Exp passive filter (4)
Upcoming SlideShare
Loading in...5
×
 

Exp passive filter (4)

on

  • 571 views

 

Statistics

Views

Total Views
571
Views on SlideShare
571
Embed Views
0

Actions

Likes
0
Downloads
9
Comments
0

0 Embeds 0

No embeds

Accessibility

Categories

Upload Details

Uploaded via as Microsoft Word

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

Exp passive filter (4) Exp passive filter (4) Document Transcript

  • NATIONAL COLLEGE OF SCIENCE AND TECHNOLOGY Amafel Bldg. Aguinaldo Highway Dasmariñas City, Cavite EXPERIMENT # 1 Passive Low-Pass and High-Pass FilterMaala, Michelle Anne C. June 28, 2011Signal Spectra and Signal Processing/ BSECE 41A1 Score: Eng’r. Grace Ramones Instructor
  • 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.
  • SAMPLE COMPUTATION(Step 4)(Step 6)
  • (Question – Step 6)Question – (Step 7 ) –Step 15(Step 17)Question (Step 17)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 specificrange of frequencies from the total frequency spectrum. This is normallyaccomplished with filters. A filter is a circuit that passes a specific range offrequencies while rejecting other frequencies. A passive filter consists of passivecircuit 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 isdesigned to pass all frequencies below the cutoff frequency and reject all frequenciesabove the cutoff frequency. A high-pass is designed to pass all frequencies above thecutoff frequency and reject all frequencies below the cutoff frequency. A band-passfilter passes all frequencies within a band of frequencies and rejects all otherfrequencies outside the band. A band-stop filter rejects all frequencies within a bandof frequencies and passes all other frequencies outside the band. A band-stop filterrejects all frequencies within a band of frequencies and passes all other frequenciesoutside the band. A band-stop filter is often is often referred to as a notch filter. Inthis experiment, you will study low-pass and high-pass filters.The most common way to describe the frequency response characteristics of a filteris to plot the filter voltage gain (Vo/Vi) in dB as a function of frequency (f). Thefrequency at which the output power gain drops to 50% of the maximum value is calledthe cutoff frequency (fC). When the output power gain drops to 50%, the voltage gaindrops 3 dB (0.707 of the maximum value). When the filter dB voltage gain is plottedas a function of frequency on a semi log graph using straight lines to approximate theactual frequency response, it is called a Bode plot. A bode plot is an ideal plot of filterfrequency response because it assumes that the voltage gain remains constant in thepassband until the cutoff frequency is reached, and then drops in a straight line. The
  • filter network voltage in dB is calculated from the actual voltage gain (A) using theequationAdB = 20 log Awhere A = Vo/ViA low-pass R-C filter is shown in Figure 1-1. At frequencies well below the cut-offfrequency, the capacitive reactance of capacitor C is much higher than the resistanceof resistor R, causing the output voltage to be practically equal to the input voltage(A=1) and constant with the variations in frequency. At frequencies well above thecut-off frequency, the capacitive reactance of capacitor C is much lower than the res-istance of resistor R and decreases with an increase in frequency, causing the outputvoltage 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 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 circuitcomponent 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-offfrequency, the capacitive reactance of capacitor C is much lower than the resistanceof resistor R causing the output voltage to be practically equal to the input voltage(A=1) and constant with the variations in frequency. At frequencies well below the cut-off frequency, the capacitive reactance of capacitor C is much higher than theresistance of resistor R and increases with a decrease in frequency, causing theoutput voltage to decrease 20 dB per decade decrease in frequency. At the cutofffrequency, the capacitive reactance of capacitor C is equal to the resistance ofresistor R, causing the output voltage to be 0.707 times the input voltage (-3dB). Theexpected cutoff frequency (fC) of the high-pass filter in Figure 1-2, based on thecircuit component value, can be calculated from
