•
Consider
=
+
+
+
+
+
Dz a0z5 a1z4 a2z3 a3z2 a4z a5
with
a
0
> 0
If
a
0
< 0, we consider –
D
(
z
), and if –
D
(
z
) is Schur, so is
D
(
z
). We form the table below. The 1
st
row is the coefficients of
D
(
z
)
arranged in descending powers of
z
. The second row is the reverse of the first row. We compute
k
1
=
a
5
/
a
0
. We then subtract
from the first row the product of the second row and
k
1
to yield the first
b
i
row. We then reverse the first
b
i
row and make
that the 2
nd
b
i
row, and calculate
k
2
=
b
4
/
b
0
, and subtract from the first
b
i
row the product of the second
b
i
row and
k
2
to obtain
the first
c
i
row. Continue until we obtain
f
0
.
The Jury Test: The polynomial of degree five with a position leading coefficient in
D
(
z
) is Schur iff the five leading
coefficients {
b
0
,
c
0
,
d
0
,
e
0
,
f
0
} in the Jury Table are all positive. If any one of them is zero or negative, then the polynomial is
not Schur. If the denominator of a transfer function
H
(
z
) is Schur, then
H
(
z
) is stable.
The Jury Table
a
0
a
1
a
2
a
3
a
4
a
5
a
5
a
4
a
3
a
2
a
1
a
0
k
1
=
a
5
/
a
0
b
0
b
1
b
2
b
3
b
4
0
b
4
b
3
b
2
b
1
b
0
k
2
=
b
4
/
b
0
c
0
c
1
c
2
c
3
0
c
3
c
2
c
1
c
0
k
3
=
c
3
/
c
0
d
0
d
1
d
2
0
d
2
d
1
d
0
k
4
=
d
2
/
d
0
e
0
e
1
0
e
1
e
0
k5
=
e
1
/
e
0
f
0
•
A polynomial is defined to be
CT stable
if all its roots have negative real parts. We discuss a method of checking
whether or not a polynomial is CT stable without computing its roots. We use the following
D
(
s
) to illustrate the procedure:
=
+
+
+
+
+
+
>
Ds a0s6 a1s5 a2s4 a3s3 a4s2 a5s a6 with a0 0
If
a
0
<
0, we apply the procedure to 
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 Fall '10
 ChiTsongChen
 Highpass filter, Lowpass filter, Row, Schur

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