Exp passive filter (2)
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Exp passive filter (2) Exp passive filter (2) Document Transcript

  • 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.
  • COMPUTATIONSolution for Step 4Solution for Step 6Solution for Question on Step 6Solution for Question on Step 7Solution for Step 15Solution for Step 17
  • Solution for Question on Step 17Solution for Question on 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 frequenciesfrom the total frequency spectrum. This is normally accomplished with filters. A filter is a circuitthat passes a specific range of frequencies while rejecting other frequencies. A passive filterconsists of passive circuit elements, such as capacitors, inductors, and resistors. There are four basictypes of filters, low-pass, high-pass, band-pass, and band-stop. A low-pass filter is designed to passall frequencies below the cutoff frequency and reject all frequencies above the cutoff frequency. Ahigh-pass is designed to pass all frequencies above the cutoff frequency and reject all frequenciesbelow the cutoff frequency. A band-pass filter passes all frequencies within a band of frequenciesand rejects all other frequencies outside the band. A band-stop filter rejects all frequencies within aband of frequencies and passes 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 filter is often is often referred to as a notch filter. In this experiment, you will study low-passand high-pass filters.The most common way to describe the frequency response characteristics of a filter is to plot thefilter voltage gain (Vo/Vi) in dB as a function of frequency (f). The frequency at which the outputpower gain drops to 50% of the maximum value is called the cutoff frequency (fC). When the outputpower gain drops to 50%, the voltage gain drops 3 dB (0.707 of the maximum value). When thefilter dB voltage gain is plotted as a function of frequency on a semi log graph using straight lines toapproximate the actual frequency response, it is called a Bode plot. A bode plot is an ideal plot offilter frequency response 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 network voltage indB is calculated from the actual voltage gain (A) using the equation AdB = 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, thecapacitive reactance of capacitor C is much higher than the resistance of resistor R, causing theoutput voltage to be practically equal to the input voltage (A=1) and constant with the variations infrequency. At frequencies well above the cut-off frequency, the capacitive reactance of capacitor Cis much lower than the resistance of resistor R and decreases with an increase in frequency, causingthe output voltage to decrease 20 dB per decade increase in frequency. At the cutoff frequency, the
  • capacitive reactance of capacitor C is equal to the resistance of resistor R, causing the output voltageto be 0.707 times the input voltage (-3dB). The expected cutoff frequency (fC) of the low-pass filterin Figure 1-1, based on the circuit component value, can be calculated from XC = 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, thecapacitive reactance of capacitor C is much lower than the resistance of resistor R causing the outputvoltage to be practically equal to the input voltage (A=1) and constant with the variations infrequency. At frequencies well below the cut-off frequency, the capacitive reactance of capacitor Cis much higher than the resistance of resistor R and increases with a decrease in frequency, causingthe output voltage to decrease 20 dB per decade decrease in frequency. At the cutoff frequency, thecapacitive reactance of capacitor C is equal to the resistance of resistor R, causing the output voltageto be 0.707 times the input voltage (-3dB). The expected cutoff frequency (fC) of the high-pass filterin Figure 1-2, based on the circuit component value, can be calculated fromFig 1-1 Low-Pass R-C FilterWhen the frequency at the input of a low-pass filter increases above the cutoff frequency, the filteroutput drops at a constant rate. When the frequency at the input of a high-pass filter decreases belowthe cutoff frequency, the filter output voltage also drops at a constant rate. The constant drop in filteroutput voltage per decade increase (x10), or decrease ( 10), in frequency is called roll-off. An ideallow-pass or high-pass filter would have an instantaneous drop at the cut-off frequency (fC), with fullsignal level on one side of the cutoff frequency and no signal level on the other side of the cutofffrequency. Although the ideal is not achievable, actual filters roll-off at -20dB/decade per pole (R-Ccircuit). A one-pole filter has one R-C circuit tuned to the cutoff frequency and rolls off at -20dB/decade. At two-pole filter has two R-C circuits tuned to the same cutoff frequency and rolls off
  • at -40dB/decade. Each additional pole (R-C circuit) will cause the filter to roll-off an additional -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 willchange by 90 degrees and over the frequency range and be 45 degrees at the cutoff frequency. In atwo-pole filter, the phase (θ) will change by 180 degrees over the frequency range and be 90 degreesat 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 Hz 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. Since capacitive reactance decreases with frequency, the R-C circuit in Fig 1-1discriminates against high frequencies. In other words, because it is a passive low-pass filter, it isexpected to reject the higher frequencies or frequencies above the cut off frequency.Step 3 Move the cursor to a flat part of the curve at a frequency of approximately 100 Hz. Record the voltage gain 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 1Question: Was the voltage gain on the flat part of the frequency response curve what you expectedfor the circuit in Fig 1-1? Explain why.Answer: Yes, because at frequencies well below the cut-off frequency or at low frequencies, theoutput voltage and the input voltage are almost equal making the voltage gain A 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 (measured) = 7.935 kHzStep 6 Calculate the expected cutoff frequency (fC) based on the circuit component values in Figure 1-1. fC (computed) = 7.958 kHz
  • Question: How did the calculated value for the cutoff frequency compare with the measured valuerecorded on the curve plot?Answer: The computed and measured values of the cut-off frequency are almost equal. Theirpercentage difference is only 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 gain 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 itwhat you expected for a single-pole (single R-C) low-pass filter?Answer: The output voltage decreases 17.11 dB per decade increase in frequency. Yes, because atwell above frequency the output voltage decreases 20dB/decade increase in frequency, therefore itis expected that the frequencies above the cutoff frequency (but not at well above frequency) havevoltage outputs almost decrease 20dB/decade increase in frequency.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 fStep 9 Move the cursor to approximately 100 Hz and 1 MHz and record the phase (θ) in degrees on the curve plot for each frequency (f). Next, move the cursor as close as possible on the curve to the cutoff frequency (fC) and phase (θ) on the curve plot. at 100 Hz: θ = –0.72o at 1MHz: θ = –89.544o at fC(7.935 kHz): θ = –44.917o
