Digital Signal Processing Reference
In-Depth Information
Example 4.8
Suppose that we are given the specification of a bandstop filter as shown in
Figure 4.15. In this example, we are given
ω
1
=
1500,
ω
2
=
2000,
ω
s
=
ω
4
=
1800,
A
p
55 dB. The passband is required to have a max-
imally flat response. With these specifications, we design the bandstop filter
following procedure given below:
=
0
.
2dB,and
A
s
=
=
2000
−
1500
=
500 and
ω
0
=
√
(
2000
)(
1500
)
1.
B
=
1732
.
1.
500[
s/(s
2
10
6
)
].
2.
The LP-BS frequency transformation is
p
=
+
3
×
3.
Let
s
=
3
.
74.
4.
Following the design procedure used in Example 4.2, we get
=
jω
s
=
j
1800. Then we get
s
=
√
10
0
.
02
−
1
=
0
.
21709, and from (4.44), we get
n
=
5
.
946 and choose
6.
5.
The six poles are calculated from (4.45) as
p
k
n
=
=−
0
.
33385
±
j
1
.
2459,
j
0
.
3329.
6.
The transfer function of the lowpass prototype filter
H(p)
is constructed
from
H(p)
−
0
.
9121
±
j
0
.
9121, and
−
1
.
246
±
H
0
/
[
k
=
1
(p
=
−
p
k
)
]as
(
1
.
664
)
3
(p
2
+
1
.
664
)(p
2
+
1
.
664
)(p
2
+
0
.
6677
p
+
1
.
824
p
+
2
.
492
p
+
1
.
664
)
(4.85)
+
3
×
10
6
)
]inthis
H(p)
and
simplify the expression to get the transfer function
H(s)
of the speci-
fied bandstop filter. This completes the design of the bandstop filter. The
magnitude response is found to exceed the given specifications.
=
500[
s/(s
2
7.
Next we have to substitute
p
The sections above briefly summarize the theory of approximating the piece-
wise constant magnitude of analog filters. This theory will be required for approx-
imating the magnitude of digital filters, which will be treated in the following
sections. The analog frequency transformations
p
g(s)
applied to the lowpass
prototype to generate the other types of filters are listed in Table 4.2.
=
TABLE 4.2 Frequency Transformations to Design HP, BP, and BS Filters
Type of
Transformation
Transformation
p
=
g(s)
Parameters Used
LP-LP
p
=
s/ω
p
ω
p
=
bandwidth-specified LP
filter
LP-HP
p
=
ω
p
/s
ω
p
=
cutoff frequency of the
specified HP filter
(
1
/B)
[
(s
2
ω
0
)/s
]
LP-BP
p
=
+
B
=
ω
2
−
ω
1
,where
B
is
bandwidth of the specified
BP filter:
ω
0
=
√
ω
1
ω
2
;
B
[
s/(s
2
ω
0
)
]
LP-BS
p
=
+
B
=
ω
2
−
ω
1
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