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NATIONAL COLLEGE OF SCIENCE AND TECHNOLOGY Amafel Building, Aguinaldo Highway Dasmariñas City, Cavite Experiment No. 3 “ACTIVE LOW-PASS and HIGH-PASS FILTERS”Pagara, Sheila Marie P. July 14, 2011Signal Spectra and Signal Processing/BSECE 41A1 Score: Engr. Grace Ramones Instructor
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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.
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SAMPLE COMPUTATIONS:Computation of voltage gain based on measured value: AdB = 20 log A 4.006 = 20 log AComputation of voltage gain based on circuit:Computation of percentage difference: Q in Step 6Computation of cutoff frequency: Step 6 and 17
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DATA SHEET:MATERIALSOne 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 specific range offrequencies from the total frequency spectrum. This is normally accomplished with filters. A filter isa circuit that passes a specific range of frequencies while rejecting other frequencies. Active filtersuse active devices such as op-amps combined with passive elements. Active filters have severaladvantages over passive filters. The passive elements provide frequency selectivity and the activedevices provide voltage gain, high input impedance, and low output impedance. The voltage gainreduces attenuation of the signal by the filter, the high input prevents excessive loading of thesource, 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 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 theop-amp. Ideally, a high-pass filter should pass all frequencies above the cutoff frequency (f c).Because op-amps have a limited open-loop bandwidth (unity-gain frequency, funity), high-pass activefilters have an upper-frequency limit on the high-pass response, making it appear as a band-passfilter with a very wide bandwidth. Therefore, active filters must be used in applications where theunity-gain frequency (funity) of the op-amp is high enough so that it does not fall within the frequencyrange of the application. For this reason, active filters are mostly used in low-frequencyapplications. The most common way to describe the frequency response characteristics of a filter is toplot the filter voltage gain (Vo/Vin) in dB as a function of frequency (f). The frequency at which theoutput power gain drops to 50% of the maximum value is called the cutoff frequency (f c). When theoutput power gain drops to 50%, the voltage gain drops 3 dB (0.707 of the maximum value). Whenthe filter dB voltage gain is plotted as a function of frequency using straight lines to approximatethe actual frequency response, it is called a Bode plot. A Bode plot is an ideal plot of filterfrequency response because it assumes that the voltage gain remains constant in the passbanduntil the cutoff frequency is reached, and then drops in a straight line. The filter network voltagegain in dB is calculated from the actual voltage gain (A) using the equation AdB = 20 log Awhere A = Vo/Vin. An ideal filter has an instantaneous roll-off at the cutoff frequency (fc), with full signal levelon one side of the cutoff frequency. Although the ideal is not achievable, actual filters roll-off at -20dB/decade or higher depending on the type of filter. The -20 dB/decade roll-off is obtained with aone-pole filter (one R-C circuit). A two-pole filter has two R-C circuits tuned to the same cutofffrequency and rolls off at -40 dB/decade. Each additional pole (R-C circuit) will cause the filter toroll off an additional -20 dB/decade. In a one-pole filter, the phase between the input and theoutput will change by 90 degrees over the frequency range and be 45 degrees at the cutofffrequency. In a two-pole filter, the phase will change by 180 degrees over the frequency range andbe 90 degrees at the cutoff frequency. Three basic types of response characteristics that can be realized with most active filtersare Butterworth, Chebyshev, and Bessel, depending on the selection of certain filter componentvalues. The Butterworth filter provides a flat amplitude response in the passband and a roll-off of -20 dB/decade/pole with a nonlinear phase response. Because of the nonlinear phase response, apulse waveshape applied to the input of a Butterworth filter will have an overshoot on the output.
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Filters with a Butterworth response are normally used in applications where all frequencies in thepassband must have the same gain. The Chebyshev filter provides a ripple amplitude response inthe passband and a roll-off better than -20 dB/decade/pole with a less linear phase response thanthe Butterworth filter. Filters with a Chebyshev response are most useful when a rapid roll-off isrequired. The Bessel filter provides a flat amplitude response in the passband and a roll-off of lessthan -20 dB/decade/pole with a linear phase response. Because of its linear phase response, theBessel filter produces almost no overshoot on the output with a pulse input. For this reason, filterswith a Bessel response are the most effective for filtering pulse waveforms without distorting thewaveshape. Because of its maximally flat response in the passband, the Butterworth filter is themost widely used active filter. A second-order (two-pole) active low-pass Butterworth filter is shown in Figure 3-1. Becauseit is a two-pole (two R-C circuits) low-pass filter, the output will roll-off -40 dB/decade at frequenciesabove the cutoff frequency. A second-order (two-pole) active high-pass Butterworth filter is shown inFigure 3-2. Because it is a two-pole (two R-C circuits) high-pass filter, the output will roll-off -40dB/decade at frequencies below the cutoff frequency. These two-pole Sallen-Key Butterworth filtersrequire a passband voltage 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 times the inputvoltage (-3 dB). The expected cutoff frequency (fc), based on the circuit component values, can becalculated from wherein, FIGURE 3 – 1 Second-order (2-pole) Sallen-Key Low-Pass Butterworth Filter
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FIGURE 3 – 2 Second-order (2-pole) Sallen-Key High-Pass Butterworth FilterPROCEDURELow-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 It allows all the frequencies below the cutoff frequency and rejects above cutoff frequencies.
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Step 3 Calculate the actual voltage gain (A) from the measured dB gain. A = 1.586Step 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? The values are the same.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 percentage difference is of 0.30%.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 kHzQuestions: 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? ROLL-OFF: It decrease 37.106 dB/decade increase in frequency. I am expecting 40 dB decrease per decade increase in frequency.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 f
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Step 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 expectedfor a two-pole low-pass filter? Yes, Two-pole filter has a 90o phase at the cutoff frequency.Step 10 Click Magnitude on the plotter. Change R to 1 kΩ in both places and C to 1 pF in bothplaces. Adjust the horizontal final frequency (F) on the Bode plotter to 20 MHz. Run the 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 the measuredvalues of the cutoff frequency when R = 1kΩ and C = 1pF. Hint: The value of the unity-gain bandwidth,funity, for the 741 op-amp is approximately 1 MHz. Due to the active filter’s upper-frequency limit. The filter cutoff frequency should not exceed the unity-gain frequency.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 dB
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Question: Is the frequency response curve that of a high-pass filter? Explain why. Yes. It will pass all the frequencies above the cutoff frequency and blocks frequencies below the cut-off frequency.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 is 2.98%.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 = 5305.16477 HzQuestion: 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 values have a percent difference of 2.89%.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? ROLL-OFF: approximately -40 dB. It is expected for a two-pole filter.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 f
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Step 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 It is the fC exceed the unity gain. And because active filters have an upper frequency limit on the high-pass response.
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CONCLUSION: Therefore I conclude that the frequency response of a active low-pass and high-pass filters arelike the passive low-pass and high-pass filters. The only difference in their response is at the high-passresponse. This is because of the high-frequency limit of the op-amp used in the active filter. Addition tothat, active high-pass filters looked like a band-pass filter at the high-frequencies. The cut-off frequencyshould never exceed to the unity gain of the op-amp. Furthermore, curve of Butterworth filter is flat in amplitude and will roll-off at -40 dB per decadeincrease in frequency. It has a voltage gain of 1.586. Lastly, a two-pole filter has a phase difference of180 degrees over the range of frequency and 90 degrees at cut-off frequency.
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