  • Question: Was the phase at the cutoff frequency what you expected for a singles-pole (single R-C)low-pass filter?Answer: Yes, It was what I expected; the input and the output change 88.824 degrees 90degrees on the frequency range and 44.917 degrees 45 degrees.Did the phase change with frequency? Is this expected for an R-C low-pass filter?Answer: Yes, the phase changes. Yes in R-C low pass filter the phase between the input andoutput changes.Step 10 Change the value of resistor R to 2 kΩ in Fig 1-1. Click “Magnitude” on the Bode plotter. Run the 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, the cut-off frequency changes. However, the dB per decade roll-off still the same.For any value of resistance of resistor R, the first order or single pole’s power roll-off will alwaysapproach 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 the new 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, compare to Fig1-1 the cut-off frequency changes. However, the dB per decade roll-off still the same. Just like what I explained earlier, for any value of capacitive reactance ofcapacitor C, the first order (single pole) power roll-off will always approach 20 dB per decade inthe 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)Step 13 Run the simulation. Notice that the gain in dB has been plotted between the frequencies of 100Hz 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, capacitive reactance decreases with frequency, the RC circuit in Fig 1-2discriminates against low frequencies. It passes all the frequencies above the cutoff frequencywhile blocking or rejecting the frequencies below the cutoff frequency.Step 14 Move the cursor to a flat part of the curve at a frequency of approximately 1 MHz Record the voltage 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 expectedfor the circuit in Figure 1-2? Explain why.Answer: Yes, because at frequencies well above the cut-off frequency, the output voltage and theinput voltage are equal therefore the voltage gain A equals 1.Step 16 Move the cursor as close as possible to the point on the curve that is 3dB down from the dB gain at 1MHz. Record the frequency (cutoff frequency, fC) on the curve plot. fC = 7.935 kHz
  • Step 17 Calculate the expected cut of frequency (fC) 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 valuerecorded on the curve plot?Answer: The values of computed and the measured cutoff frequency are almost equal having apercentage 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 dB gain 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 you expected for a single-pole (single R-C) high-pass filter?Answer: The output voltage decreases 18.161 dB per decrease in frequency. Yes, it is expected that the frequencies below the cutoff frequency have voltage outputs 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 (I) is 0o and the final value (f) is 90o. 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
  • Step 20 Move the cursor to approximately 100 Hz and 1 MHz and record the phase (θ) in degrees on the curve plot for each frequency (f). Next, move the cursor as close as possible on the curve to the cutoff frequency (fC). Record the frequency (fC) and phase (θ). at 100 Hz: θ = 89.28 at 1MHz: θ = 0.456o at 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 pass filter?Yes, it was what I expected; the input and the output change 89.824 degrees 90 degrees on thefrequency range and 44.738 degrees 45 degrees.Did the phase change with frequency? Is this expected for an R-C high pass filter?Answer: Yes, the phase changes. Yes in R-C low pass filter the phase between the input andoutput changes.Step 21 Change the value of resistor R to 2 kΩ in Figure 1-2. Click “Magnitude” on the Bode plotter. Run the 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.Answer: Yes, the cut-off frequency changes. Meanwhile, the roll-off did not change. For anyvalue of resistance of resistor R, the first order (single pole) dB per decade roll-off will alwaysapproach 20 dB per decade in the limit of low frequency.Step 22 Change the value of the capacitor C to 0.04µF in Figure 1-2. Run the simulation/ measure the cutoff frequency (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, compare to Figure 1-2 the cutoff frequency changes. Meanwhile, the roll-off did notchange. For any value of capacitive reactance of capacitor C, the first order (single pole) powerroll-off will always approach 20 dB per decade in the limit of low frequency.
  • CONCLUSIONAfter performing the experiment I concluded that the cutoff frequency is the basis if the filter willpass a frequency. In R-C one-pole low-pass filter is a filter that passes low-frequency signals butreduces the amplitude of signals with frequencies higher than the cutoff frequency. On the otherhand, the high-pass filter reduces the amplitude of signals with frequencies lower than the cut-offfrequency and passes all the frequencies higher than the cutoff frequency.Moreover, I observe that the voltage gain of low-pass filter at well below the cutoff frequency (fC) isalmost equal to 1, while the voltage gain in high-pass filter becomes 1 if it is well above the fC. I also took notice in low-pass filter that at well above fC (1 MHz) the dB per decade roll-offdecreases by 20 dB per decade increase in frequency; yet my answer in Step 7 Question is 17.11 dBper decade decrease in frequency. It is because there is no 79.35 kHz on the Bode Plotter. The BodePlotter already had assigned values so I only move the cursor as close as possible to 79.35 kHz.Anyway, the output voltages decrease 20 dB per decade increase in frequency at frequencies abovethe cut-off frequency. Meanwhile, in high-pass filter the frequencies below fC, the output voltagedecrease 20 dB/decade decrease in frequency.Furthermore, the phase between the input and the output changes approximately equal to 90 degreesover the frequency range, and approximately equal to 45 degrees at the cutoff frequency.Lastly, the value of the components affects the value of the cutoff frequency. As the capacitance orresistance increases, the cutoff frequency decreases. Therefore the resistance and the capacitance areinversely proportional to the cutoff frequency. However, roll-off frequency was not affected by thevalue of the resistance and the capacitance.