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    3 (1) 3 (1) Document Transcript

    • NATIONAL COLLEGE OF SCIENCE AND TECHNOLOGY Amafel Building, Aguinaldo Highway Dasmariñas City, Cavite Experiment No. 3 ACTIVE LOW-PASS and HIGH-PASS FILTERSAgdon, Berverlyn B. July 14, 2011Signal Spectra and Signal Processing/BSECE 41A1 Score: Engr. Grace Ramones Instructor
    • Objectives Plot the gain-frequency response and determine the cutoff frequency of a second- order (two-pole) low-pass active filter. Plot the gain-frequency response and determine the cutoff frequency of a second- order (two-pole) high-pass active filter. Determine the roll-off in dB per decade for a second-order (two-pole) filter. Plot the phase-frequency response of a second-order (two-pole) filter.
    • Sample ComputationsStep 3 AdB = 20 log A 4.006 = 20 log AStep 4Question Step 4Step 6Question Step 6Q uestion Step 7 -36.146 dB – 0.968 dB = -37.106 dBStep 14 A = 1.54
    • Data SheetMATERIALSOne function generatorOne dual-trace oscilloscopeOne LM741 op-ampCapacitors: two 0.001 µF, one 1 pFResistors: one 1kΩ, one 5.86 kΩ, two 10kΩ, two 30 kΩTHEORY In electronic communications systems, it is often necessary to separate a specificrange of frequencies from the total frequency spectrum. This is normally accomplishedwith filters. A filter is a circuit that passes a specific range of frequencies while rejectingother frequencies. Active filters use active devices such as op-amps combined withpassive elements. Active filters have several advantages over passive filters. The passiveelements provide frequency selectivity and the active devices provide voltage gain, highinput impedance, and low output impedance. The voltage gain reduces attenuation ofthe signal by the filter, the high input prevents excessive loading of the source, and thelow output impedance prevents the filter from being affected by the load. Active filtersare 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. Ideally, a high-pass filter should pass allfrequencies above the cutoff frequency (fc). Because op-amps have a limited open-loopbandwidth (unity-gain frequency, funity), high-pass active filters have an upper-frequencylimit on the high-pass response, making it appear as a band-pass filter with a very widebandwidth. Therefore, active filters must be used in applications where the unity-gainfrequency (funity) of the op-amp is high enough so that it does not fall within thefrequency range of the application. For this reason, active filters are mostly used in low-frequency applications. The most common way to describe the frequency response characteristics of afilter is to plot the filter voltage gain (Vo/Vin) in dB as a function of frequency (f). Thefrequency at which the output power gain drops to 50% of the maximum value is calledthe cutoff frequency (fc). When the output power gain drops to 50%, the voltage gaindrops 3 dB (0.707 of the maximum value). When the filter dB voltage gain is plotted as afunction of frequency using straight lines to approximate the actual frequency response,it is called a Bode plot. A Bode plot is an ideal plot of filter frequency response because itassumes that the voltage gain remains constant in the passband until the cutofffrequency is reached, and then drops in a straight line. The filter network voltage gain indB is calculated from the actual voltage gain (A) using the equation AdB = 20 log A
    • where A = Vo/Vin. An ideal filter has an instantaneous roll-off at the cutoff frequency (fc), with fullsignal level on one side of the cutoff frequency. Although the ideal is not achievable,actual filters roll-off at -20 dB/decade or higher depending on the type of filter. The -20dB/decade roll-off is obtained with a one-pole filter (one R-C circuit). A two-pole filterhas two R-C circuits tuned to the same cutoff frequency and rolls off at -40 dB/decade.Each additional pole (R-C circuit) will cause the filter to roll off an additional -20dB/decade. In a one-pole filter, the phase between the input and the output will changeby 90 degrees over the frequency range and be 45 degrees at the cutoff frequency. In atwo-pole filter, the phase will change by 180 degrees over the frequency range and be 90degrees at the cutoff frequency. Three basic types of response characteristics that can be realized with most activefilters are Butterworth, Chebyshev, and Bessel, depending on the selection of certainfilter component values. The Butterworth filter provides a flat amplitude response in thepassband and a roll-off of -20 dB/decade/pole with a nonlinear phase response. Becauseof the nonlinear phase response, a pulse