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Exp passive filter (6)

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  • 1. NATIONAL COLLEGE OF SCIENCE AND TECHNOLOGY Amafel Bldg. Aguinaldo Highway Dasmariñas City, Cavite EXPERIMENT # 1 Passive Low-Pass and High-Pass FilterAgdon, Berverlyn B. June 28, 2011Signal Spectra and Signal Processing/ BSECE 41A1 Score: Eng’r. Grace Ramones Instructor
  • 2. OBJECTIVES1. Plot the gain frequency response of a first-order (one-pole) R-C low-pass filter.2. Determine the cutoff frequency and roll-off of an R-C first-order (one-pole) low-pass filter.3. Plot the phase-frequency of a first-order (one-pole) low-pass filter.4. Determine how the value of R and C affects the cutoff frequency of an R-C low-pass filter.5. Plot the gain-frequency response of a first-order (one-pole) R-C high pass filter.6. Determine the cutoff frequency and roll-off of a first-order (one-pole) R-C high pass filter.7. Plot the phase-frequency response of a first-order (one-pole) high-pass filter.8. Determine how the value of R and C affects the cutoff frequency of an R-C high pass filter.COMPUTATIONStep 4 Step 15Step 6 Step 17Question – Step 6 Question – Step 17Question – Step 7 Question – Step 18 –
  • 3. DATA SHEETMATERIALSOne function generatorOne dual-trace oscilloscopeCapacitors: 0.02 µF, 0.04µFResistors: 1 kΩ, 2 kΩTHEORYIn electronic communication systems, it is often necessary to separate a specific range of frequencies fromthe total frequency spectrum. This is normally accomplished with filters. A filter is a circuit that passes aspecific range of frequencies while rejecting other frequencies. A passive filter consists of passive circuitelements, such as capacitors, inductors, and resistors. There are four basic types of filters, low-pass, high-pass, band-pass, and band-stop. A low-pass filter is designed to pass all frequencies below the cutofffrequency and reject all frequencies above the cutoff frequency. A high-pass is designed to pass allfrequencies above the cutoff frequency and reject all frequencies below the cutoff frequency. A band-passfilter passes all frequencies within a band of frequencies and rejects all other frequencies outside the band. Aband-stop filter rejects all frequencies within a band of frequencies and passes all other frequencies outsidethe band. A band-stop filter rejects all frequencies within a band of frequencies and passes all otherfrequencies outside the band. A band-stop filter is often is often referred to as a notch filter. In thisexperiment, you will study low-pass and high-pass filters.The most common way to describe the frequency response characteristics of a filter is to plot the filter voltagegain (Vo/Vi) in dB as a function of frequency (f). The frequency at which the output power gain drops to 50% ofthe maximum value is called the cutoff frequency (f C). When the output power gain drops to 50%, the voltagegain drops 3 dB (0.707 of the maximum value). When the filter dB voltage gain is plotted as a function offrequency on a semi log graph 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 that the voltage gainremains constant in the passband until the cutoff frequency is reached, and then drops in a straight line. Thefilter network voltage in dB is calculated from the actual voltage gain (A) using the equationAdB = 20 log Awhere A = Vo/ViA low-pass R-C filter is shown in Figure 1-1. At frequencies well below the cut-off frequency, the capacitivereactance of capacitor C is much higher than the resistance of resistor R, causing the output voltage to bepractically equal to the input voltage (A=1) and constant with the variations in frequency. At frequencies wellabove the cut-off frequency, the capacitive reactance of capacitor C is much lower than the resistance ofresistor R and decreases with an increase in frequency, causing the output voltage to decrease 20 dB perdecade increase in frequency. At the cutoff frequency, the capacitive reactance of capacitor C is equal to theresistance of resistor R, causing the output voltage to be 0.707 times the input voltage (-3dB). The expectedcutoff frequency (fC) of the low-pass filter in Figure 1-1, based on the circuit component value, can becalculated fromXC = RSolving for fC produces the equationA high-pass R-C filter is shown in figure 1-2. At frequencies well above the cut-off frequency, the capacitivereactance of capacitor C is much lower than the resistance of resistor R causing the output voltage to be
  • 4. practically equal to the input voltage (A=1) and constant with the variations in frequency. At frequencies wellbelow the cut-off frequency, the capacitive reactance of capacitor C is much higher than the resistance ofresistor R and increases with a decrease in frequency, causing the output voltage to decrease 20 dB perdecade decrease in frequency. At the cutoff frequency, the capacitive reactance of capacitor C is equal to theresistance of resistor R, causing the output voltage to be 0.707 times the input voltage (-3dB). The expectedcutoff frequency (fC) of the high-pass filter in Figure 1-2, based on the circuit component value, can becalculated fromFig 1-1 Low-Pass R-C FilterWhen the frequency at the input of a low-pass filter increases above the cutoff frequency, the filter outputdrops at a constant rate. When the frequency