
We see that the same quotient of the Hankel determinants H
k+1
(c
n
)/H
k
(c
n
)
appears in the Shanks transform e
k
(c
n
) and in the amplitude d
k+1
. There
fore, the necessary and su
cient condition (2.193) that c
n
has precisely K
components can equivalently be written as d
K+1
= 0, or more generally
d
K+m
= 0
(m = 1, 2, 3,...).
(2.209)
An important implication of this result is that, whenever the input time sig
nal c
n
has K harmonics{z
k
}(1≤k≤K), the exact computation which
eventually finds some higherorder transients{z
K+m
,d
K+m
}(m = 1, 2, 3,...)
must produce zerovalued amplitude d
K
′
= 0 (K
′
> K), as per (2.209). Let a
computation find an estimate c
n
with K +m (m > 0) harmonics for the input
time signal c
n
which, however, possesses only K harmonics
K+m
d
k
z
k
c
n
=
(m = 1, 2, 3,...).
(2.210)
k=1
Then, the result (2.209) will apply to yield the reduction formula which, in
turn, reconstructs the exact c
n
K+m
d
k
z
k
c
n
=
k=1
K
d
k
z
k
d
K+1
z
K+1
+ d
K+2
z
K+2
++ d
K+m
z
K+m
=
+
k=1
K
d
k
z
n
= c
n
=
(QED).
(2.211)
k=1
In other words, the higherorder harmonics{z
K+m
,d
K+m
}(m = 1, 2, 3,...)
in the estimate c
n
ought to be interpreted as being spurious in the sense
Search WWH ::
Custom Search