Filter Circuits
RC High-Pass Filter
RC Low-Pass Filter
Homework
RC High-Pass Filter
C
v out
R
v in
'
(
*
+
)
&
$
"
$
"
$
"
$
"
%
%
#
!
#
%
#
%
#
!
(
(
*
+
)
!
*
+
)
'
,
RC High-Pass Filter (cont’d)
It is common practice to express the voltage gain (or attenuation) in decibels (dB) defined as
/
3
21
0
4
5
.-
89:
;<
.-
.-
6
;<
0.5
0.707
1.414
2.0
4.0
6 6
-40
-20
0
+20
+40
;<
89:
67
6
89:
6 ;<
89:
67
6
0.01
0.1
1.0
10.0
100.0
-6
-3
+3
+6
+12
RC High-Pass Filter (cont’d)
1.0, so we can write the high-frequency approximation
@
89:
=
>
=
;<
@
C
B
F
E
D
6?
6
,
Note that as
A
I
89:
6 J
1
;<
G
G
6
of the gain as
H
I
N
M
L
;<
89:
6K
6
Also note that the low frequency approximation of the gain is
89:
6 If we make a log-log graph of
versus , we see that the actual gain can be approximated quite
and
with a sharp break or "knee" given
well by two straight lines corresponding to
by
6
;<
;<
89:
6K
6
89:
6 G
M
G
;<
L
6
H
M
G
G
L
H
.
.
At this break point, the magnitude of the gain is
.
O
1
O
J1
0
+
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)
A
.
(
,
'
R
QP
.-
RC High-Pass Filter (cont’d)
20 log
v out
v in
0 dB
log ω
ωB
−3 dB
High frequency approximation
Gain
Low frequency approximation (6 dB/octave or 20 dB/decade slope)
v out
vin
1.0
0.707
ωB
ω
RC High-Pass Filter (cont’d)
j
I
V C = IX C =
ωC
φ
V out = IR
Real
90
o
φ
V in
45
o
ωB
ω
RC Low-Pass Filter
R
YZ
ST
WX
UV
&
$
"
$
"
"
$
"
$
%
#
%
#
%
#
%
#
!
!
(
*+)
'
,
>
=
=
0, and goes to zero as
Note that this goes to unity as
v out
C
v in
RC Low-Pass Filter (cont’d)
20 log
v out
v in
Low frequency approximation
ωB
0 dB
−3 dB
log ω
High frequency approximation (6 dB/octave slope)
Gain
v out
vin
1.0
0.707
ωB
ω
RC Low-Pass Filter (cont’d)
φ
j
I
V out = IX C =
ωC
V R = IR
Real
o
φ
−45
V in
o
−90
ωB
ω
Homework Set 18 - Due Wed. Feb. 25
Design a high-pass RC filter with a breakpoint at 100 kHz. Use a 1-k resistor. Explain in words
why the high-pass filter attenuates the low frequencies.
[
1.
Design a low-pass RC filter that will attenuate a 60-Hz sinusoidal voltage by 12 dB relative to
the dc gain. Use a 100- resistor. Explain in words why the low-pass filter attenuates the high
frequencies.
3. For a low-pass RC filter prove that at the frequency
= 1/RC the voltage gain equals 0.707.
2.
[