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  • 1. NATIONAL COLLEGE OF SCIENCE AND TECHNOLOGY Amafel Building, Aguinaldo Highway Dasmariñas City, Cavite Experiment No. 3 ACTIVE LOW-PASS and ACTIVE HIGH-PASS FILTERSBalane, Maycen M. July 14, 2011Signal Spectra and Signal Processing/BSECE 41A1 Score: Engr. Grace Ramones Instructor
  • 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. SAMPLE COMPUTATIONS: Computation for Voltage Gain based on measured AdB: AdB = 20 log A 4.006 = 20 log A AdB = 20 log A 3.776 = 20 log A A = 1.5445 Computation for Voltage Gain based circuit components: Computation for percentage difference: Computation cut-off frequency based on circuit components:
  • 4. 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 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 with passiveelements. Active filters have several advantages over passive filters. The passive elementsprovide frequency selectivity and the active devices provide voltage gain, high inputimpedance, and low output impedance. The voltage gain reduces attenuation of the signalby the filter, the high input prevents excessive loading of the source, and the low outputimpedance 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 ofactive filters is the upper-frequency limit due to the limited open-loop bandwidth (funity) ofop-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(fc). Because op-amps have a limited open-loop bandwidth (unity-gain frequency, funity),high-pass active filters have an upper-frequency limit on the high-pass response, making itappear as a band-pass filter with a very wide bandwidth. Therefore, active filters must beused in applications where the unity-gain frequency (funity) of the op-amp is high enough sothat it does not fall within the frequency range of the application. For this reason, activefilters are mostly used in low-frequency applications. The most common way to describe the frequency response characteristics of a filteris to plot the filter voltage gain (Vo/Vin) in dB as a function of frequency (f). The frequencyat which the output power gain drops to 50% of the maximum value is called the cutofffrequency (fc). When the output power gain drops to 50%, the voltage gain drops 3 dB(0.707 of the maximum value). When the filter dB voltage gain is plotted as a function offrequency using straight lines to approximate the actual frequency response, it is called aBode plot. A Bode plot is an ideal plot of filter frequency response because it assumes thatthe voltage gain remains constant in the passband until the cutoff frequency is reached, andthen drops in a straight line. The filter network voltage gain in dB is calculated from theactual voltage gain (A) using the equation AdB = 20 log A
  • 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, actualfilters 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. Eachadditional pole (R-C circuit) will cause the filter to roll off an additional -20 dB/decade. In aone-pole filter, the phase between the input and the output will change by 90 degreesover the frequency range and be 45 degrees at the cutoff frequency. In a two-pole filter,the phase will change by 180 degrees over the frequency range and be 90 degrees at thecutoff frequency. Three basic types of response characteristics that can be realized with most activefilters are Butterworth, Chebyshev, and Bessel, depending on the selection of certain filtercomponent 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. Because ofthe nonlinear phase response, a pulse waveshape applied to the input of a Butterworthfilter will have an overshoot on the output. Filters with a Butterworth response arenormally used in applications where all frequencies in the passband must have the samegain. The Chebyshev filter provides a ripple amplitude response in the passband and aroll-off better than -20 dB/decade/pole with a less linear phase response than theButterworth filter. Filters with a Chebyshev response are most useful when a rapid roll-offis required. The Bessel filter provides a flat amplitude response in the passband and a roll-off of less than -20 dB/decade/pole with a linear phase response. Because of its linearphase response, the Bessel filter produces almost no overshoot on the output with a pulseinput. For this reason, filters with a Bessel response are the most effective for filteringpulse waveforms without distorting the waveshape. Because of its maximally flat responsein the passband, the Butterworth filter 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 passband voltagegain of 1.586 to produce the Butterworth response. Therefore,and
  • 6. At the cutoff frequency of both filters, the capacitive reactance of each capacitor (C)is equal 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 FilterFIGURE 3 – 2 Second-order (2-pole) Sallen-Key High-Pass Butterworth Filter
  • 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 expected for the high-pass filter to allow the frequencies below the fc and blocks all frequencies above the cut-off region.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. Av = 1.586Question: How did the measured voltage gain in Step 3 compared with the calculated voltage gain in Step 4? The % difference of the two values is 0 %. They are equal.
  • 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 = 5305.16477 HzQuestion: 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 2.98 %. The values of the two 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 = -36.202 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? The roll off decrease 37.167dB/decade which is -40 dB. This is expected for a two-pole filter.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
  • 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 = -91.293oQuestion: Was the phase shift between input and output at the cutoff frequency what you expected for a two-pole low-pass filter? Yes because it is expected to have 90 degrees phase shift 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 both places. 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 measured values 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. There was such a large difference between the calculated and measured values of the cutoff frequency because the fC exceeds the unity-gain bandwidth, funity of the opamp.
  • 10. 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, it is expected for the high-pass filter to allow the frequencies above the fc and blocks all frequencies below the cut-off region.Step 14 Calculate the actual voltage gain (A) from the measured dB gain. A = 1.5445Step 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 measured and calculated voltage percentage difference is 00.26%
  • 11. 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 % and yet still approximately the same.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? The roll-off is 37.23dB decrease per decade which is -40 dB. This 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
  • 12. Step 20 Measure the upper cutoff frequency (fc2) and record the value on the curve plot. fc = 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 This is because op-amps have a limited open-loop bandwidth (unity- gain frequency, funity), high-pass active filters have an upper- frequency limit on the high-pass response, making it appear as a band-pass filter with a very wide bandwidth
  • 13. CONCLUSION: After performing the experiment I conclude that active filters uses op-amps (orother active devices) combined with other passive elements. This filter also providesvoltage gain, high input impedance, and low output impedance that is why it is better thanthe passive filter. However, because of the limited value of the unity-gain bandwidth funityof the op-amp it is not recommended in high frequencies. Moreover, a two-pole low-pass filter is expected to have 40 dB decrease per decadeincrease while a two-pole high-pass filter is expected to have 40 dB decrease per decadedecrease. Also, the phase shift of these kind of filters has 90O- phase shift. Finally, I notice that with a very wide bandwidth, high-pass response appears as aband-pass filter. This is because of the limited open-loop bandwidth of the op-amps.