You know how to evaluate the first-order active low-pass filter of Figure 4 (Lab 12)
because it is essentially identical to the circuit of HW12, problem 12-4.
R1 = 2 4.7kΩ resistors in series
R2 = 47kΩ
R3 = 1kΩ
C2 = 0.0047µf
We can evaluate the transfer function for this circuit by simply using the formula for the
gain of an OP AMP amplifier and recognizing that R2-C are in parallel and form a
1
feedback impedance Z 2 = R2 ||
jω C2
1
47000
jω 0.0047 × 10 −6
1
1
+ 47000
|| R2
jω 0.0047 × 10 −6
Z2
jω C2
TV ( jω ) = −
=−
=−
Z1
R1
4700 + 4700
(
)
(
)
1
jω 0.0047 × 10 −6
47000
1
TV ( jω ) = −
= −5
1
9400
1 + jω 0.0047 × 10 −6 47000
+ 47000
−6
jω 0.0047 × 10
(
)
(
TV ( jω ) =
−5
1 + jω 2.209 × 10 −4
(
)
)
(
)
Computing the gain in decibels gives
(
20 log10 TV ( jω ) = 20 log10 −5 − 20 log10 1 + jω 2.209 × 10 −4
)
By inspection this is a low pass filter with a passband gain of –5 since TV ( 0 ) = −5 and
lim TV ( jω ) = 0 . This can also be evaluated as ω → ∞ to give TV ( j∞ ) = 0 . The cutoff
ω →∞
(
)
1
= 4526.94 radians/sec.
2.209 × 10 −4
4526.94
This can be converted to a linear frequency fC =
= 720.5 Hz.
2π
frequency is given by ω C 2.209 × 10 −4 = 1 or ω C =
We can plot this filter response as shown below.
+20
+10
gain
(db)
0
-10
-20
1
10
100
1000
f (Hz)
104
105
The second order filter is more complicated to analyze because there is no simple
impedance for Z1 and we need to analyze the circuit from basic principles.
R1 = 4.7kΩ
R2 = 47kΩ
R3 = 1kΩ
C1= 0.1µf
C2 = 0.0047µf
The voltage at the non-inverting input of the OP AMP is zero because there is no current
into the OP AMP and consequently no voltage drop across R3 to ground. This requires
the voltage at the inverting input to be zero. We then proceed to apply KCL at the
inverting input. To do this we need to determine the current entering the node through
the 4.7kΩ resistor.
We first compute the impedance Z EQ as seen by V1 ( jω ) :
⎛ 1 ⎞
R1 ⎜
⎝ jω C1 ⎟⎠
1
R1
jω C1 R12 + R1 + R1
Z EQ = R1 + R1 ||
= R1 +
= R1 +
=
1
jω C1
jω C1 R1 + 1
jω C1 R1 + 1
R1 +
jω C1
Rearranging gives
1 + jω 1 2 R1C1
Z EQ = 2R1
1 + jω R1C1
(
)
The current iTOT is given by
V ( jω ) V1 ( jω )
V1 ( jω )
V ( jω ) 1 + jω R1C1
iTOT = 1
=
=
= 1
Z EQ
Z EQ
2R1 1 + jω 1 R1C1
1 + jω 1 2 R1C1
2
2R1
1 + jω R1C1
(
)
The current i1 entering the node at the inverting input of the OP AMP is can be
determined using a current divider as
1
V ( jω )
V1 ( jω ) 1 + jω R1C1
1
1
jω C1
i1 =
i =
= 1
1 TOT 1 + jω R1C1 2R1 1 + jω 1 R C
2R1 1 + jω 1 R1C1
R1 +
2 1 1
2
jω C1
The current i1 entering the node at the inverting input of the OP AMP must be equal to the
current through the feedback resistor, i.e.,
V1 ( jω )
0 − V2 ( jω )
−V2 ( jω )
1
1 + jω R2C2
= if =
=
= −V2 ( jω )
R2
2R1 1 + jω 1 R1C1
⎛ 1 ⎞
R2
2
R2 ⎜
⎟
1
+
j
ω
R
C
2 2
⎝ jω C2 ⎠
1
R2 +
jω C2
Rearranging this gives
R
− 2
V ( jω )
2R1
TV ( jω ) = 2
=
V1 ( jω ) 1 + jω 1 R1C1 (1 + jω R2C2 )
2
i1 =
(
We substitute numbers to get
)
−
47000
2 ( 4700 )
TV ( jω ) =
V2 ( jω )
=
V1 ( jω ) 1 + jω 1 [ 4700 ] ⎡ 0.1 × 10 −6 ⎤ 1 + jω [ 47000 ] ⎡ 0.0047 × 10 −6 ⎤
⎣
⎦
⎣
⎦
2
TV ( jω ) =
V2 ( jω )
−5
=
V1 ( jω ) 1 + jω ⎡⎣ 2.35 × 10 −4 ⎤⎦ 1 + jω ⎡⎣ 2.209 × 10 −4 ⎤⎦
(
)(
(
)(
)
)
Computing the gain in decibels gives
20 log10 TV ( jω ) = 20 log10 −5 − 20 log10 1 + jω 2.35 × 10 −4 − 20 log10 1 + jω 2.209 × 10 −4
(
(
)
)
(
(
20 log10 TV ( jω ) = 13.97dB − 20 log10 1 + jω 2.35 × 10 −4 − 20 log10 1 + jω 2.209 × 10 −4
)
This is a low pass filter with a passband gain of –5 since TV ( 0 ) = −20 and
lim TV ( jω ) = 0 . However, the slope relationship is more complicated since there are
ω →∞
two frequencies of interest: f1 =
f2 =
1
= 677.25 Hz and
2π 2.35 × 10 −4
(
)
1
= 720.48 Hz.
2π 2.209 × 10 −4
(
)
What this means is that we plot a horizontal line of 14 db until we reach 677.2 Hz. At
this frequency TV ( jω ) starts decreasing at -20db/decade. This continues until we reach
720.5 Hz when we have a second term which decreases at another –20db/decade giving
an overall decrease of –40 db/decade. This is plotted below.
)
The basic difference between these two very similar first and second order filter circuits
is that the gain decreases much faster with frequency beginning at about 720 Hz. The
change in slope from –20db/decade between 677 and 720 Hz to –40db/decade for
frequencies above 720 Hz is hard to see on the above plot and very difficult to see in
measured data.