at the input of a high-pass filter decreases below the cutofffrequency, the filter output voltage also drops at a constant rate. The constant drop in filter output voltage perdecade increase (x10), or decrease ( 10), in frequency is called roll-off. An ideal low-pass or high-pass filterwould have an instantaneous drop at the cut-off frequency (fC), with full signal level on one side of the cutofffrequency and no signal level on the other side of the cutoff frequency. Although the ideal is not achievable,actual filters roll-off at -20dB/decade per pole (R-C circuit). A one-pole filter has one R-C circuit tuned to thecutoff frequency and rolls off at -20dB/decade. At two-pole filter has two R-C circuits tuned to the same cutofffrequency and rolls off at -40dB/decade. Each additional pole (R-C circuit) will cause the filter to roll-off anadditional -20dB/decade. Therefore, an R-C filter with more poles (R-C circuits) more closely approaches anideal filter.In a pole filter, as shown the Figure 1-1 and 1-2 the phase (θ) between the input and the output will change by90 degrees and over the frequency range and be 45 degrees at the cutoff frequency. In a two-pole filter, thephase (θ) will change by 180 degrees over the frequency range and be 90 degrees at the cutoff frequency.Fig 1-2 High-Pass R-C Filter
  • 5. PROCEDURELow-Pass FilterStep 1 Open circuit file FIG 1-1. Make sure that the following Bode plotter settings are selected: Magnitude,Vertical (Log, F=0 dB, I=–40dB), Horizontal (Log, F=1 MHz, I=100 Hz)Step 2 Run the simulation. Notice that the voltage gain in dB has been plotted between the frequencies 200Hz and 1 MHz by the Bode plotter. Sketch the curve plot in the space provided. AdB fQuestion: Is the frequency response curve that of a low-pass filter? Explain why.Answer: Yes. by definition, a low-pass filter is a circuit offering easy passage to low-frequency signals anddifficult passage to high-frequency signals.Step 3 Move the cursor to a flat part of the curve at a frequency of approximately 100 Hz. Record the voltagegain in dB on the curve plot.AdB = -0.001 dBStep 4 Calculate the actual voltage gain (A) from the dB voltage gain (AdB)A = 0.99988 1
  • 6. Question: Was the voltage gain on the flat part of the frequency response curve what you expected for thecircuit in Fig 1-1? Explain why.Answer: Yes, at frequencies below the cut-off frequency VO and Vi are almost equal therefore the voltage gainA equals 1.Step 5 Move the cursor as close as possible to a point on the curve that is 3dB down from the dB at 100 Hz.Record the frequency (cut-off frequency, fC) on the curve plot.fC = 7.935 kHzStep 6 Calculate the expected cutoff frequency (f C) based on the circuit component values in Figure 1-1.fC = 7.958 kHzQuestion: How did the calculated value for the cutoff frequency compare with the measured value recordedon the curve plot?Answer: The % difference of the measure and the computed fC is 0.29%. Therefore, there is only littledifference.Step 7 Move the cursor to a point on the curve that is as close as possible to ten times f C. Record the dBgain and frequency (f2) on the curve plot. AdB = -20.108 dBQuestion: How much did the dB gain decrease for a one decade increase (x10) in frequency? Was it whatyou expected for a single-pole (single R-C) low-pass filter?Answer: The roll-off of the circuit is 17.11 dB per decade increase in frequency. Yes it is what I expected,because above frequency the output voltage decreases 20dB/decade increase in frequency. And 17.11 dB isalmost equal to 20 dB per decade.Step 8 Click “Phase” on the Bode plotter to plot the phase curve. Make sure that the vertical axis initial value(1) is -90 and the final value (F) is 0. Run the simulation again. You are looking at the phase difference (θ)between the filter input and output as a function of frequency (f). Sketch the curve plot in the space provided. θ f
  • 7. Step 9 Move the cursor to approximately 100 Hz and 1 MHz and record the phase (θ) in degrees on thecurve plot for each frequency (f). Next, move the cursor as close as possible on the curve to the cutofffrequency (fC) and phase (θ) on the curve plot. o100 Hz: θ = –0.72 o1MHz: θ = –89.544 ofC: θ = –44.917Question: Was the phase at the cutoff frequency what you expected for a singles-pole (single R-C) low-passfilter? Did the phase change with frequency? Is this expected for an R-C low-pass filter?Answer: Yes, the input and the output change 88.824 degrees or approximately 90 degrees on the frequencyrange and 44.917 degrees or almost equal to 45 degrees. Yes, the phase between the input and outputchanges.Step 10 Change the value of resistor R to 2 kΩ in Fig 1-1. Click “Magnitude” on the Bode plotter. Runthe simulation. Measure the cutoff frequency (fC) and record your answer.fC = 4.049 kHzQuestion: Did the cutoff frequency changes? Did the dB per decade roll-off changes? Explain.Answer: Yes it decreases. But, the dB per decade roll-off did not change. For any value of resistance ofresistor R, the single pole’s roll-off will always approach 20 dB per decade in the limit of high frequency.Step 11 Change the value of capacitor C is 0.04 µF in Figure 1-1. Run the simulation. Measure thenew cutoff frequency (fC) and record your answer.fC = 4.049 kHzQuestion: Did the cutoff frequency change? Did the dB per decade roll-off change? Explain.Answer: Yes it decreases, compare to Fig1-1 the cut-off frequency changes.. The dB per decade roll-off didnot change. It is because for any value of capacitive reactance of capacitor C, the first order power roll-off willalways approach 20 dB per decade in the limit of high frequency.High-Pass FilterStep 12 Open circuit file FIG 1-2. Make sure that the following Bode plotter settings are selected:Magnitude, Vertical (Log, F=0 dB, I=–40dB), Horizontal (Log, F=1 MHz, I=100 Hz)
