This document describes an experiment to characterize active band-pass and band-stop filters. The objectives are to determine the gain-frequency response, center frequency, bandwidth, quality factor, and phase response. For the band-pass filter, the measured and calculated results for center frequency, gain, bandwidth, and quality factor agree to within 5%. For the band-stop filter, the measured and calculated results for center frequency, gain, bandwidth, and quality factor agree to within 1%. The phase response of the band-pass filter shows the output is approximately 180 degrees out of phase with the input at the center frequency.

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

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

4. 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 3
0.05µF
C3 0.05µF
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
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)

6. 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.
 It is a frequency response of a band-pass filter because the filter only let the
frequencies from 100.219 Hz to 10 kHz to pass 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?
 They have a difference of 0.42. The percentage difference of the measured and
calculated value is 0.84%
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

7. 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 difference of 0.02 kHz. The calculated and measured center frequencies
have 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?
 The percentage difference of the two is only 5.26% The difference is only 0.25.
Therefore, they are almost equal.
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 (fo), 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

8. Question: What does this result tell you about the relationship between the filter output and input at
the center frequency?
 The phase result tells that the filters output is 173.987o (approximately 180o) out
of phase with input.
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)
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, because the center frequency Is located at the lowest gain. Moreover, it blocks the
frequencies lying in 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.

9. 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
 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?
 The percentage difference of the calculated and measured center frequency is 0.51%.
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?
 The calculated and measure quality factor have 0.46% difference.