1. NATIONAL COLLEGE OF SCIENCE AND TECHNOLOGY
Amafel Building, Aguinaldo Highway Dasmariñas City, Cavite
Experiment No. 3
ACTIVE LOW-PASS and HIGH-PASS FILTERS
Maala, Michelle Anne C. July 14, 2011
Signal Spectra and Signal Processing/BSECE 41A1 Score:
Engr. Grace Ramones
Instructor

2. OBJECTIVES:
1. Plot the gain-frequency response and determine the cutoff frequency of a
second-order (two-pole) low-pass active filter.
2. Plot the gain-frequency response and determine the cutoff frequency of a
second-order (two-pole) high-pass active filter.
3. Determine the roll-off in dB per decade for a second-order (two-pole) filter.
4. Plot the phase-frequency response of a second-order (two-pole) filter.

3. VOLTAGE GAIN (MEASURED VALUE):
Step 3
AdB = 20 log A
4.006 = 20 log A
Step 14
AdB = 20 log A
3.776 = 20 log A
A = 1.54
VOLTAGE GAIN(CALCULATED):
Step 4
PERCENTAGE DIFFERENCE:
Q Step 4
Q Step 6
Q Step 15
CUTOFF FREQUENCY (CALCULATED):
Step 6 and 17
Step 11

4. DATA SHEET:
MATERIALS
One function generator
One dual-trace oscilloscope
One LM741 op-amp
Capacitors: two 0.001 µF, one 1 pF
Resistors: one 1kΩ, one 5.86 kΩ, two 10kΩ, two 30 kΩ
THEORY
In electronic communications systems, it is often necessary to separate a specific
range of frequencies from the total frequency spectrum. This is normally accomplished
with filters. A filter is a circuit that passes a specific range of frequencies while rejecting
other frequencies. Active filters use active devices such as op-amps combined with
passive elements. Active filters have several advantages over passive filters. The passive
elements provide frequency selectivity and the active devices provide voltage gain,
high input impedance, and low output impedance. The voltage gain reduces
attenuation of the signal by the filter, the high input prevents excessive loading of the
source, and the low output impedance prevents the filter from being affected by the
load. Active filters are also easy to adjust over a wide frequency range without altering
the desired response. The weakness of active filters is the upper-frequency limit due to
the limited open-loop bandwidth (funity) of op-amps. The filter cutoff frequency cannot
exceed the unity-gain frequency (funity) of the op-amp. Ideally, a high-pass filter should
pass all frequencies above the cutoff frequency (fc). Because op-amps have a limited
open-loop bandwidth (unity-gain frequency, funity), high-pass active filters have an
upper-frequency limit on the high-pass response, making it appear as a band-pass filter
with a very wide bandwidth. Therefore, active filters must be used in applications where
the unity-gain frequency (funity) of the op-amp is high enough so that it does not fall
within the frequency range of the application. For this reason, active filters are mostly
used in low-frequency applications.
The most common way to describe the frequency response characteristics of a
filter is to plot the filter voltage gain (Vo/Vin) in dB as a function of frequency (f). The
frequency at which the output power gain drops to 50% of the maximum value is called
the cutoff frequency (fc). When the output power gain drops to 50%, the voltage gain
drops 3 dB (0.707 of the maximum value). When the filter dB voltage gain is plotted as a
function of frequency using straight lines to approximate the actual frequency
response, it is called a Bode plot. A Bode plot is an ideal plot of filter frequency
response because it assumes that the voltage gain remains constant in the passband
until the cutoff frequency is reached, and then drops in a straight line. The filter network
voltage gain in dB is calculated from the actual voltage gain (A) using the equation
AdB = 20 log A

