1. NATIONAL COLLEGE OF SCIENCE AND TECHNOLOGY
Amafel Building, Aguinaldo Highway Dasmariñas City, Cavite
Experiment No. 4
ACTIVE BAND-PASS and BAND-STOP FILTERS
Maala, Michelle Anne C. July 19, 2011
Signal Spectra and Signal Processing/BSECE 41A1 Score:
Engr. Grace Ramones
Instructor

2. OBJECTIVES
1. Plot the gain-frequency response curve and determine the center frequency for an
active band-pass filter.
2. Determine the quality factor (Q) and bandwidth of an active band-pass filter
3. Plot the phase shift between the input and output for a two-pole active band-pass
filter.
4. Plot the gain-frequency response curve and determine the center frequency for an
active band-stop (notch) filter.
5. Determine the quality factor (Q) and bandwidth of an active notch filter.

3. DATA SHEET
MATERIALS
One function generator
One dual-trace oscilloscope
Two LM741 op-amps
Capacitors: two 0.001 µF, two 0.05 µF, one 0.1 µF
Resistors: one 1 kΩ, two 10 kΩ, one 13 kΩ, one 27 kΩ, two 54 kΩ, and one 100kΩ
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 impedance 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. 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.
A band-pass filter passes all frequencies lying within a band of frequencies and rejects all
other frequencies outside the band. The low cut-off frequency (fC1) and the high-cutoff
frequency (fC2) on the gain-frequency plot are the frequencies where the voltage gain
has dropped by 3 dB (0.707) from the maximum dB gain. A band-stop filter rejects a band
of frequencies and passes all other frequencies outside the band, and of then referred to
as a band-reject or notch filter. The low-cutoff frequency (fC1) and high-cutoff frequency
(fC2) on the gain frequency plot are the frequencies where the voltage gain has dropped
by 3 dB (0.707) from the passband dB gain.
The bandwidth (BW) of a band-pass or band-stop filter is the difference between the high-
cutoff frequency and the low-cutoff frequency. Therefore,
BW = fC2 – fC1
The center frequency (fo)of the band-pass or a band-stop filter is the geometric mean of
the low-cutoff frequency (fC1) and the high-cutoff frequency (fC2). Therefore,

4. The quality factor (Q) of a band-pass or a band-stop filter is the ratio of the center
frequency (fO) and the bandwidth (BW), and is an indication of the selectivity of the filter.
Therefore,
A higher value of Q means a narrower bandwidth and a more selective filter. A filter with a
Q less than one is considered to be a wide-band filter and a filter with a Q greater than
ten is considered to be a narrow-band filter.
One way to implement a band-pass filter is to cascade a low-pass and a high-pass filter.
As long as the cutoff frequencies are sufficiently separated, the low-pass filter cutoff
frequency will determine the low-cutoff frequency of the band-pass filter and a high-pass
filter cutoff frequency will determine the high-cutoff frequency of the band-pass filter.
Normally this arrangement is used for a wide-band filter (Q 1) because the cutoff
frequencies need to be sufficient separated.
A multiple-feedback active band-pass filter is shown in Figure 4-1. Components R1 and C1
determine the low-cutoff frequency, and R2 and C2 determine the high-cutoff frequency.
The center frequency (fo) can be calculated from the component values using the
equation
Where C = C1 = C2. The voltage gain (AV) at the center frequency is calculated from
and the quality factor (Q) is calculated from
Figure 4-1 Multiple-Feedback Band-Pass Filter
XBP1
XFG1
IN OUT
10nF
C1
100kΩ
R2
741
3
Vo
6
Vin 1kΩ 2 10kΩ
10nF
R1 RL
C2

5. Figure 4-2 shows a second-order (two-pole) Sallen-Key notch filter. The expected center
frequency (fO) can be calculated from
At this frequency (fo), the feedback signal returns with the correct amplitude and phase to
attenuate the input. This causes the output to be attenuated at the center frequency.
The notch filter in Figure 4-2 has a passband voltage gain
and a quality factor
The voltage gain of a Sallen-Key notch filter must be less than 2 and the circuit Q must be
less than 10 to avoid oscillation.
Figure 4-2 Two pole Sallen-Key Notch Filter
XBP1
XFG1 IN OUT
27kΩ
27kΩ
R52
R/2
50nF 50nF
0.05µF 3
0.05µF
C3 C
Vin C C 6
2 741 Vo
RL
54kΩ 54kΩ 10kΩ
54kΩ 54kΩ
R
R3 0
R R
R2
100nF
2C R1 10kΩ
13kΩ
0
0

6. PROCEDURE
Active Band-Pass Filter
Step 1 Open circuit file FIG 4-1. Make sure that the following Bode plotter settings
are selected. Magnitude, Vertical (Log, F = 40 dB, I = 10 dB), Horizontal (Log,
F = 10 kHz, I = 100 Hz)
Step 2 Run the simulation. Notice that the voltage gain has been plotted between
the frequencies of 100 Hz and 10 kHz. Draw the curve plot in the space
provided. Next, move the cursor to the center of the curve. Measure the
center frequency (fo) and the voltage gain in dB. Record the dB gain and
center frequency (fo) on the curve plot.
fo = 1.572 kHz
AdB = 33.906 dB
Question: Is the frequency response curve that of a band-pass filters? Explain why.
 Yes, because the filter only allows the frequencies from 100.219 Hz to 10
kHz and block the other frequency.
Step 3 Based on the dB voltage gain at the center frequency, calculate the actual
voltage gain (AV)
 AV = 49.58
Step 4 Based on the circuit component values, calculate the expected voltage
gain (AV) at the center frequency (fo)
 AV = 50
Question: How did the measured voltage gain at the center frequency compare with
the voltage gain calculated from the circuit values?
 The percentage difference of the measured and calculated value is
0.84%

