Your SlideShare is downloading. ×
Exp passive filter (7)
Upcoming SlideShare
Loading in...5
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×
Saving this for later? Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime – even offline.
Text the download link to your phone
Standard text messaging rates apply

Exp passive filter (7)

222

Published on

Published in: Technology, Business
0 Comments
1 Like
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total Views
222
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
Downloads
1
Comments
0
Likes
1
Embeds 0
No embeds

Report content
Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
No notes for slide

Transcript

  • 1. NATIONAL COLLEGE OF SCIENCE AND TECHNOLOGY Amafel Bldg. Aguinaldo Highway Dasmariñas City, Cavite EXPERIMENT # 1 Passive Low-Pass and High-Pass FilterBani, Arviclyn C. 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 4Step 6Question – Step 6Question – Step 7 –Step 15
  • 3. Step 17Question – Step 17Question – Step 18DATA 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 offrequencies from the total frequency spectrum. This is normally accomplished with filters. A filter isa circuit that passes a specific range of frequencies while rejecting other frequencies. A passivefilter consists of passive circuit elements, such as capacitors, inductors, and resistors. There arefour basic types 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 frequencies above thecutoff frequency. A high-pass is designed to pass all frequencies above the cutoff frequency and
  • 4. reject all frequencies below the cutoff frequency. A band-pass filter passes all frequencies within aband 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. Aband-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 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 (f C). 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 thepassband until the cutoff frequency is reached, and then drops in a straight line. The filter networkvoltage 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, 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,causing the output voltage to decrease 20 dB per decade increase in frequency. At the cutofffrequency, the capacitive reactance of capacitor C is equal to the resistance of resistor R, causingthe output voltage to be 0.707 times the input voltage (-3dB). The expected cutoff frequency (fC) ofthe 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, thecapacitive reactance of capacitor C is much lower 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 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,the capacitive reactance of capacitor C is equal to the resistance of resistor R, causing the output
  • 5. voltage to be 0.707 times the input voltage (-3dB). The expected cutoff frequency (fC) of the high-pass filter in 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 decreasesbelow the cutoff frequency, the filter output voltage also drops at a constant rate. The constantdrop in filter output voltage per decade increase (x10), or decrease ( 10), in frequency is calledroll-off. An ideal low-pass or high-pass filter would have an instantaneous drop at the cut-offfrequency (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 filters roll-off at -20dB/decade per pole (R-C circuit). A one-pole filter has one R-C circuit tuned to the cutofffrequency and rolls off at -20dB/decade. At two-pole filter has two R-C circuits tuned to the samecutoff frequency and rolls off at -40dB/decade. Each additional pole (R-C circuit) will cause the filterto roll-off an additional -20dB/decade. Therefore, an R-C filter with more poles (R-C circuits) moreclosely approaches an ideal 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 90degrees at the cutoff frequency.
  • 6. 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 thefrequencies 200 Hz and 1 MHz by the Bode plotter. Sketch the curve plot in the space provided. AdB f
  • 7. Question: Is the frequency response curve that of a low-pass filter? Explain why.=I expected it. This filter pass the low frequency and blocks the high frequency depending on thecutoff frequency.Step 3 Move the cursor to a flat part of the curve at a frequency of approximately 100 Hz. Recordthe 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 youexpected for the circuit in Fig 1-1? Explain why.= I expected it, at below cutoff frequency, the VI is approximately equal to Vo making the voltagegain 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 at100 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 Figure1-1.fC = 7.958 kHzQuestion: How did the calculated value for the cutoff frequency compare with the measured valuerecorded on the curve plot?= Almost the same and only has 0.29% difference.Step 7 Move the cursor to a point on the curve that is as close as possible to ten times f C. Recordthe 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?= The circuits roll-off is at rate of 17.11 dB decrease per decade increase in frequency. I expected itbecause, above frequency the output voltage decreases 20dB/decade increase in frequency; 17.11dB is approximately equal to 20 dB per decade.
