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# Cauan (2)

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### Cauan (2)

1. 1. NATIONAL COLLEGE OF SCIENCE AND TECHNOLOGY Amafel Building, Aguinaldo Highway Dasmariñas City, Cavite Experiment No. 3 ACTIVE LOW-PASS and HIGH-PASS FILTERSCauan, Sarah Krystelle P. July 14, 2011Signal Spectra and Signal Processing/BSECE 41A1 Score: Engr. Grace Ramones Instructor
2. 2. 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.
3. 3. SAMPLE COMPUTATIONS:Step 3 Computation of voltage gain based on measured value: AdB = 20 log A 4.006 = 20 log AStep 4 Computation of voltage gain based on circuit:Q in Step 4 Percentage DifferenceStep 6 Computation of cutoff frequency:Q in Step 6 Percentage DifferenceQ in Step 7 Roll –Off -36.146 dB – 0.968 dB = -37.106 dBStep 14 Calculated the actual voltage gain (A) from the dB gain A = 1.54
4. 4. Step 15 Computation of expected voltage gain based on circuit:Q in Step 15 Computation of percentage difference:Step 17 Computation of expected cutoff frequency:Q in Step 17 Computation of percentage difference:Q in Step 18 Roll –Off -36.489 dB – 0.741 dB = -37.23 dB
5. 5. 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 Awhere A = Vo/Vin.