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 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 4-1 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
0.05µF
50nF
0.05µF 3
C3 C
Vin C C 6
2 741 Vo
RL
54kΩ 54kΩ 10kΩ
54kΩ R
R3 54kΩ 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
AdB
40dB
10 dB F (Hz)
100 10k
Question: Is the frequency response curve that of a band-pass filters? Explain why.
 Yes. From 100.219 Hz to 10 kHz the filter allows the frequency to pass
however it blocked the other frequencies that is why it is a band-pass filter.
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 (A V) 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?
 It is only 0.42 differences. The measure voltage gain is 0.84% different with
the calculated value.

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 = 1.59 kHz
Question: How did the calculated value of the center frequency compare with the measured
value?
 The calculated and measured center frequencies have a difference of 1.14%.
Step 8 Based on the measured center frequency (fo) and the bandwidth (BW), calculate
the quality factor (Q) of the band-pass filter.
 Q = 4.8
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 calculated value compared to the measured
value is 4.167%.

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.
θ
o
270
o
-270 f (Hz)
100 10k
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 (f o) and the phase (θ) on the
phase curve plot.
 fo = 1.572 kHz
 θ = 173.987o
Question: What does this result tell you about the relationship between the filter output and
input at the center frequency?
 The phase output is 173.987 out of phase in 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)

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
AdB
10
-20 f (Hz)
2 500
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 = 1.769
Question: How did the measured voltage gain in the passband compare with the voltage
gain calculated from the circuit values?
 There is a 0.057% difference.

10. 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 = 58.946Hz
Question How did the calculated value of the center frequency compare with the measured
value?
 There is a percentage difference of 0.506%. It is only a small difference.
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 a 0.46% difference. Thus, they are almost the same.

11. CONCLUSION
Therefore, I conclude that a band-pass filter allows the frequencies lying within
the passband and block all the frequencies outside the passband. I notice that the center
frequency of an active band-pass frequency is at the highest dB voltage gain of the filter.
On the contrary, band-stop filter, which is also referred to as band-reject or notch filter,
blocks the frequencies lying within the passband, passes all the frequencies outside the
passband, and having a center frequency at the lowest dB voltage gain of the filter
Moreover, the quality factor is inversely proportional to the bandwidth. The
higher the quality factor the narrower the response will be.
Lastly, the bandwidth is determined by subtracting the high and low cutoff
frequency of the filter, while the center frequency is the geometric mean of both cutoffs.