7. Step 5 Move the cursor as close as possible to a point on the left of the curve that is
3 dB down from the dB gain at the center frequency (fo). Record the
frequency (low-cutoff frequency, fC1) on the curve plot. Next, move the
cursor as close as possible to a point on the right side of the curve that is 3 dB
down from the center frequency (fo). Record the frequency (high-cutoff
frequency, fC2) on the curve plot.
 fC1 = 1.415 kHz
 fC2 = 1.746 kHz
Step 6 Based on the measured values of fC1 and fC2, calculate the bandwidth (BW)
of the band-pass filter.
 BW = 0.331 kHz
Step 7 Based on the circuit component values, calculate the expected center
frequency (fo)
 fo = 1.592 kHz
Question: How did the calculated value of the center frequency compare with the
measured value?
 They have a small difference. The calculated and measured center
frequency has a difference of 1.27%.
Step 8 Based on the measured center frequency (fo) and the bandwidth (BW),
calculate the quality factor (Q) of the band-pass filter.
 Q = 4.75
Step 9 Based on the component values, calculate the expected quality factor (Q)
of the band-pass filter.
 Q=5
Question: How did your calculated value of Q based on the component values
compare with the value of Q determined from the measured fo and BW?
 They have a difference of 0.25. The percentage difference of the two is
only 5.26%

8. Step 10 Click Phase on the Bode plotter to plot the phase curve. Change the vertical
initial value (I) to -270o and the final value (F) to +270o. 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.
Step 11 Move the cursor as close as possible to the curve center frequency (f o),
recorded on the curve plot in Step 2. Record the frequency (fo) and the
phase (θ) on the phase curve plot.
 fo = 1.572 kHz
 θ = 50.146o
Question: What does this result tell you about the relationship between the filter output
and input at the center frequency?
 It tells that the output is 173.987o out of phase with input in the center
frequency.
Active Band-Pass (Notch) Filter
Step 12 Open circuit file FIG 4-2. Make sure that the following Bode plotter settings
are selected. Magnitude, Vertical (Log, F = 10 dB, I = -20 dB), Horizontal (Log,
F = 500 Hz, I = 2 Hz)

9. Step 13 Run the simulation. Notice that the voltage gain has been plotted between
the frequencies of 2 Hz and 500 Hz. Draw the curve plot in the space
provided. Next, move the cursor to the center of the curve at its center
point. Measure the center frequency (fo) and record it on the curve plot.
Next, move the cursor to the flat part of the curve in the passband. Measure
the voltage gain in dB and record the dB gain on the curve plot.
fo = 58.649 Hz
AdB = 4. dB
Question: Is the frequency response curve that of a band-pass filters? Explain why.
 Yes, at the bandpass the frequencies are blocked. It only allow the
frequencies outside the band.
Step 14 Based on the dB voltage gain at the center frequency, calculate the actual
voltage gain (AV)
 AV = 1.77
Step 15 Based on the circuit component values, calculate the expected voltage
gain in the passband.
 AV = 1.77
Question: How did the measured voltage gain in the passband compare with the
voltage gain calculated from the circuit values?
 They are the same. There is a 0% difference.
Step 16 Move the cursor as close as possible to a point on the left of the curve that is
3 dB down from the dB gain in the bandpass Record the frequency (low-
cutoff frequency, fC1) on the curve plot. Next, move the cursor as close as
possible to a point on the right side of the curve that is 3 dB down from dB
gain in the passband. Record the frequency (high-cutoff frequency, fC2) on
the curve plot.
 fC1 = 46.743 Hz

10.  fC2 = 73.588 Hz
Step 17 Based on the measured values of fC1 and fC2, calculate the bandwidth (BW)
of the notch filter.
 BW = 26.845 Hz
Step 18 Based on the circuit component values, calculate the expected center
frequency (fo)
 fo = 58.95Hz
Question How did the calculated value of the center frequency compare with the
measured value?
 There is 0.51% difference between the values of the calculated and
measured center frequency.
Step 19 Based on the measured center frequency (fo) and bandwidth (BW) ,
calculate the quality factor (Q) of the notch filter.
 Q = 2.18
Step 20 Based on the calculated passband voltage gain (Av), calculate the
expected quality factor (Q) of the notch filter.
 Q = 2.17
Question: How did your calculated value of Q based on the passband voltage gain
compare with the value of Q determined from the measured fo and BW?
 There is 0.46% difference between the calculated and measure quality
factor. Their difference is 0.01

11. CONCLUSION
After conducting the experiment, I conclude the response of active filter is still the
same with the mountain-like response of a passive band-pass filter having the center
frequency at the peak dB gain. On the contrary, the center frequency of a band-stop is at
the lowest dB gain. Like in the passive filter, active band-pass allows the frequency inside
the band and active band-stop allows the frequencies outside the band range.
The quality factor (Q) and bandwidth of an active filter are inversely proportion. The
higher the value of Q the narrower the response becomes. The center frequency is the
geometric mean of the cutoff frequencies. And also, the bandwidth of the response curve
is the difference between the high and the low cutoff frequencies. The response of a two-
pole active band-pass filter’s output is 180O out of phase with the input.

12. SAMPLE COMPUTATION
Step 3
Step 4
Step 7
Step 7 Q
Step 8
Step 9
Step 9 Q
Step 14

13. Step 15
Step 18
Step 18 Q
Step 19 (
Step 20