5. where A = Vo/Vin.
An ideal filter has an instantaneous roll-off at the cutoff frequency (fc), with full
signal level on one side of the cutoff frequency. Although the ideal is not achievable,
actual filters roll-off at -20 dB/decade or higher depending on the type of filter. The -20
dB/decade roll-off is obtained with a one-pole filter (one R-C circuit). A two-pole filter
has two R-C circuits tuned to the same cutoff frequency and rolls off at -40 dB/decade.
Each additional pole (R-C circuit) will cause the filter to roll off an additional -20
dB/decade. In a one-pole filter, the phase between the input and the output will
change by 90 degrees over the frequency range and be 45 degrees at the cutoff
frequency. In a two-pole filter, the phase will change by 180 degrees over the
frequency range and be 90 degrees at the cutoff frequency.
Three basic types of response characteristics that can be realized with most
active filters are Butterworth, Chebyshev, and Bessel, depending on the selection of
certain filter component values. The Butterworth filter provides a flat amplitude response
in the passband and a roll-off of -20 dB/decade/pole with a nonlinear phase response.
Because of the nonlinear phase response, a pulse waveshape applied to the input of a
Butterworth filter will have an overshoot on the output. Filters with a Butterworth
response are normally used in applications where all frequencies in the passband must
have the same gain. The Chebyshev filter provides a ripple amplitude response in the
passband and a roll-off better than -20 dB/decade/pole with a less linear phase
response than the Butterworth filter. Filters with a Chebyshev response are most useful
when a rapid roll-off is required. The Bessel filter provides a flat amplitude response in
the passband and a roll-off of less than -20 dB/decade/pole with a linear phase
response. Because of its linear phase response, the Bessel filter produces almost no
overshoot on the output with a pulse input. For this reason, filters with a Bessel response
are the most effective for filtering pulse waveforms without distorting the waveshape.
Because of its maximally flat response in the passband, the Butterworth filter is the most
widely used active filter.
A second-order (two-pole) active low-pass Butterworth filter is shown in Figure 3-
1. Because it is a two-pole (two R-C circuits) low-pass filter, the output will roll-off -40
dB/decade at frequencies above the cutoff frequency. A second-order (two-pole)
active high-pass Butterworth filter is shown in Figure 3-2. Because it is a two-pole (two R-
C circuits) high-pass filter, the output will roll-off -40 dB/decade at frequencies below
the cutoff frequency. These two-pole Sallen-Key Butterworth filters require a passband
voltage gain of 1.586 to produce the Butterworth response. Therefore,
and

6. At the cutoff frequency of both filters, the capacitive reactance of each
capacitor (C) is equal to the resistance of each resistor (R), causing the output voltage
to be 0.707 times the input voltage (-3 dB). The expected cutoff frequency (fc), based
on the circuit component values, can be calculated from
wherein,
FIGURE 3 – 1 Second-order (2-pole) Sallen-Key Low-Pass Butterworth Filter
FIGURE 3 – 2 Second-order (2-pole) Sallen-Key High-Pass Butterworth Filter

7. PROCEDURE
Low-Pass Active Filter
Step 1 Open circuit file FIG 3-1. Make sure that the following Bode plotter settings are
selected: Magnitude, Vertical (Log, F = 10dB, I = -40dB), Horizontal (Log, F = 100
kHz, I = 100 Hz).
Step 2 Run the simulation. Notice that the voltage gain has been plotted between the
frequencies of 100 Hz and 100 kHz by the Bode plotter. Draw the curve plot in
the space provided. Next, move the cursor to the flat part of the curve at a
frequency of approximately 100 Hz and measure the voltage gain in dB.
Record the dB gain on the curve plot.
AdB
f
dB gain = 4.006 dB
Question: Is the frequency response curve that of a low-pass filter? Explain why.
Yes it is a low-pass filter. This is because it allows all the frequencies
below the cutoff frequency and rejects other frequencies.
Step 3 Calculate the actual voltage gain (A) from the measured dB gain.
A = 1.586
Step 4 Based on the circuit component values in Figure 3-1, calculate the expected
voltage gain (A) on the flat part of the curve for the low-pass Butterworth filter.
A = 1.586
Question: How did the measured voltage gain in Step 3 compared with the calculated
voltage gain in Step 4?
Measured voltage gain in Step 3 and the calculated voltage gain in
Step 4 are the same

8. Step 5 Move the cursor as close as possible to a point on the curve that is 3dB down
from the dB gain at the low frequencies. Record the dB gain and the
frequency (cutoff frequency, fc) on the curve plot.
dB gain = 0.968 dB
fc = 5.321 kHz
Step 6 Calculate the expected cutoff frequency (fc) based on the circuit component
values.
fc = 5.305 kHz
Question: How did the calculated value for the cutoff frequency compare with the
measured value recorded on the curve plot for the two-pole low-pass active
filter?
The values are almost equal. There is a difference of 0.30%.
Step 7 Move the cursor to a point on the curve where the frequency is as close as
possible to ten times fc. Record the dB gain and frequency (fc) on the curve
plot.
dB gain = -36.146 dB
fc = 53.214 kHz
Questions: Approximately how much did the dB gain decrease for a one-decade increase
in frequency? Was this what you expected for a two-pole filter?
-37.106 dB/decade and is almost -40 dB per decade which is the ideal
for a 2-pole filter . It is a two-pole filter so I am expecting a 40 dB
decrease per decade increase and our result is almost equal to my
expectation
Step 8 Click Phase on the Bode plotter to plot the phase curve. Change the vertical
axis initial value (I) to 180 degrees and the final value (F) to 0 degree. Run the
simulation again. You are looking at the phase difference (θ) between the filter
input and output wave shapes as a function of frequency (f). Draw the curve
plot in the space provided.
θ
f

