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Comm008 e4 balane

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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 FILTERSBalane, Maycen M. July 21, 2011Signal 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 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 of frequenciesfrom the total frequency spectrum. This is normally accomplished with filters. A filter is a circuit thatpasses a specific range of frequencies while rejecting other frequencies. Active filters use activedevices such as op-amps combined with passive elements. Active filters have several advantages overpassive filters. The passive elements provide frequency selectivity and the active devices providevoltage gain, high input impedance, and low output impedance. The voltage gain reduces attenuationof the signal by the filter, the high input impedance prevents excessive loading of the source, and thelow output impedance prevents the filter from being affected by the load. Active filters are also easy toadjust over a wide frequency range without altering the desired response. The weakness of activefilters is the upper-frequency limit due to the limited open-loop bandwidth (funity) of op-amps. Thefilter cutoff frequency cannot exceed the unity-gain frequency (funity) of the op-amp. Therefore, activefilters must be used in applications where the unity-gain frequency (funity) of the op-amp is high enoughso that it does not fall within the frequency range of the application. For this reason, active filters aremostly used in low-frequency applications.A band-pass filter passes all frequencies lying within a band of frequencies and rejects all otherfrequencies outside the band. The low cut-off frequency (fC1) and the high-cutoff frequency (fC2) on thegain-frequency plot are the frequencies where the voltage gain has dropped by 3 dB (0.707) from themaximum dB gain. A band-stop filter rejects a band of frequencies and passes all other frequenciesoutside 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 gainhas 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-cutofffrequency 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 of the low-cutofffrequency (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) andthe 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 thanone is considered to be a wide-band filter and a filter with a Q greater than ten is considered to be anarrow-band filter.One way to implement a band-pass filter is to cascade a low-pass and a high-pass filter. As long as thecutoff frequencies are sufficiently separated, the low-pass filter cutoff frequency will determine thelow-cutoff frequency of the band-pass filter and a high-pass filter cutoff frequency will determine thehigh-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 determinethe 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 equationWhere 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 center frequency (fO)can be calculated fromAt this frequency (fo), the feedback signal returns with the correct amplitude and phase to attenuatethe 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 be less than 10to 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 0PROCEDUREActive 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)
  • 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 (f o) 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.  It is a frequency response curve 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 = 49.58Step 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 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 kHzStep 6 Based on the measured values of fC1 and fC2, calculate the bandwidth (BW) of the band- pass filter.
  • 7.  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 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?  The percentage difference is only 5.26% 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 = 50.146oQuestion: What does this result tell you about the relationship between the filter output and input at the center frequency?
  • 8.  The output is 173.987o out of phase with input.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)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, it attenuates all the frequencies lying within the bandpass.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.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
  • 9. 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 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?  The percentage difference 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 = 2.18Step 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?  They have 0.46% difference.
  • 10. CONCLUSION I conclude that an active filter is using active devices such as operational amplifiers combinedwith passive elements. It allows the frequencies within range and blocks the frequencies outside thatrange. While active band-stop filter only passes the frequencies outside the band and blocks thefrequencies lying within that band. The center frequency of the band-pass is at the peak, while in notchfilter, it is at the lowest dB gain of the curve. The bandwidth is the difference between the cutofffrequencies. The quality factor is inversely proportional to the bandwidth. The higher the qualityfactor, the more selective the filter and the less bandwidth it will be. The phase shift between the inputand output for a two-pole active band-pass filter is 180o.
  • 11. SAMPLE COMPUTATIONStep 3Step 4Step 4 QuestionStep 7Step 8 (Measured quality factor)Step 9 (Calculated quality factor)Step 15 (Expected Voltage Gain)Step 18 (Expected center frequency)Step 19 (Measure Quality Factor)Step 20 (Expected Quality Factor)

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