Comm008 e4 pula

535 views
475 views

Published on

Published in: Business, Technology
0 Comments
0 Likes
Statistics
Notes
  • Be the first to comment

  • Be the first to like this

No Downloads
Views
Total views
535
On SlideShare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
Downloads
12
Comments
0
Likes
0
Embeds 0
No embeds

No notes for slide

Comm008 e4 pula

  1. 1. NATIONAL COLLEGE OF SCIENCE AND TECHNOLOGY Amafel Building, Aguinaldo Highway Dasmariñas City, Cavite EXPERIMENT NO. 4 ACTIVE BAND-PASS AND BAND-STOP FILTERSPula, Rolando A. July 21, 2011Signal Spectra and Signal Processing/BSECE 41A1 Score: Engr. Grace Ramones Instructor
  2. 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. 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 100kΩTHEORY In electronic communications systems, it is often necessary to separate a specific range offrequencies from the total frequency spectrum. This is normally accomplished with filters. A filteris a circuit that passes a specific range of frequencies while rejecting other frequencies. Activefilters use active devices such as op-amps combined with passive elements. Active filters haveseveral advantages over passive filters. The passive elements provide frequency selectivity and theactive devices provide voltage gain, high input impedance, and low output impedance. Thevoltage gain reduces attenuation of the signal by the filter, the high input impedance preventsexcessive loading of the source, and the low output impedance prevents the filter from beingaffected by the load. Active filters are also easy to adjust over a wide frequency range withoutaltering the desired response. The weakness of active filters is the upper-frequency limit due to thelimited open-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 in applicationswhere the unity-gain frequency (funity) of the op-amp is high enough so that it does not fall withinthe 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 allother 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 allother 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 frequencieswhere 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 thelow-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. 4. 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 than ten isconsidered 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. Aslong as the cutoff frequencies are sufficiently separated, the low-pass filter cutoff frequency willdetermine the low-cutoff frequency of the band-pass filter and a high-pass filter cutoff frequencywill determine the high-cutoff frequency of the band-pass filter. Normally this arrangement is usedfor 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 C1determine the low-cutoff frequency, and R2 and C2 determine the high-cutoff frequency. The centerfrequency (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 from Figure 4-2 shows a second-order (two-pole) Sallen-Key notch filter. The expected centerfrequency (fO) can be calculated from At 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 factor The 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 4-1 Multiple-Feedback Band-Pass Filter
  5. 5. XBP1 XFG1 IN OUT 10nF C1 100kΩ R2 741 3 Vo 6 Vin 1kΩ 2 10kΩ 10nF R1 RL C2Figure 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 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)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. 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 = 49.58Step 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 difference is 0.42 or 0.84% percentage differenceStep 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 kHz
  7. 7. Step 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? = 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 = 4.75Step 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? = 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 phasedifference (θ) between the filter input and output wave shapes as a function of frequency (f). Drawthe 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. 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) 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. 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. 9. Step 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? = 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 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? = They have 0.51% difference. They are almost the sameStep 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? = The calculated value and the measured value has a difference of 0.01 or percentage difference of 0.46%
  10. 10. CONCLUSION After performing the experiment, I can able to say that the active filter response is alikewith the passive filter response. It band-pass only allows the frequencies lying within the band andblocks other frequency. As for the band-stop, the frequencies lying in the band are blocked whilepassing the frequencies outside the band. The center frequency of the band-pass is at the peak gain, while the center frequency of theband-stop is at the lowest dB gain. In a two-pole circuit, the phase response of the output 180o outof phase with the input. The quality factor determines the selectivity of the response curve. Thehigher the quality factor, the more selective the graph is. It is inversely proportional to thebandwidth of the curve. Bandwidth is the difference of the high and the low cutoff frequencies.
  11. 11. SAMPLE COMPUTATION(Actual Voltage Gain)Step 3Step 14(Expected voltage gain)Step 4Step 15 (Expected Voltage Gain)(Percentage Difference)Step 4 Question
  12. 12. Step 7 QuestionStep 9 Question(Center frequency)Step 7Step 18(Measured quality factor)Step 8Step 19(Expected quality factor)Step 9Step 20

×