  • 8. Step 13 Run the simulation. Notice that the gain in dB has been plotted between the frequencies of100Hz and 1 MHz by the Bode plotter. Sketch the curve plot in the space provided. AdB fQuestion: Is the frequency response curve that of a high-pass filter? Explain why.Answer: Yes, it passes all the frequencies above the cutoff frequency and blocks or rejects the frequenciesbelow the cutoff frequency.Step 14 Move the cursor to a flat part of the curve at a frequency of approximately 1 MHz Record the voltagegain in dB on the curve plot. AdB = 0 dBStep 15 Calculate the actual voltage gain (A) from the dB voltage gain (AdB). A=1Question: Was the voltage gain on the flat part of the frequency response curve what you expected for thecircuit in Figure 1-2? Explain why.Answer: Yes, because at frequencies well above the cut-off frequency, the voltage gain A equals 1 becauseVO = Vi.Step 16 Move the cursor as close as possible to the point on the curve that is 3dB down from the dB gain at1MHz. Record the frequency (cutoff frequency, fC) on the curve plot. fC = 7.935 kHzStep 17 Calculate the expected cut of frequency (f C) based on the circuit component value in Figure 1-2 fC = 7.958 kHzQuestion: How did the calculated value of the cutoff frequency compare with the measured value recorded onthe curve plot?Answer: The values of computed and the measured cutoff frequency has percentage difference of 0.29%.Step 18 Move the cursor to a point on the curve that is as close as possible to one-tenth fC. Record the dBgain and frequency (f2) on the curve plot. AdB = -20.159 dB
  • 9. Question: How much did the dB gain decrease for a one-decade decrease ( ) in frequency? Was it what youexpected for a single-pole (single R-C) high-pass filter?Answer: The output voltage decreases 18.161 dB per decrease in frequency. Yes, the frequencies below thecutoff frequency have VO almost decrease 20dB/decade decrease in frequency.Step 19 Click “Phase” on the Bode plotter to plot the phase curve. Make sure that the vertical axis initial value o o(I) is 0 and the final value (f) is 90 . Run the simulation again. You are looking at the phase difference (θ)between the filter input and output as a function of frequency (f). Sketch the curve plot in the space provided θ fStep 20 Move the cursor to approximately 100 Hz and 1 MHz and record the phase (θ) in degrees on thecurve plot for each frequency (f). Next, move the cursor as close as possible on the curve to the cutofffrequency (fC). Record the frequency (fC) and phase (θ).at 100 Hz: θ = 89.28 oat 1MHz: θ = 0.456 oat fC(7.935 kHz): θ = 44.738Question: Was the phase at the cutoff frequency (f C) what you expected for a single-pole (single R-C) highpass filter?Yes, the input and the output change 89.824 degrees almost 90 degrees on the frequency range and 44.738degrees almost degrees.Did the phase change with frequency? Is this expected for an R-C high pass filter?Answer: Yes. Yes the phase between the input and output changes.Step 21 Change the value of resistor R to 2 kΩ in Figure 1-2. Click “Magnitude” on the Bode plotter. Run thesimulation. Measure the cutoff frequency (f C) and record your answer.fC = 4.049 kHz
  • 10. Question: Did the cutoff frequency change? Did the dB per decade roll-off change? Explain.Answer: Yes cutoff decreases. The roll-off did not change. Roll-off is still the same for any value of resistance.Step 22 Change the value of the capacitor C to 0.04µF in Figure 1-2. Run the simulation/ measure the cutofffrequency (fC) and record you answer.fC = 4.049 kHzQuestion: Did the cutoff frequency change? Did the dB per decade roll-off change? Explain.Answer: Yes, cutoff frequency changes. The roll-off did not change. The roll-off is still 20dB per decade forany value of capacitance.CONCLUSIONAfter the experiment, I concluded the following:1. fC is the frequency where the filter decides to pass or reject a frequency.2. In low-pass filter, it allows the frequencies below the cut-off frequency and rejects the frequenciesabove the cu-off frequency.3. The AV of low-pass filter at well below the cutoff frequency (f C) is almost equal to 1. The voltage gainin high-pass filter is1 if it is well above the fC.4. In low-pass filter frequencies at well above cutoff the roll-off is dB per decade roll-off decreases by 20dB per decade increase in frequency. In high-pass filter the frequencies below f C, the output voltage decrease20 dB/decade decrease in frequency.5. The phase of low-pass and high-pass between the input and the output changes 90 degrees over thefrequency range and 45 degrees at the cutoff frequency.6. The cutoff frequency is inversely proportional to the resistance and capacitance. So if the resistanceor capacitance changes, the cutoff frequency also changes.7. The roll-off is not affected by the resistance and the capacitance.