Exercise 18
(a). Construct a filter using the
zeros
for
"zeroing out"
,
, and
.
Solution 18 (a).
See text and/or instructor's solution manual.
Answer. Compute
the product
.
The desired filter is
.
Solution. Use
the conjugate pair of zeros
and
,
and the "zero
out factors"
and
.
Then calculate
This is a basic filter and we can use
Property (i) Zeroing Out
Filter.
The transfer function for part (a) has the form
and corresponds to the filter
.
For this exercise,
we use
and
in
these equations to get the desired recursive formula
.
Therefore, the desired filter for part (a) is
.
We are done.
Aside. We can let Mathematica double check our work.
Aside. The Maple
commands are similar
We are really done.
Aside. We can graph
the amplitude response for the filter
.
![[Graphics:../Images/ZTransformFilterModHome_gr_2141.gif]](../Images/ZTransformFilterModHome_gr_2141.gif)
Amplitude
response
, and
zero-pole plot of
,
for
the filter
.
We
can see that some of the high-range frequencies are attenuated,
and
for
.
Aside. In Exercise
12 (b) we investigated the filter
.
We are really really done.
Aside. For
illustration, we can graph the causal input
sequence
,
and the corresponding causal output
sequence
.
The signal component
will
be amplified by the factor
, and
The signal component
will
be attenuated by the factor
.
![[Graphics:../Images/ZTransformFilterModHome_gr_2161.gif]](../Images/ZTransformFilterModHome_gr_2161.gif)
The
causal input sequence
and
the corresponding causal output sequence.
Here
we have
and ![]()
This
filter reduces the proportion of the signal
component
by
a factor of
.
This solution is complements of the authors.
(c) 20098 John H. Mathews, Russell W. Howell