1. NATIONAL COLLEGE OF SCIENCE AND TECHNOLOGY
Amafel Building, Aguinaldo Highway Dasmariñas City, Cavite
EXPERIMENT NO. 4
ACTIVE BAND-PASS AND BAND-STOP FILTERS
Pula, Rolando A. July 21, 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,
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 4-2 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 4-1 Multiple-Feedback Band-Pass Filter

5. XBP1
XFG1
IN OUT
10nF
C1
100kΩ
R2
741
3
Vo
6
Vin 1kΩ 2 10kΩ
10nF
R1 RL
C2
Figure 4-2 Two pole Sallen-Key Notch Filter
XBP1
XFG1 IN OUT
27kΩ
27kΩ R/2
R52
50nF 50nF 3
0.05µF
C3 0.05µF
C
Vin C C 6
2 741 Vo
RL
54kΩ 54kΩ 10kΩ
54kΩ
R3 54kΩ
R
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)
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

6. AdB
F (Hz)
Question: Is the frequency response curve that of a band-pass filters? Explain why.
= Yes, the bode plotter only shows the signal of the passband that is from 100.219
Hz to 10 kHz.
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 difference is 0.42 or 0.84% percentage difference
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

7. 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?
= The difference is 0.02 kHz or 1.27% percentage difference.
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?
= There is a difference of 0.25 or 5.26% difference.
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.
θ
f (Hz)
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 = 173.987o

8. Question: What does this result tell you about the relationship between the filter output and
input at the center frequency?
= This only shows that the output is 173.987o out of phase with input at 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)
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
AdB
f (Hz)
Question: Is the frequency response curve that of a band-pass filters? Explain why.
= Yes, the response is a band-pass because the response curve in the bode plotter
block the frequency at the passband and allows 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

9. 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?
= The calculated voltage gain is exactly the same as the measured voltage gain.
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?
= They have 0.51% difference. They are almost the same
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 value and the measured value has a difference of 0.01 or
percentage difference of 0.46%

10. CONCLUSION
After performing the experiment, I can able to say that the active filter response is alike
with the passive filter response. It band-pass only allows the frequencies lying within the band and
blocks other frequency. As for the band-stop, the frequencies lying in the band are blocked while
passing the frequencies outside the band.
The center frequency of the band-pass is at the peak gain, while the center frequency of the
band-stop is at the lowest dB gain. In a two-pole circuit, the phase response of the output 180o out
of phase with the input. The quality factor determines the selectivity of the response curve. The
higher the quality factor, the more selective the graph is. It is inversely proportional to the
bandwidth of the curve. Bandwidth is the difference of the high and the low cutoff frequencies.

11. SAMPLE COMPUTATION
(Actual Voltage Gain)
Step 3
Step 14
(Expected voltage gain)
Step 4
Step 15 (Expected Voltage Gain)
(Percentage Difference)
Step 4 Question

12. Step 7 Question
Step 9 Question
(Center frequency)
Step 7
Step 18
(Measured quality factor)
Step 8
Step 19
(Expected quality factor)
Step 9
Step 20