  • Fig 1-1 Low-Pass R-C FilterWhen the frequency at the input of a low-pass filter increases above the cutofffrequency, the filter output drops at a constant rate. When the frequency at theinput of a high-pass filter decreases below the cutoff frequency, the filter outputvoltage 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 filter would have an instantaneous drop at the cut-off frequency(fC), with full signal level on one side of the cutoff frequency and no signal level on theother side of the cutoff frequency. Although the ideal is not achievable, actual filtersroll-off at -20dB/decade per pole (R-C circuit). A one-pole filter has one R-C circuittuned to the cutoff frequency and rolls off at -20dB/decade. At two-pole filter hastwo 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 closelyapproaches an ideal filter.In a pole filter, as shown the Figure 1-1 and 1-2 the phase (θ) between the input andthe output will change by 90 degrees and over the frequency range and be 45 degreesat the cutoff frequency. In a two-pole filter, the phase (θ) will change by 180 degreesover 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 plottersettings 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 plottedbetween the frequencies 200 Hz and 1 MHz by the Bode plotter. Sketch the curveplot in the space provided. AdB f
  • Question: Is the frequency response curve that of a low-pass filter? Explain why.=Yes, Low-pass allow the frequencies below cutoff frequency and blocks thefrequencies above the cutoff frequency.Step 3 Move the cursor to a flat part of the curve at a frequency ofapproximately 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 whatyou expected for the circuit in Fig 1-1? Explain why.= I expected it, at below cutoff frequency, the VI is approximately equal to Vo makingthe voltage gain approximately equal to 1.Step 5 Move the cursor as close as possible to a point on the curve that is 3dBdown from the dB at 100 Hz. Record the frequency (cut-off frequency, fC) on thecurve plot.fC = 7.935 kHzStep 6 Calculate the expected cutoff frequency (fC) based on the circuitcomponent values in Figure 1-1.fC = 7.958 kHzQuestion: How did the calculated value for the cutoff frequency compare with themeasured value recorded on the curve plot?= Almost equal and has 0.29% difference.Step 7 Move the cursor to a point on the curve that is as close as possible to tentimes fC. Record the dB gain and frequency (f2) on the curve plot.AdB = -20.108 dB
  • Question: How much did the dB gain decrease for a one decade increase (x10) infrequency? Was it what you expected for a single-pole (single R-C) low-pass filter?= The circuit’s roll-off is 17.11 dB decrease per decade increase in frequency.It is expected because above frequency the output voltage decreases 20dB/decadeincrease in frequency; 17.11 dB is almost 20 dB per decade.Step 8 Click “Phase” on the Bode plotter to plot the phase curve. Make sure thatthe vertical axis initial value (1) is -90 and the final value (F) is 0. Run the simulationagain. You are looking at the phase difference (θ) between the filter input and outputas 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 thephase (θ) in degrees on the curve plot for each frequency (f). Next, move the cursoras close as possible on the curve to the cutoff frequency (fC) and phase (θ) on thecurve plot.@100 Hz: θ = –0.72o@1MHz: θ = –89.544o@fC: θ = –44.917o
  • Question: Was the phase at the cutoff frequency what you expected for a singles-pole (single R-C) low-pass filter? Did the phase change with frequency? Is thisexpected for an R-C low-pass filter?= The phase at the cut-off frequency is expected. The phase changes between theinput and output. It is expected for R-C low-pass filter because the input and theoutput change 88.824 degrees or 90 degrees on the frequency range and 44.917degrees or 45 degrees.Step 10 Change the value of resistor R to 2 kΩ in Fig 1-1. Click “Magnitude” onthe Bode plotter. Run the simulation. Measure the cutoff frequency (fC) and recordyour answer.fC = 4.049 kHzQuestion: Did the cutoff frequency changes? Did the dB per decade roll-off changes?Explain.= The cutoff changes it decreases as the resistance increases. The dB per decaderoll-off did not change. The single pole’s roll-off will always approach 20 dB perdecade 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 thesimulation. 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.= The cutoff changes, it decreases as the capacitance increases The dB per decaderoll-off did not change. The single pole’s roll-off will always approach 20 dB perdecade in the limit of high frequency even if the capacitance changes.
