Digital Signal Processing Reference
In-Depth Information
to 0
-
,|
z
| will vary from 0 to 1
-
. Hence, poles
inside
the unit circle within region
2 in the
z
-plane will yield a stable system. The response of such system will be a
decaying exponential if the poles are real or a decaying sinusoid if the poles are
complex.
-•
s>
0
Poles on the right side of the
j
axis (region 3) in the
s
-plane represent an unstable
system, and (4.12) yields a magnitude of |
z
|
w
>
1, because
e
s
T
>
1. As
s
varies from 0
+
to
. Hence, poles
outside
the unit circle within region 3
in the
z
-plane will yield an unstable system. The response of such system will be an
increasing exponential if the poles are real or a growing sinusoid if the poles are
complex.
•
,|
z
| will vary from 1
+
to
•
s=
0
Poles on the
j
axis (region 1) in the
s
-plane represent a marginally stable system,
and (4.12) yields a magnitude of |
z
|
w
1, which corresponds to region 1. Hence, poles
on
the unit circle in region 1 in the
z
-plane will yield a sinusoid. In Chapter 5 we
implement a sinusoidal signal by programming a difference equation with its poles
on
the unit circle. Note that from Exercise 4.2 the poles of
X
(
s
)
=
=
sin
n
w
T
in (4.9)
or
X
(
s
)
=
cos
n
w
T
in (4.11) are the roots of
z
2
-
2
z
cos
w
T
+
1, or
2
cos
w
T
±
4
cos
2
w
T
-
4
p
=
12
,
2
=
cos
w
T
±
-
sin
2
w
T
=
cos
w
T
±
j
sin
w
T
(4.13)
The magnitude of each pole is
2
2
pp
==
cos
w
T
+
sin
w
T
=
1
(4.14)
1
2
The phase of
z
is
q=w
T
=
2
p
f
/
F
s
. As the frequency
f
varies from zero to
±
F
s
/2, the
phase
q
will vary from 0 to
p
.
4.1.2 Difference Equations
A digital filter is represented by a difference equation in a similar fashion as an
analog filter is represented by a differential equation. To solve a difference equa-
tion, we need to find the
z
-transform of expressions such as
x
(
n
k
), which corre-
sponds to the
k
th derivative
d
k
x
(
t
)/
dt
k
of an analog signal
x
(
t
). The order of the
difference equation is determined by the largest value of
k
. For example,
k
-
=
2
represents a second-order derivative. From (4.5)
•
Â
0
()
=
()
()
+
()
()
Xz
xnz
-
n
=
x
0
x
1
z
-
1
+
x
2
z
-
2
+◊◊◊
(4.15)
n
=
Then the
z
-transform of
x
(
n
-
1), which corresponds to a first-order derivative
dx
/
dt
,
is
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