NATIONAL COLLEGE OF SCIENCE AND TECHNOLOGY            Amafel Building, Aguinaldo Highway Dasmariñas City, Cavite          ...
OBJECTIVES:  1. Plot the gain-frequency response and determine the cutoff frequency of a     second-order (two-pole) low-p...
VOLTAGE GAIN (MEASURED VALUE):     Step 3                AdB = 20 log A                 4.006 = 20 log A      Step 14     ...
DATA SHEET:MATERIALS     One function generator     One dual-trace oscilloscope     One LM741 op-amp     Capacitors: two 0...
where A = Vo/Vin.        An ideal filter has an instantaneous roll-off at the cutoff frequency (fc), with fullsignal level...
At the cutoff frequency of both filters, the capacitive reactance of eachcapacitor (C) is equal to the resistance of each ...
PROCEDURELow-Pass Active FilterStep 1       Open circuit file FIG 3-1. Make sure that the following Bode plotter settings ...
Step 5       Move the cursor as close as possible to a point on the curve that is 3dB down             from the dB gain at...
Step 9       Move the cursor as close as possible on the curve to the cutoff frequency (fc).             Record the freque...
Step 13     Run the simulation. Notice that the voltage gain has been plotted between the            frequencies of 100 Hz...
Step 17        Calculate the expected cutoff frequency (fc) based on the circuit component               values.          ...
CONCLUSION:       I conclude that the gain-frequency response of a second-order low-pass activeappears as a passive filter...
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  1. 1. NATIONAL COLLEGE OF SCIENCE AND TECHNOLOGY Amafel Building, Aguinaldo Highway Dasmariñas City, Cavite Experiment No. 3 ACTIVE LOW-PASS and HIGH-PASS FILTERSMaala, Michelle Anne C. July 14, 2011Signal Spectra and Signal Processing/BSECE 41A1 Score: Engr. Grace Ramones Instructor
  2. 2. OBJECTIVES: 1. Plot the gain-frequency response and determine the cutoff frequency of a second-order (two-pole) low-pass active filter. 2. Plot the gain-frequency response and determine the cutoff frequency of a second-order (two-pole) high-pass active filter. 3. Determine the roll-off in dB per decade for a second-order (two-pole) filter. 4. Plot the phase-frequency response of a second-order (two-pole) filter.
  3. 3. VOLTAGE GAIN (MEASURED VALUE): Step 3 AdB = 20 log A 4.006 = 20 log A Step 14 AdB = 20 log A 3.776 = 20 log A A = 1.54VOLTAGE GAIN(CALCULATED): Step 4PERCENTAGE DIFFERENCE: Q Step 4 Q Step 6 Q Step 15CUTOFF FREQUENCY (CALCULATED): Step 6 and 17 Step 11
  4. 4. DATA SHEET:MATERIALS One function generator One dual-trace oscilloscope One LM741 op-amp Capacitors: two 0.001 µF, one 1 pF Resistors: 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,high input impedance, and low output impedance. The voltage gain reducesattenuation 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 theload. Active filters are also easy to adjust over a wide frequency range without alteringthe desired response. The weakness of active filters is the upper-frequency limit due tothe limited open-loop bandwidth (funity) of op-amps. The filter cutoff frequency cannotexceed the unity-gain frequency (funity) of the op-amp. Ideally, a high-pass filter shouldpass all frequencies above the cutoff frequency (fc). Because op-amps have a limitedopen-loop bandwidth (unity-gain frequency, funity), high-pass active filters have anupper-frequency limit on the high-pass response, making it appear as a band-pass filterwith a very wide bandwidth. Therefore, active filters must be used in applications wherethe unity-gain frequency (funity) of the op-amp is high enough so that it does not fallwithin the frequency range of the application. For this reason, active filters are mostlyused 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 frequencyresponse, it is called a Bode plot. A Bode plot is an ideal plot of filter frequencyresponse 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 networkvoltage gain in dB is calculated from the actual voltage gain (A) using the equation AdB = 20 log A
  5. 5. 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 willchange 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 thefrequency range and be 90 degrees at the cutoff frequency. Three basic types of response characteristics that can be realized with mostactive filters are Butterworth, Chebyshev, and Bessel, depending on the selection ofcertain filter component values. The Butterworth filter provides a flat amplitude responsein the passband and a roll-off of -20 dB/decade/pole with a nonlinear phase response.Because of 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 Bessel responseare the most effective for filtering pulse waveforms without distorting the waveshape.Because of its maximally flat response in the passband, the Butterworth filter is the mostwidely used active filter. A second-order (two-pole) active low-pass Butterworth filter is shown in Figure 3-1. 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)active high-pass Butterworth filter is shown in Figure 3-2. Because it is a two-pole (two R-C circuits) high-pass filter, the output will roll-off -40 dB/decade at frequencies belowthe cutoff frequency. These two-pole Sallen-Key Butterworth filters require a passbandvoltage gain of 1.586 to produce the Butterworth response. Therefore,and
  6. 6. At the cutoff frequency of both filters, the capacitive reactance of eachcapacitor (C) is equal to the resistance of each resistor (R), causing the output voltageto be 0.707 times the input voltage (-3 dB). The expected cutoff frequency (fc), based on the circuit component 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
  7. 7. 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 it is a low-pass filter. This is because it allows all the frequencies below the cutoff frequency and rejects other frequencies.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? Measured voltage gain in Step 3 and the calculated voltage gain in Step 4 are the same
  8. 8. 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 values are almost equal. There is a difference 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? -37.106 dB/decade and is almost -40 dB per decade which is the ideal for a 2-pole filter . It is a two-pole filter so I am expecting a 40 dB decrease per decade increase and our result is almost equal to my expectationStep 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
  9. 9. 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 expected for a two-pole low-pass filter? Yes, at cutoff frequency the phase shift is expected to have a 90 degreesStep 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. The weakness of active filter is the upper-frequency limit that is why there is a large difference between the two. The filter cutoff frequency must 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).
  10. 10. 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 it is a high-pass filter. If the frequency is within the limit of funity it will allow all the frequencies above the cutoff frequency and rejects 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%. They are 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 kHz
  11. 11. Step 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? Almost the same. 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? -37.23. The roll-off is -40 dB. This what I am expecting 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 fStep 20 Measure the upper cutoff frequency (fc2) and record the value on the curveplot. 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 The curve response is like a band-pass response because active filters have an upper frequency limit on the high-pass response.
  12. 12. CONCLUSION: I conclude that the gain-frequency response of a second-order low-pass activeappears as a passive filter and high-pass active filter is like high-pass passive filter. However,they differ in roll-off and its phase frequency response. Second-order low-pass filter filters has 40dB gain decrease for a one-decade increasein frequency while Second-order high-pass filter filters has 40dB gain decrease for a one-decade decrease in frequency. Two-pole filter response on cutoff frequency is 90 degreeswhile one-pole has only 45 degrees. Active filters have lot of advantages, however, active filters has an upper-frequencylimit because of the open-loop bandwidth of op-amps. That is why active filters appears asband-pass filter at the higher frequencies.

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