  • High-Pass FilterStep 12 Open circuit file FIG 1-2. Make sure that the following Bode plottersettings 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 betweenthe frequencies of 100Hz and 1 MHz by the Bode plotter. Sketch the curve plot in thespace provided. AdB fQuestion: Is the frequency response curve that of a high-pass filter? Explain why.= I expected it. This filter pass the high frequency and blocks the low frequencydepending on the cutoff frequency.Step 14 Move the cursor to a flat part of the curve at a frequency ofapproximately 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=1
  • Question: Was the voltage gain on the flat part of the frequency response curve whatyou expected for the circuit in Figure 1-2? Explain why.= Yes voltage gain on the flat part of the frequency response curve what I expectedfrequencies well above the cut-off frequency VO equal to Vi so the voltage gain Aequals 1Step 16 Move the cursor as close as possible to the point on the curve that is 3dBdown from the dB gain at 1MHz. Record the frequency (cutoff frequency, fC) on thecurve plot.fC = 7.935 kHzStep 17 Calculate the expected cut of frequency (fC) based on the circuitcomponent value in Figure 1-2fC = 7.958 kHzQuestion: How did the calculated value of the cutoff frequency compare with themeasured value recorded on the curve plot?= The circuit has 0.29% difference.Step 18 Move the cursor to a point on the curve that is as close as possible toone-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 ( ) infrequency? Was it what you expected for a single-pole (single R-C) high-pass filter?= The dB gain decreases 18.161 dB per decrease in frequency. It is expected for asingle-pole (single R-C) high-pass filter because the frequencies below the cutofffrequency 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 thatthe vertical axis initial value (I) is 0o and the final value (f) is 90o. Run the simulationagain. You are looking at the phase difference (θ) between the filter input and outputas 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 thephase (θ) in degrees on the curve plot for each frequency (f). Next, move the cursoras close as possible on the curve to the cutoff frequency (fC). Record the frequency(fC) and phase (θ).@ 100 Hz: θ = 89.28@ 1MHz: θ = 0.456o@ 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, the phase at the cutoff frequency (fC) what I expected for a single-pole (singleR-C) high pass filter the input and the output change 89.824 degrees almost 90degrees on the frequency range 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 in R--C highpass filter.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 recordyour answer.
  • fC = 4.049 kHzQuestion: Did the cutoff frequency change? Did the dB per decade roll-off change?Explain.= The cutoff changes it decreases as the resistance increases. The dB per decaderoll-off did not change. The single pole’s roll-off will always approach 20 dB perdecade in the limit of low frequency even if the resistance changes.Step 22 Change the value of the capacitor C to 0.04µF in Figure 1-2. Run thesimulation/ 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.= The cutoff changes it decreases as the resistance increases. The dB per decaderoll-off did not change. The single pole’s roll-off will always approach 20 dB perdecade in the limit of low frequency even if the resistance changes.
  • CONCLUSIONI conclude that filters are named according to the frequency of signals they allow topass through them. There are Low-pass filters that allow only low frequency signals topass, High-pass filters that allow only high frequency signals to pass through.Moreover, in Low-Pass Filter and High-Pass Filter the voltage gain of 1. When it comesto roll-off, in Low-Pass Filter frequencies at well above cutoff the dB per decade roll-off decreases by 20 dB per decade increase in frequency but in High-Pass Filterfrequencies below fC the dB per decade roll-off decreases by 20 dB per decadedecrease in frequency. The phase response of low-pass and high-pass between theinput and the output changes 90 degrees and 45 degrees at the cutoff frequency. Ifthe resistance or capacitance changes, the cutoff frequency also changes. Cutoff isinversely proportional to the resistance and capacitance. So if the resistance orcapacitance decreases the cutoff increases and when resistance and capacitanceincreases the cut-off decreases. Roll-off for Low-Pass and High-Pass Filter is notaffected by the resistance and the capacitance. It remain constant even theresistance and capacitance changes.