9. Step 9 Move the cursor as close as possible on the curve to the cutoff frequency (fc).
Record the frequency (fc) and phase (θ) on the curve.
fc = 5.321 kHz
θ = -90.941
Question: Was the phase shift between input and output at the cutoff frequency what
you expected for a two-pole low-pass filter?
Yes, at cutoff frequency the phase shift is expected to have a 90 degrees
Step 10 Click Magnitude on the plotter. Change R to 1 kΩ in both places and C to 1 pF
in both places. Adjust the horizontal final frequency (F) on the Bode plotter to 20 MHz. Run
the simulation. Measure the cutoff frequency (fc) and record your answer.
fc = 631.367 kHz
Step 11 Based on the new values for resistor R and capacitor C, calculate the new
cutoff frequency (fc).
fc = 159.1549 MHz
Question: Explain why there was such a large difference between the calculated and the
measured values of the cutoff frequency when R = 1kΩ and C = 1pF. Hint: The value of the
unity-gain bandwidth, funity, for the 741 op-amp is approximately 1 MHz.
The weakness of active filter is the upper-frequency limit that is why there
is a large difference between the two. The filter cutoff frequency must not
exceed the unity-gain frequency.
High-Pass Active Filter
Step 12 Open circuit file FIG 3-2. Make sure that the following Bode plotter settings are
selected: Magnitude, Vertical (Log, F = 10dB, I = -40dB), Horizontal (Log, F = 100
kHz, I = 100 Hz).

10. Step 13 Run the simulation. Notice that the voltage gain has been plotted between the
frequencies of 100 Hz and 100 kHz by the Bode plotter. Draw the curve plot in
the space provided. Next, move the cursor to the flat part of the curve at a
frequency of approximately 100 kHz and measure the voltage gain in dB.
Record the dB gain on the curve plot.
AdB
f
dB gain = 3.776 dB
Question: Is the frequency response curve that of a high-pass filter? Explain why.
Yes it is a high-pass filter. If the frequency is within the limit of funity it will
allow all the frequencies above the cutoff frequency and rejects
frequencies below the cut-off frequency.
Step 14 Calculate the actual voltage gain (A) from the measured dB gain.
A = 1.54
Step 15 Based on the circuit component values in Figure 3-2, calculate the expected
voltage gain (A) on the flat part of the curve for the high -pass Butterworth filter.
Av = 1.586
Question: How did the measured voltage gain in Step 14 compare with the calculated
voltage gain in Step 15?
The percentage difference is 2.98%. They are almost equal.
Step 16 Move the cursor as close as possible to a point on the curve that is 3dB down
from the dB gain at the high frequencies. Record the dB gain and the
frequency (cutoff frequency, fc) on the curve plot.
dB gain = 0.741 dB
fc = 5.156 kHz

11. Step 17 Calculate the expected cutoff frequency (fc) based on the circuit component
values.
fc = 5305.16477 Hz
Question: How did the calculated value of the cutoff frequency compare with the
measured value recorded on the curve plot for the two-pole low-pass active
filter?
Almost the same. The values have a percent difference of 2.89%.
Step 18 Move the cursor to a point on the curve where the frequency is as close as
possible to one-tenth fc. Record the dB gain and frequency (fc) on the curve
plot.
dB gain = -36.489 dB
fc = 515.619 Hz
Questions: Approximately how much did the dB gain decrease for a one-decade
decrease in frequency? Was this what you expected for a two-pole filter?
-37.23. The roll-off is -40 dB. This what I am expecting for a two-pole
filter.
Step 19 Change the horizontal axis final setting (F) to 50 MHz on the Bode plotter. Run
the simulation. Draw the curve plot in the space provided.
AdB
f
Step 20 Measure the upper cutoff frequency (fc2) and record the value on the curve
plot.
fC2 = 92.595 kHz
Question: Explain why the filter frequency response looked like a band-pass response
when frequencies above 1 MHz were plotted. Hint: The value of the unity-gain
bandwidth, funity, for the 741 op-amp is approximately 1 MHz
The curve response is like a band-pass response because active filters
have an upper frequency limit on the high-pass response.

12. CONCLUSION:
I conclude that the gain-frequency response of a second-order low-pass active
appears as a passive filter and high-pass active filter is like high-pass passive filter. However,
they differ in roll-off and its phase frequency response.
Second-order low-pass filter filters has 40dB gain decrease for a one-decade increase
in frequency while Second-order high-pass filter filters has 40dB gain decrease for a one-
decade decrease in frequency. Two-pole filter response on cutoff frequency is 90 degrees
while one-pole has only 45 degrees.
Active filters have lot of advantages, however, active filters has an upper-frequency
limit because of the open-loop bandwidth of op-amps. That is why active filters appears as
band-pass filter at the higher frequencies.