waveshape applied to the input of aButterworth filter will have an overshoot on the output. Filters with a Butterworthresponse are normally used in applications where all frequencies in the passband musthave the same gain. The Chebyshev filter provides a ripple amplitude response in thepassband and a roll-off better than -20 dB/decade/pole with a less linear phaseresponse than the Butterworth filter. Filters with a Chebyshev response are most usefulwhen a rapid roll-off is required. The Bessel filter provides a flat amplitude response inthe passband and a roll-off of less than -20 dB/decade/pole with a linear phaseresponse. Because of its linear phase response, the Bessel filter produces almost noovershoot on the output with a pulse input. For this reason, filters with a Besselresponse are the most effective for filtering pulse waveforms without distorting thewaveshape. Because of its maximally flat response in the passband, the Butterworthfilter is the most widely used active filter. A second-order (two-pole) active low-pass Butterworth filter is shown in Figure3-1. Because it is a two-pole (two R-C circuits) low-pass filter, the output will roll-off -40dB/decade at frequencies above the cutoff frequency. A second-order (two-pole) activehigh-pass Butterworth filter is shown in Figure 3-2. Because it is a two-pole (two R-Ccircuits) high-pass filter, the output will roll-off -40 dB/decade at frequencies below thecutoff frequency. These two-pole Sallen-Key Butterworth filters require a passbandvoltage gain of 1.586 to produce the Butterworth response. Therefore,and
    • At the cutoff frequency of both filters, the capacitive reactance of each capacitor (C) isequal to the resistance of each resistor (R), causing the output voltage to be 0.707 timesthe input voltage (-3 dB). The expected cutoff frequency (fc), based on the circuitcomponent values, can be calculated from wherein, FIGURE 3 – 1 Second-order (2-pole) Sallen-Key Low-Pass Butterworth Filter FIGURE 3 – 2 Second-order (2-pole) Sallen-Key High-Pass Butterworth Filter
    • PROCEDURELow-Pass Active FilterStep 1 Open circuit file FIG 3-1. Make sure that the following Bode plotter settings are selected: Magnitude, Vertical (Log, F = 10dB, I = -40dB), Horizontal (Log, F = 100 kHz, I = 100 Hz).Step 2 Run the simulation. Notice that the voltage gain has been plotted between the frequencies of 100 Hz and 100 kHz by the Bode plotter. Draw the curve plot in the space provided. Next, move the cursor to the flat part of the curve at a frequency of approximately 100 Hz and measure the voltage gain in dB. Record the dB gain on the curve plot. AdB f dB gain = 4.006 dBQuestion: Is the frequency response curve that of a low-pass filter? Explain why. Yes. Low-pass filter only allow the frequencies below the cutoff frequency and block the frequencies above the fc.Step 3 Calculate the actual voltage gain (A) from the measured dB gain. A = 1.586
    • Step 4 Based on the circuit component values in Figure 3-1, calculate the expected voltage gain (A) on the flat part of the curve for the low-pass Butterworth filter. A = 1.586Question: How did the measured voltage gain in Step 3 compared with the calculated voltage gain in Step 4? They are equal.Step 5 Move the cursor as close as possible to a point on the curve that is 3dB down from the dB gain at the low frequencies. Record the dB gain and the frequency (cutoff frequency, fc) on the curve plot. dB gain= 0.968 dB fc = 5.321 kHzStep 6 Calculate the expected cutoff frequency (fc) based on the circuit component values. fc = 5.305 kHzQuestion: How did the calculated value for the cutoff frequency compare with the measured value recorded on the curve plot for the two-pole low-pass active filter The difference is only 0.30%. They are almost equal.Step 7 Move the cursor to a point on the curve where the frequency is as close as possible to ten times fc. Record the dB gain and frequency (fc) on the curve plot. dB gain = -36.146 dB fc = 53.214 kHz
    • Questions: Approximately how much did the dB gain decrease for a one-decade increase in frequency? Was this what you expected for a two-pole filter? The dB gain decrease 37.106 dB for a one-decade increase. I expected 40 dB decrease per decade increase.Step 8 Click Phase on the Bode plotter to plot the phase curve. Change the vertical axis initial value (I) to 180 degrees and the final value (F) to 0 degree. 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 fStep 9 Move the cursor as close as possible on the curve to the cutoff frequency (fc). Record the frequency (fc) and phase (θ) on the curve. fc = 5.321 kHz θ = -90.941Question: Was the phase shift between input and output at the cutoff frequency what you expected for a two-pole low-pass filter? The result is what I expected because yhe phase at cutoff frequency should be 90o