  • 8. Step 8 Click “Phase” on the Bode plotter to plot the phase curve. Make sure that the vertical axisinitial value (1) is -90 and the final value (F) is 0. Run the simulation again. You are looking at thephase difference (θ) between the filter input and output as a function of frequency (f). Sketch thecurve 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 onthe curve plot for each frequency (f). Next, move the cursor as close as possible on the curve to thecutoff frequency (fC) 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-pass filter? Did the phase change with frequency? Is this expected for an R-C low-pass filter?= I expected it. The phase changes between the input and output. I expected it because the inputand the output change 88.824 degrees or 90 degrees on the frequency range and 44.917 degrees or45 degrees.Step 10 Change the value of resistor R to 2 kΩ in Fig 1-1. Click “Magnitude” on the Bodeplotter. Run the simulation. Measure the cutoff frequency (fC) and record your answer.fC = 4.049 kHz
  • 9. Question: Did the cutoff frequency changes? Did the dB per decade roll-off changes? Explain.= The cutoff changes, as a matter of fact it decreases. The dB per decade roll-off did not change. Thesingle pole’s roll-off will always approach 20 dB per decade in the limit of high frequency even if theresistance changes.Step 11 Change the value of capacitor C is 0.04 µF in Figure 1-1. Run the simulation. Measurethe 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, as a matter of fact it decreases. The dB per decade roll-off did not change. Thesingle pole’s roll-off will always approach 20 dB per decade in the limit of high frequency even if thecapacitance changes.High-Pass FilterStep 12 Open circuit file FIG 1-2. Make sure that the following Bode plotter settings areselected: 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 thefrequencies 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.= I expected it. This filter pass the high frequency and blocks the low frequency depending on thecutoff frequency.
  • 10. Step 14 Move the cursor to a flat part of the curve at a frequency of approximately 1 MHzRecord 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.= Yes I expected it, frequencies 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 3dB down fromthe dB gain at 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 inFigure 1-2fC = 7.958 kHzQuestion: How did the calculated value of the cutoff frequency compare with the measured valuerecorded 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 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 itwhat you expected for a single-pole (single R-C) high-pass filter?= The dB gain decreases 18.161 dB per decrease in frequency. I expected it, because the frequenciesbelow the cutoff frequency have output voltage almost decrease 20dB/decade decrease infrequency.Step 19 Click “Phase” on the Bode plotter to plot the phase curve. Make sure that thevertical axis initial value (I) is 0o and the final value (f) is 90o. Run the simulation again. You arelooking 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
  • 11. θ fStep 20 Move the cursor to approximately 100 Hz and 1 MHz and record the phase (θ) indegrees on the curve plot for each frequency (f). Next, move the cursor as close as possible on thecurve to the cutoff frequency (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 pass filter?It is what I expected, the input and the output change 89.824 degrees almost 90 degrees on thefrequency 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 high pass filterStep 21 Change the value of resistor R to 2 kΩ in Figure 1-2. Click “Magnitude” on the Bodeplotter. Run the simulation. Measure the cutoff frequency (fC) and record your answer.fC = 4.049 kHz
  • 12. Question: Did the cutoff frequency change? Did the dB per decade roll-off change? Explain.= The cutoff changes, as a matter of fact it decreases. The dB per decade roll-off did not change. Thesingle pole’s roll-off will always approach 20 dB per decade in the limit of low frequency even if theresistance changes.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.= The cutoff changes, as a matter of fact it decreases. The dB per decade roll-off did not change. Thesingle pole’s roll-off will always approach 20 dB per decade in the limit of low frequency even if thecapacitance changes.
  • 13. CONCLUSIONThe following conclusions are made after experimenting. This will compare the low-pass filter andthe high-pass filter:The frequencies that allowed by the filter: (Low-Pass Filter) Allows the frequencies below the cut-off frequency and blocks the frequencies above the cut-off frequency. (High-Pass Filter) Allows the frequencies above the cut-off frequency and blocks the frequencies below the cut-off frequency.Voltage Gain: Both Low Pass and High Pass has a voltage gain of 1. VO = VI (Low-Pass Filter) The voltage gain at well below the cutoff frequency is almost equal to 1; (High-Pass Filter) The voltage gain becomes 1 if it is well above the fCRoll-off: (Low-Pass Filter) decreases by 20 dB per decade increase in frequency. (High-Pass Filter) decreases by 20 dB per decade decrease in frequency.Phase (Low-Pass and High Pass Frequency) The phase of low-pass and high-pass between the input and the output changes 90 degrees over the frequency range and 45 degrees at the cutoff frequency.When Resistance and Capacitance Changes: (Effect on Cutoff for Low-Pass and High-Pass Filter) If the resistance or capacitance changes, the cutoff frequency also changes. Cutoff is inversely proportional to the resistance and capacitance. (Effect on Roll-off for Low-Pass and High-Pass Filter) Roll-off is not affected by the resistance and the capacitance.

×