Exercise 12
(b). Construct a filter using the
zeros
. What
signals are "zeroed out" ?
Solution 12 (b).
See text and/or instructor's solution manual.
Answer. Compute
the product
.
The desired filter is
.
Solution. Use
the zeros in part (a) and the additional conjugate pair of
zeros
,
and the "zero
out factors"
,
and
.
Then calculate
Now use the
General
Filter Equation (9-29) and
the corresponding Transfer
Function (9-34) in the
case
.
Get the new fact that the transfer function
,
corresponds to the filter
.
For this exercise, we
use
,
and
in
these equations to get the desired recursive formula
.
Therefore, the desired filter for (b) is
.
for "zeroing out" the signals
.
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_1194.gif]](../Images/ZTransformFilterModHome_gr_1194.gif)
Amplitude
response
and
zero-pole plot of
,
for
the filter
.
We
can see that the mid-range frequencies are attenuated,
and
for
.
Aside. In Exercise
18 (a) we will investigate the filter
.
We are really really done.
Aside. Let us
investigate how well the filter works to eliminate
signals
which are close to the "zero-out" frequencies.
For illustration
purposes we will explore the casual input
signal
.
The signal component
will
be amplified by the factor
.
The signal component
will
be attenuated by the factor
, and
The signal component
will
be attenuated by the factor
, and
The signal component
will
be attenuated by the factor
.
![[Graphics:../Images/ZTransformFilterModHome_gr_1222.gif]](../Images/ZTransformFilterModHome_gr_1222.gif)
The
causal input sequence
and
the corresponding causal output sequence.
We
can see that the filter practically eliminates the
signals
which are close to the "zero-out" frequencies.
This solution is complements of the authors.
(c) 20098 John H. Mathews, Russell W. Howell