Digital Signal Processing Reference
In-Depth Information
∏
K
tot
=
K
m
=
22 943
.
m
If the desired output resistance is 10 kΩ, the following values result for the
voltage divider with the final filter shown in Figure 4.13.
R
=
229 4
.
k
Ω
,
R
=
10 46
.
k
Ω
x
y
Figure 4.13
Inverse Chebyshev active lowpass filter for Example 4.5.
Example 4.6 Elliptic Bandpass Active Filter Design
Problem:
Determine the resistor and capacitor values to implement an elliptic
bandpass active filter to meet the following specifications:
a
pass
= −1 dB,
a
stop
= −60 dB,
f
pass1
= 250 Hz,
f
pass2
= 400 Hz,
f
stop1
= 100 Hz, and
f
stop2
= 1,000 Hz
Solution:
The order of the equivalent lowpass filter is 3, which indicates that
a sixth-order bandpass approximation function will be necessary. The transfer
function is shown below:
−
3
2
6
2
3
20
.
82
⋅
10
⋅
469
.
⋅
(
s
+
50
.
07
⋅
10
)
⋅
(
s
+
311
.
3
⋅
10
)
H
a
(
s
)
=
2
6
2
6
2
6
(
s
+
469
.
8
⋅
s
+
3
.
948
⋅
10
)(
s
+
178
.
5
s
+
2
.
503
⋅
10
)(
s
+
281
.
s
+
6
.
227
⋅
10
)
These three quadratic terms must be matched to three active filter stages. The
stage related to the first-order factor in the
normalized
LP filter is implemented by
the standard Sallen-Key bandpass filter. The other two stages are implemented
using the twin-tee filters. By picking
C
= 0.01 µF and
R
A
= 10 kΩ, the remaining
values can be calculated as shown below. One of the twin-tee stages will have a
R
o
= 18.86 kΩ and the other will have
C
o
= 95.04 nF.
