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• 1. NATIONAL COLLEGE OF SCIENCE AND TECHNOLOGY Amafel Building, Aguinaldo Highway Dasmariñas City, Cavite Experiment No. 4 ACTIVE BAND-PASS and BAND-STOP FILTERSMaala, Michelle Anne C. July 19, 2011Signal Spectra and Signal Processing/BSECE 41A1 Score: Engr. Grace Ramones Instructor
• 2. OBJECTIVES1. 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 filter3. 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 SHEETMATERIALSOne function generatorOne dual-trace oscilloscopeTwo LM741 op-ampsCapacitors: two 0.001 µF, two 0.05 µF, one 0.1 µFResistors: one 1 kΩ, two 10 kΩ, one 13 kΩ, one 27 kΩ, two 54 kΩ, and one 100kΩTHEORYIn electronic communications systems, it is often necessary to separate a specific range offrequencies from the total frequency spectrum. This is normally accomplished with filters. Afilter is a circuit that passes a specific range of frequencies while rejecting otherfrequencies. Active filters use active devices such as op-amps combined with passiveelements. Active filters have several advantages over passive filters. The passive elementsprovide frequency selectivity and the active devices provide voltage gain, high inputimpedance, and low output impedance. The voltage gain reduces attenuation of thesignal 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. Activefilters are also easy to adjust over a wide frequency range without altering the desiredresponse. The weakness of active filters is the upper-frequency limit due to the limitedopen-loop bandwidth (funity) of op-amps. The filter cutoff frequency cannot exceed theunity-gain frequency (funity) of the op-amp. Therefore, active filters must be used inapplications where the unity-gain frequency (funity) of the op-amp is high enough so that itdoes not fall within the frequency range of the application. For this reason, active filtersare mostly used in low-frequency applications.A band-pass filter passes all frequencies lying within a band of frequencies and rejects allother frequencies outside the band. The low cut-off frequency (fC1) and the high-cutofffrequency (fC2) on the gain-frequency plot are the frequencies where the voltage gainhas dropped by 3 dB (0.707) from the maximum dB gain. A band-stop filter rejects a bandof frequencies and passes all other frequencies outside the band, and of then referred toas 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 droppedby 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 – fC1The center frequency (fo)of the band-pass or a band-stop filter is the geometric mean ofthe 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 centerfrequency (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 aQ less than one is considered to be a wide-band filter and a filter with a Q greater thanten 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 cutofffrequency will determine the low-cutoff frequency of the band-pass filter and a high-passfilter 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 cutofffrequencies need to be sufficient separated.A multiple-feedback active band-pass filter is shown in Figure 4-1. Components R1 and C1determine 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 theequationWhere C = C1 = C2. The voltage gain (AV) at the center frequency is calculated fromand the quality factor (Q) is calculated fromFigure 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 centerfrequency (fO) can be calculated fromAt this frequency (fo), the feedback signal returns with the correct amplitude and phase toattenuate the input. This causes the output to be attenuated at the center frequency.The notch filter in Figure 4-2 has a passband voltage gainand a quality factorThe voltage gain of a Sallen-Key notch filter must be less than 2 and the circuit Q must beless 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. PROCEDUREActive Band-Pass FilterStep 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 dBQuestion: 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 = 49.58Step 4 Based on the circuit component values, calculate the expected voltage gain (AV) at the center frequency (fo)  AV = 50Question: 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 kHzStep 6 Based on the measured values of fC1 and fC2, calculate the bandwidth (BW) of the band-pass filter.  BW = 0.331 kHzStep 7 Based on the circuit component values, calculate the expected center frequency (fo)  fo = 1.592 kHzQuestion: 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 = 4.75Step 9 Based on the component values, calculate the expected quality factor (Q) of the band-pass filter.  Q=5Question: 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 = 50.146oQuestion: 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) FilterStep 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. dBQuestion: 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 = 1.77Step 15 Based on the circuit component values, calculate the expected voltage gain in the passband.  AV = 1.77Question: 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 HzStep 17 Based on the measured values of fC1 and fC2, calculate the bandwidth (BW) of the notch filter.  BW = 26.845 HzStep 18 Based on the circuit component values, calculate the expected center frequency (fo)  fo = 58.95HzQuestion 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 = 2.18Step 20 Based on the calculated passband voltage gain (Av), calculate the expected quality factor (Q) of the notch filter.  Q = 2.17Question: 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 thesame with the mountain-like response of a passive band-pass filter having the centerfrequency at the peak dB gain. On the contrary, the center frequency of a band-stop is atthe lowest dB gain. Like in the passive filter, active band-pass allows the frequency insidethe 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. Thehigher the value of Q the narrower the response becomes. The center frequency is thegeometric mean of the cutoff frequencies. And also, the bandwidth of the response curveis 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 15Step 18Step 18 QStep 19 (Step 20