Biomedical Engineering Reference
In-Depth Information
Substituting
V
L
into the previous equation yields
0
@
1
A
V
0
¼
R
b
H
R
a
H
j
o
1
R
a
L
C
L
1
V
s
1
R
a
H
C
H
1
R
b
L
C
L
j
o
þ
j
o
þ
The form of the solution is simply the product of each filter. The magnitude of the filter is
1
R
a
L
C
L
¼
R
b
H
R
a
H
V
0
V
s
o
s
s
2
2
1
R
a
H
C
H
1
R
b
L
C
L
o
2
þ
o
2
þ
Since there are two cutoff frequencies, two equations evolve:
1
R
a
H
C
H
¼
100
rad
s
o
c
H
¼
and
1
R
b
L
C
L
¼
500
rad
s
o
c
L
¼
5
100
rad
s
p
At either cutoff frequency, the magnitude is
, such that at o
c
H
¼
1
R
a
L
C
L
p ¼
R
b
H
5
o
c
H
t
t
0
@
1
A
0
@
1
A
R
a
H
2
2
1
R
a
H
C
H
1
R
b
L
C
L
o
2
c
H
þ
o
2
C
H
þ
1
R
a
L
C
L
100
2
¼
R
b
H
R
a
H
100
100
2
p
p
100
2
500
2
þ
þ
Therefore,
2
p
¼
R
b
H
R
a
H
R
a
L
C
L
The other cutoff frequency gives the same result as the previous equation. There are now
three equations
1
500
!
, and six unknowns. For
500, and 500
p
1
R
b
L
C
L
¼
R
b
H
R
a
H
R
a
L
C
L
R
a
H
C
H
¼
100,
26
¼
1
500
convenience,
set
R
b
L
¼
100 k
O
and
R
a
H
¼
100 k
O
, which gives
C
L
¼
R
b
L
¼
20 nF and
F. Now from 500
p
1
100
¼
R
b
H
R
a
H
1
R
a
L
C
L
C
H
¼
R
a
H
¼
0
:
1
m
26
,
500
p
R
b
H
R
a
L
¼
26
C
L
R
a
H
¼
5
:
099
Once again, one can specify one of the resistors—say,
R
a
L
¼
10 k
O
—giving
R
b
H
¼
50
:
.
Continued
099 k
O
