Digital Signal Processing Reference
In-Depth Information
9
=
;
K = 4
−u
2
u
3
u
4
c
0
+ (u
2
u
3
+ u
2
u
4
+ u
3
u
4
)c
1
−(u
2
+ u
3
+ u
4
)c
2
+ c
3
(u
1
−u
2
)(u
1
−u
3
)(u
1
−u
4
)
d
1
=
−u
1
u
3
u
4
c
0
+ (u
1
u
3
+ u
1
u
4
+ u
3
u
4
)c
1
−(u
1
+ u
3
+ u
4
)c
2
+ c
3
(u
2
d
2
=
−u
1
)(u
2
−u
3
)(u
2
−u
4
)
−u
1
u
2
u
4
c
0
+ (u
1
u
2
+ u
1
u
4
+ u
2
u
4
)c
1
−(u
1
+ u
2
+ u
4
)c
2
+ c
3
(u
3
−u
1
)(u
3
−u
2
)(u
3
−u
4
)
d
3
=
−u
1
u
2
u
3
c
0
+ (u
1
u
2
+ u
1
u
3
+ u
2
u
3
)c
1
+ (u
1
+ u
2
+ u
3
)c
2
+ c
3
(u
4
d
4
=
−u
1
)(u
4
−u
2
)(u
4
−u
3
)
(2.144)
etc. We see that, in general, irrespective of the actual value of the number K,
the numerator of d
k
is obtained by first taking all the roots of the characteristic
equation L
K
(u
k
) = 0 except u
k
and forming all possible products of the
K−1 roots{u
r
}(r = k). Afterwards, the sums of these products are formed
and multiplied by c
0
,c
1
,c
2
,...,c
K−2
, respectively. Subsequently, each of the
resulting terms is multiplied by−1 or +1 dependent on whether K is even or
odd. Finally, all these terms are summed up and the constant c
K−1
is added
at the end of the procedure to give the numerator of d
k
, as per (2.141)-(2.144).
Likewise, to obtain the denominator of d
k
, we first subtract successively from
u
k
all the other roots{u
r
}(r = k). Then, the product of all these differences
of the roots u
r
−u
s
(r = s) yields the denominator of d
k
, as is clear from
(2.141)-(2.144).
Returning now to the K equations of the string (2.135), we see that they
represent the terms that are of the general form
K−1
c
r
=
a
r−1
b
r
−
1
b
r
b
k
c
k−1
.
(2.145)
k=1
To gain an insight into this relationship, we introduce the following charac
teristic polynomial
J
K
(u) = b
0
u
K
+ b
1
u
K−1
+ b
2
u
K−2
+ b
3
u
K−3
++ b
K
.
(2.146)
Comparing the polynomials L
K
(u) and J
K
(u), it follows
L
K
(u) = J
K
(u
−1
).
(2.147)
This means that if u
k
is the k th root of L
K
(u), then the k th root of J
K
(u
−1
) =
0 is u
−1
k
L
K
(u) = b
K
(u−u
1
)(u−u
2
)(u−u
K
) =
b
0
b
K
+
b
1
b
K
u +
b
2
b
K
u
2
++ u
K
(2.148)
1
u
1
1
u
2
1
u
K
J
K
(u) = b
0
−u
−u
−u
= u
K
+
b
1
b
0
u
K−1
+
b
2
b
0
u
K−2
++
b
K
b
0
.
(2.149)
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