Digital Signal Processing Reference
In-Depth Information
Solution:
Using the material of Section 2.5, the important values for this
example and the normalized transfer function are listed below:
Ω
r
= 1.5
n
= 4.61 (5th order)
ω
o
= 3,000.0 rad/sec
2
2
0
.
04507
⋅
0
4260
⋅
(
S
+
5
438
)
⋅
(
S
+
2
.
426
)
H
E
(
S
)
=
,
5
2
2
(
S
+
0
.
4260
)
⋅
(
S
+
0
.
5702
⋅
S
+
0
5760
)
⋅
(
S
+
0
.
1635
⋅
S
+
1
.
032
)
The unnormalized transfer function is then determined by making the
substitution
S
= ω
o
/
s
and using the relationships just developed:
2
6
2
6
s
⋅
(
s
+
1
.
655
⋅
10
)
⋅
(
s
+
3
.
711
⋅
10
)
H
E
(
s
)
=
,
5
2
7
2
6
(
s
+
7
,
043
)
⋅
(
s
+
2
970
⋅
s
+
1
.
562
⋅
10
)
⋅
(
s
+
475
.
⋅
s
+
8
.
722
⋅
10
)
Example 3.4 Unnormalized Chebyshev Highpass Filter
Problem:
Determine the transfer function for a Chebyshev highpass filter to
satisfy the following specifications:
a
pass
= −1.5 dB,
a
stop
= −40 dB,
f
pass
= 2,000 Hz,
f
stop
= 800 Hz
Solution:
Using the material of Section 2.3, the important values for this
example and the normalized transfer function are listed below:
Ω
r
= 2.5
n
= 3.66 (4th order)
ω
o
= 12,566.4 rad/sec
0
8414
⋅
0
9505
⋅
0
.
2434
H
C
(
S
)
=
,
4
2
2
(
S
+
0
.
2383
⋅
S
+
0
9505
)
⋅
(
S
+
0
5752
⋅
S
+
0
2434
)
The unnormalized transfer function is then determined by making the
substitution
S
= ω
o
/
s
and using the relationships just developed:
2
2
0
.
8414
⋅
s
⋅
s
H
C
(
s
)
=
,
4
2
8
2
8
(
s
+
3
150
⋅
s
+
1
.
661
⋅
10
)
⋅
(
s
+
29
,
703
⋅
s
+
6
.
489
⋅
10
)
The results of using WFilter to design the filter of Example 3.4 are illustrated
in Figures 3.4 and 3.5, which show the coefficients and magnitude response.