    • Step 10 Click Magnitude on the plotter. Change R to 1 kΩ in both places and C to 1 pFin both places. Adjust the horizontal final frequency (F) on the Bode plotter to 20 MHz. Runthe simulation. Measure the cutoff frequency (fc) and record your answer. fc = 631.367 kHzStep 11 Based on the new values for resistor R and capacitor C, calculate the new cutoff frequency (fc). fc = 159.1549 MHzQuestion: Explain why there was such a large difference between the calculated and themeasured values of the cutoff frequency when R = 1kΩ and C = 1pF. Hint: The value of theunity-gain bandwidth, funity, for the 741 op-amp is approximately 1 MHz. Because the cutoff frequency exceed the unity-gain frequency of the op- amp the curve shown was like a band-pass filter. And because of limited open-loop bandwidth of the op-amp, active filter have an upper- frequency limit.High-Pass Active FilterStep 12 Open circuit file FIG 3-2. Make sure that the following Bode plotter settings are selected: Magnitude, Vertical (Log, F = 10dB, I = -40dB), Horizontal (Log, F = 100 kHz, I = 100 Hz).Step 13 Run the simulation. Notice that the voltage gain has been plotted between the frequencies of 100 Hz and 100 kHz by the Bode plotter. Draw the curve plot in the space provided. Next, move the cursor to the flat part of the curve at a frequency of approximately 100 kHz and measure the voltage gain in dB. Record the dB gain on the curve plot.
    • AdB f dB gain = 3.776 dBQuestion: Is the frequency response curve that of a high-pass filter? Explain why. Yes. High-pass filter only allow the frequencies above the cutoff frequency and reject the frequencies below the fc.Step 14 Calculate the actual voltage gain (A) from the measured dB gain. A = 1.54Step 15 Based on the circuit component values in Figure 3-2, calculate the expected voltage gain (A) on the flat part of the curve for the high -pass Butterworth filter. Av = 1.586Question: How did the measured voltage gain in Step 14 compare with the calculated voltage gain in Step 15? The percentage difference of 2.98%. They are still almost equal
    • Step 16 Move the cursor as close as possible to a point on the curve that is 3dB down from the dB gain at the high frequencies. Record the dB gain and the frequency (cutoff frequency, fc) on the curve plot. dB gain = 0.741 dB fc = 5.156 kHzStep 17 Calculate the expected cutoff frequency (fc) based on the circuit component values. fc = 5.305 kHzQuestion: How did the calculated value of the cutoff frequency compare with the measured value recorded on the curve plot for the two-pole low-pass active filter? The percentage difference between the two is 2.89%. They are almost equal.Step 18 Move the cursor to a point on the curve where the frequency is as close as possible to one-tenth fc. Record the dB gain and frequency (fc) on the curve plot. dB gain = -36.489 dB fc = 515.619 HzQuestions: Approximately how much did the dB gain decrease for a one-decade decrease in frequency? Was this what you expected for a two-pole filter? It rolls-off at -37.23 dB per decade.Step 19 Change the horizontal axis final setting (F) to 50 MHz on the Bode plotter. Run the simulation. Draw the curve plot in the space provided.
    • AdB fStep 20 Measure the upper cutoff frequency (fc2) and record the value on the curve plot. fc2 = 92.595 kHzQuestion: Explain why the filter frequency response looked like a band-pass response when frequencies above 1 MHz were plotted. Hint: The value of the unity-gain bandwidth, funity, for the 741 op-amp is approximately 1 MHz Because of the limited value of the unity-gain bandwidth of the op-amp, the response appears like a band-pass filter.
    • Conclusion Therefore, the active filter’s frequency response appears the same as the passive filter;however, active filter has several advantages over the passive filter. It provides voltage gain,high input impedance, and low output impedance. But the disadvantage of this filter ishaving a upper-frequency limit due to the limited open-loop of the bandwidth. At high-passresponse, active filter’s response curve is like the response curve of a band-pass filter.Moreover, a two-pole filter or second order low-pass filter rolls-off at -40 dB per decadeincrease in frequency while second order high-pass filter rolls-off at -40 dB per decrease infrequency. The phase shift at the cutoff frequency is 90 degrees while there is 180 degreesphase difference over the frequency range.