Digital Signal Processing Reference
In-Depth Information
Example 3.7 Unnormalized Chebyshev Bandstop Filter
Problem:
Determine the transfer function for a Chebyshev bandstop filter to
satisfy the following specifications:
ω
pass1
= 3,000 rad/sec, ω
pass2
= 24,000 rad/sec, ω
stop1
= 6,000 rad/sec,
ω
stop2
= 12,000 rad/sec,
a
pass
= −1.0 dB,
a
stop
= −35 dB
Solution:
Using the material of Section 2.3, the important values for this
example and the normalized transfer function are listed below. The transfer
function is shown with quadratics in factored form as shown by the (2)
superscript.
Ω
r
= 3.5
n
= 2.80 (3rd order)
ω
o
= 8485.3 rad/sec
BW
= 21,000 rad/sec
0
49417
⋅
0
99421
H
C
(
S
)
=
,
3
(
2
)
(
S
+
0
49417
)
⋅
(
S
+
0
24709
±
j
0
96600
)
After making the substitution of (3.39) and factoring again, the following
equation emerges, which can be simplified into the final result:
2
7
3
(
s
+
7
.
200
⋅
10
)
H
C
(
s
)
=
,
6
2
7
(
2
)
(
2
)
(
s
+
42
,
500
⋅
s
+
7
.
200
⋅
10
)
⋅
(
s
+
587
.
6
±
j
2
965
)
⋅
(
s
+
4
632
±
j
23
,
370
)
H
C
(
s
)
=
,
6
2
7
3
(
s
+
7
200
⋅
10
)
2
7
2
6
2
8
(
s
+
42
,
495
⋅
s
+
7
.
200
⋅
10
)(
s
+
1
175
⋅
s
+
9
134
⋅
10
)(
s
+
9
263
⋅
s
+
5
.
676
⋅
10
)
Example 3.8 Unnormalized Elliptic Bandstop Filter
Problem:
Determine the transfer function for an elliptic bandstop filter to
satisfy the following specifications:
a
pass
= −0.3 dB,
a
stop
= −50 dB,
f
pass1
= 50 Hz,
f
pass2
= 72 Hz,
f
stop1
= 58 Hz,
f
stop2
= 62 Hz
Solution:
Using the material of Section 2.5, the important values for this
example and the normalized transfer function are listed below. In this case,
f
pass2
