1. COMPUTATION (Step 4) (Step 15) (Step 6) (Step 17) (Question – Step 6) Question (Step 17) Question – (Step 7 ) – Question (Step 18 )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 from thetotal frequency spectrum. This is normally accomplished with filters. A filter is a circuit that passes a specificrange of frequencies while rejecting other frequencies. A passive filter consists of passive circuit elements,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 cutoff frequency and reject allfrequencies above the cutoff frequency. A high-pass is designed to pass all frequencies above the cutofffrequency and reject all frequencies below the cutoff frequency. A band-pass filter passes all frequencies withina band of frequencies and rejects all other frequencies outside the band. A band-stop filter rejects allfrequencies within a band of frequencies and passes all other frequencies outside the band. A band-stop filterrejects all frequencies within a band of frequencies and passes all other frequencies outside the band. A band-stop filter is often is often referred to as a notch filter. In this experiment, 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 a
2. Bode 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 per decadeincrease in frequency. At the cutoff frequency, the capacitive reactance of capacitor C is equal to the resistanceof resistor R, causing the output voltage to be 0.707 times the input voltage (-3dB). The expected cutofffrequency (fC) of the low-pass filter in Figure 1-1, based on the circuit component value, can be calculated 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 bepractically 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 per decadedecrease 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 an ideal
3. filter.In a pole filter, as shown the Figure 1-1 and 1-2 the phase (θ) between the input and the output will changeby 90 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 FilterPROCEDURELow-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 200 Hzand 1 MHz by the Bode plotter. Sketch the curve plot in the space provided. Ad B fQuestion: Is the frequency response curve that of a low-pass filter? Explain why.= Yes it is. This filter allows the low frequency to pass and blocks the high frequency based on it cutofffrequency.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 dStep 4 Calculate the actual voltage gain (A) from the dB voltage gain (AdB)A = 0.99988 1Question: Was the voltage gain on the flat part of the frequency response curve what you expected for thecircuit in Fig 1-1? Explain why.= Yes, because VI approximately equal to Vo making the voltage gain approximately equal to 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 (fC) based on the circuit component values in Figure 1-1.fC = 7.958 kHz
4. Question: How did the calculated value for the cutoff frequency compare with the measured value recorded onthe curve plot?= The difference is exactly 0.29%.Step 7 Move the cursor to a point on the curve that is as close as possible to ten times f C. Record the dB gainand 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 what youexpected for a single-pole (single R-C) low-pass filter?= It is what I expected. dB gain decreases 17.11 dB per decade increase in frequency. Above frequency theoutput voltage decreases 20dB/decade increase in frequency; 17.11 dB is approximately equal to 20 dB perdecade.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) betweenthe filter input and output as a function of frequency (f). Sketch the curve plot in the space provided. θ fStep 9 Move the cursor to approximately 100 Hz and 1 MHz and record the phase (θ) in degrees on the curveplot for each frequency (f). Next, move the cursor as close as possible on the curve to the cutoff frequency (f C)and phase (θ) on the curve plot.100 Hz: θ = –0.72o1MHz: θ = –89.544ofC: θ = –44.917oQuestion: 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?= It is what I expected. The phase between the input and output changes. The input and the output change88.824 degrees or 90 degrees on the frequency range and 44.917 degrees or 45 degrees.Step 10 Change the value of resistor R to 2 kΩ in Fig 1-1. Click “Magnitude” on the Bode plotter. Run thesimulation. 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.= Yes,because it decreases. The dB per decade roll-off did not change. The single pole’s roll-off will alwaysapproach 20 dB per decade in the limit of high frequency even if the resistance changes.Step 11 Change the value of capacitor C is 0.04 µF in Figure 1-1. Run the simulation. Measure the newcutoff frequency (fC) and record your answer.fC = 4.049 kHzQuestion: Did the cutoff frequency change? Did the dB per decade roll-off change? Explain.= Yes because it decreases. The dB per decade roll-off did not change. The single pole’s roll-off will alwaysapproach 20 dB per decade in the limit of high frequency even if the capacitance changes.
5. 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)Step 13 Run the simulation. Notice that the gain in dB has been plotted between the frequencies of 100Hzand 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.= It is what I expected, it passes all the frequencies above the cutoff frequency and rejects all 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 thevoltage gain 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.= At frequencies well above the cut-off frequency VO = Vi therefore the voltage gain A equals 1Step 16 Move the cursor as close as possible to the point on the curve that is 3dB down from the dB gainat 1MHz. Record the frequency (cutoff frequency, fC) on the curve plot.fC = 7.935 kHzStep 17 Calculate the expected cut of frequency (fC) based on the circuit component value in Figure 1-2fC = 7.958 kHzQuestion: How did the calculated value of the cutoff frequency compare with the measured value recorded onthe curve plot?= There is a 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 dBQuestion: 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?= The dB gain decreases 18.161 dB per decrease in frequency. It is what I expected, the frequencies below thecutoff frequency have output voltage 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 initialvalue (I) is 0 and the final value (f) is 90 o. Run the simulation again. You are looking at the phase difference (θ) obetween the filter input and output as a function of frequency (f). Sketch the curve plot in the space provided
6. 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 = 89.28at 1MHz: θ = 0.456oat fC(7.935 kHz): θ = 44.738oQuestion: Was the phase at the cutoff frequency (fC) what you expected for a single-pole (single R-C) high passfilter?It is what I expected, the input and the output change 89.824 degrees almost 90 degrees on the frequencyrange and 44.738 degrees almost degrees.Did the phase change with frequency? Is this expected for an R-C high pass filter?= Yes the phase between the input and output changes. It is expected.Step 21 Change the value of resistor R to 2 kΩ in Figure 1-2. 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 change? Did the dB per decade roll-off change? Explain.= Yes ,it decreases. The roll-off did not change. Roll-off is still the same even if resistance changes.Step 22 Change the value of the capacitor C to 0.04µF in Figure 1-2. Run the simulation/ measure thecutoff frequency (fC) and record you answer.fC = 4.049 kHzQuestion: Did the cutoff frequency change? Did the dB per decade roll-off change? Explain.= Yes it decreases. The roll-off did not change. Roll-off is still the same even if capacitance changes.CONCLUSION After this experiment, I could say that the cut-off fC is the basis of a filter.It filters the frequency ifit will ve allowed or rejected. In low-pass filter it only allows the frequencies below the cutoff frequency. On theother hand, the high-pass filter only allows the frequencies above the cutoff frequency.The voltage gain of low-pass filter below the cutoff frequency is almost equal to 1 because V o = Vi. The voltagegain in high-pass filter becomes 1 if it is well above the fC because Vo = Vi. Frequencies above cutoff (for the low-pass filter) the dB per decade roll-off decreases by 20 dB perdecade increase in frequency. Frequencies below f C (for high-pass filter) ) the dB per decade roll-off decreasesby 20 dB per decade decrease in frequency. The Phase response for a first-order low-pass filter and high-passfilter, vOUT always lags vIN by some phase angle betweeen 0 and 90°. If the resistance or capacitance changes, the cutoff frequency also changes. Cutoff is inverselyproportional to the resistance and capacitance. However, the roll-off is not affected by the resistance and thecapacitance.
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