Digital Signal Processing Reference
In-Depth Information
Example 2.37.
Determine the frequency response of the system defined by the difference equation
shown below by use of the
z
-transform. Assume that
y
= 0 for
n<
0. Note that the frequency
response is the response to constant, unity-amplitude complex exponentials.
[
n
]
=
x
[
n
]
y
[
n
]=
x
[
n
]+
1
.
1
y
[
n
−
1
]
This system is a single-pole IIR with a growing impulse response. The
z
-transform is only conver-
gent for values of
z
having a magnitude greater than 1.1. Since the frequency response is determined by
evaluating the
z
-transform for values of
z
on the unit circle (i.e., magnitudes of 1.0), we cannot evaluate
the frequency response since the unit circle is not in the Region of Convergence.
2.6.5 CASCADED SINGLE-POLE FILTERS
If we were to cascade two identical single-pole IIRs, their equivalent
z
-transform would be the product
of the two
z
-transforms, or
1
−
pz
−
1
)
2
Supposing we cascaded two IIRs with poles related as complex conjugates. We would have
Y (z)/X(z)
=
(
1
1
1
Y (z)/X(z)
=
(
−
pz
−
1
)(
−
p
cc
z
−
1
)
1
1
where
p
cc
denotes the complex conjugate of
p
. Doing the algebra, we get
1
Y (z)/X(z)
=
p)z
−
1
(p
cc
p)z
−
2
1
−
(p
cc
+
+
which reduces to
1
Y (z)/X(z)
=
(2.20)
2
z
−
2
real
(p))z
−
1
1
−
(
2
·
+ |
p
|
Example 2.38.
Evaluate the magnitude of the z-transform for a cascaded connection of two single-pole
IIR filters each having a pole at 0.9.
Using Eq. (2.20), the
z
-transform would be:
1
H(z)
=
(2.21)
1
.
8
z
−
1
0
.
81
z
−
2
1
−
+
The
z
-transform of Eq. (2.21), evaluated along a constant radius contour having radius 1.0 (i.e.,
the unit circle) is shown in Fig. 2.18. It was obtained by making the call
LVxPlotZXformMagCoeff([1],[1,-1.8,0.81],[1],[1],256)
