The transfer function of the fourth order lowpass filter is given by
Section 1
a1 = 16978.1110, a0 = 8.44283×107
? 0 = a 0= 9188.4874 rad/s
The Q value is given by
Q = ? a1
0= 0.541196
The normalized component values are given by
C1 = 1 F, C2 = 1 F, R2 = 1 ?, RB = 1 ? , RA = 2 – 1/Q = 0.152241 ?
Ra = 3 – 1/Q = 1.152241 ?, Rb = (3Q-1)/(2Q-1) = 7.5685 ?
Application of the magnitude scaling (km = 1000) and frequency scaling (kf = ?0 = 9253.1355) yields
C1 = 1F
k m k× f
= 0.1088 ?F
C2 = 1F
k m k× = 0.1088 ?F
f
R2 = 1×1000 ? = 1 k?
RB = 1×1000 ? = 1 k?
RA = 0.152241 ×1000 ? = 152.240915 ?
Ra = 1.152241 ×1000 ? = 1.152241 k?
Rb = 7.5685 ×1000 ? = 7.5685 k?
Section 2
a1 = 7032.5639, a0 = 8.44283×107
? 0 = a 0= 9188.4874 rad/s
The Q value is given by
Q = ? a1
0= 1.306563
The normalized component values are given by
C1 = 1 F, C2 = 1 F, R2 = 1 ?, RB = 1 ? , RA = 2 – 1/Q = 1.2346 ?
Ra = 3 – 1/Q = 2.2346 ?, Rb = (3Q-1)/(2Q-1) = 1.80996 ?
Application of the magnitude scaling (km = 1000) and frequency scaling (kf = ?0 = 9253.1355) yields
C1 = 1F
k m k× f
= 0.1088 ?F
C2 = 1F
k m k× = 0.1088 ?F
f
R2 = 1×1000 ? = 1 k?
RB = 1×1000 ? = 1 k?
RA = 1.2346 ×1000 ? = 1.2346 k?
Ra = 2.2346 ×1000 ? = 2.2346 k?
Rb = 1.80996 ×1000 ? = 1.80996 k?
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