Digital Signal Processing Reference
In-Depth Information
a
pass
= −0.5 dB,
a
stop
= −30 dB,
f
pass
= 1000 Hz, and
f
stop
= 400 Hz
Solution:
A fourth-order transfer function as shown below is required. Since
this is an even-order Chebyshev, there is an adjustment factor of 0.94406 included
in the numerator.
4
.
94406
⋅
s
H
a
(
s
)
=
2
6
2
6
(
s
+
2
,
071
.
9
⋅
s
+
37
.
121
⋅
10
)
⋅
(
s
+
14
,
926
⋅
s
+
110
.
77
⋅
10
)
The two quadratic terms can be matched to two active filter stages by again
picking
C
= 0.01 µF and
R
A
= 10 kΩ. The other circuit values can be calculated as
shown below:
m
R
m
K
m
R
Bm
0
16.41 kΩ
2.6599
16.6 kΩ
1
9.502 kΩ
1.5818
5.82 kΩ
In order to achieve a gain of exactly 1 at ω = ∞, we can use a voltage divider
at the output. First, we determine the gain adjustment factor for the filter that in
this case includes not only the
K
m
factors for each quadratic, but also the
approximation function's gain of 0.94406:
=
m
GA
(
K
)
/
0
.
94406
=
4
4567
m
Figure 4.7
Chebyshev highpass active filter for Example 4.2.
Assuming an equivalent resistance of 10 kΩ, the resistor divider values can be
determined with the resulting filter shown in Figure 4.7. (The gain adjustment
could be factored into the overall gain of the system instead of using the voltage
divider.)
R
=
44 57
.
k,
Ω
R
=
12 89
.
k
Ω
x
y
