Upcoming SlideShare
×

# Exp2 passive band pass and band-stop filter

5,186 views

Published on

Published in: Business, Technology
• Full Name
Comment goes here.

Are you sure you want to Yes No
Your message goes here
• Be the first to comment

### Exp2 passive band pass and band-stop filter

1. 1. NATIONAL COLLEGE OF SCIENCE AND TECHNOLOGY Amafel Bldg. Aguinaldo Highway Dasmariñas City, Cavite EXPERIMENT # 2 Passive Band-Pass and Band-Stop FilterCauan, Sarah Krystelle P. July 05, 2011Signal Spectra and Signal Processing/BSECE 41A1 Score: Eng’r. Grace Ramones Instructor
2. 2. OBJECTIVES1. Plot the gain-frequency response of an L-C series resonant and an L-C parallel resonant band-pass filter.2. Determine the center frequency and the bandwidth of the L-C band pass filter.3. Determine how the circuit resistance affects the bandwidth of an L-C band-pass filter.4. Plot the gain-frequency response of an L-C series resonant and an L-C parallel resonant band-stop (notch) filter.5. Determine the center frequency and the bandwidth of the L-C band-stop filter.6. Determine how the circuit resistance affects the bandwidth of an L-C band-stop filter.
3. 3. SAMPLE COMPUTATIONSSolution for Step 4Solution for Step 6:Solution for Step 7Solution for Question in Step 7Solution for Step 8
4. 4. Solution for Question in Step 8Solution for Step 9Solution for Step 10Solution for Question in Step 10Solution for Step 17
5. 5. Solution for Step 19Solution for Step 20Solution for Question in Step 20Solution for Step 21Solution for Question in Step 21Solution for Step 22Solution for Step 23
6. 6. Solution for Step 24Solution for Step 25Solution for Step 26Solution for Question in Step 26Solution for Step 33Solution for Step 34
7. 7. Solution for Question in Step 34Solution for Step 35Solution for Question in Step 35Solution for Step 36Solution for Step 37Solution for Step 38
8. 8. Solution for Step 43Solution for Step 44Solution for Question in Step 44Solution for Step 45Solution for Question in Step 45
9. 9. Solution for Step 46Solution for Step 47Solution for Step 48Solution for Step 49Solution for Step 50Solution for Step Question in 50
10. 10. DATA SHEETMATERIALSOne function generatorOne dual-trace oscilloscopeCapacitors: one 0.1µF, one 0.25 µFInductors: one 50 mH, one 100 mHResistors: 100 Ω, 200 Ω, 2 kΩ, 4 kΩ, 5kΩ, 200kΩTHEORY In electronic communications systems, it is often necessary to separate a specificrange of frequency from the total frequency spectrum. This is normally accomplished withfilters. A filter is circuit that passes a specific range of frequencies while rejecting otherfrequencies. 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 cutofffrequency and reject all frequencies above the cutoff frequency. A high-pass filter isdesigned to pass all frequencies above the cutoff frequency. A high-pass filter is designed topass all frequencies above the cutoff frequency and reject all frequencies below the cutofffrequency. A band-pass filter passes all frequencies within a band of frequencies and rejectsall other frequencies outside the band. A band-stop filter rejects all frequencies within aband of frequencies outside the band. A band-stop filter is often referred to as a notch filter.In this experiment, you will study band-pass and band-stop (notch) 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 function of frequency (f). The frequency atwhich the output power gain drops is 50% of the maximum value is called the cutofffrequency (fC). When the output power gain drops to 50%, the voltage gain drops 3 db(0.707 of the maximum value). When the filter dB voltage gain is plotted as a function offrequency on semi log graph using straight lines to approximately the actual frequencyresponse, it is called a Bode plot. A bode plot is an ideal plot of filter frequency responsebecause it assumes that the voltage gain remains constant in the passband until the cutofffrequency is reached, and then drops in straight line. The filter network voltage gain in dBis calculated from the actual voltage gain (A) using the equation AdB = 20 log A Where A = Vo/Vi An L-C series resonant band-pass filter is shown in Figure 2-1. The impedance of theseries L-C circuit is lowest at the resonant frequency and increases on both sides of theresonant frequency. This will cause the output voltage to be highest at the resonantfrequency and decrease on both sides of the resonant frequency. At L-C parallel resonantband-pas filter is shown in Figure 2-2. The impedance of the parallel L-C circuit is highest atthe resonant frequency and decreases on both sides of the resonant frequency. This will
11. 11. also cause the output voltage to be highest at the resonant frequency and decrease on bothsides of the resonant. Figure 2-1 L-C Series Resonant Band-Pass Filter ` Figure 2-2 L-C Parallel Resonant Band-Pass Filter An L-C series resonant band-stop (notch) filter is shown in Figure 2-3. Theimpedance at the series L-C circuit is lowest at the resonant frequency and increase on bothsides of the resonant frequency. This will cause the output voltage to be lowest at theresonant frequency and increase on both side of the resonant frequency. An L-C parallelresonant band-stop (notch) filter is shown in Figure 2-4. The impedance of the parallel L-C
12. 12. circuit is highest at the resonant frequency and decreases on both sides of the resonantfrequency. This will also cause the output voltage to the lowest at the resonant frequencyand increase on both sides of the resonant frequency. Figure 2-3 L-C Series Resonant Band-Stop (Notch) Filter Figure 2-4 L-C Parallel Resonant Band-Stop (Notch) Filter
13. 13. The center frequency (fO) for the L-C series resonant and the L-C parallel resonantband-pass and band-stop (notch) filter is equal to the resonant frequency of the L-C circuit,which can be calculated from For an L-C parallel resonant filter, the equation is accurate only for a high Q inductorcoil (Qf 10) where QL is calculated from and XL is the inductive reactance at the resonant frequency (center frequency, fO)and Rw is the inductor coil resistance. In the band-pass and band-stop (notch) filters, the low-cutoff frequency (fC1) and thehigh-cutoff frequency (fC2) on the gain-frequency plot are the frequencies where the wholevoltage gain has dropped 3dB (0.707) from the highest dB gain. The filter bandwidth (BW)is the difference between the cutoff frequency (fC2) and the low-cutoff frequency (fC1)Therefore, BW = fC2 – fC1 The center frequency (f0) is the geometric mean of the low-cutoff frequency and thehigh-cutoff frequency. Therefore, The quality factor (Q) of the band-pass and the band-stop (notch) filters is the ratioof the center frequency (fO) and the bandwidth (BW), and it is an indication of the activityof the filter. Therefore, A higher value of Q means a narrower bandwidth and a more selective filter. The quality factor (QS) of a series resonant filter is determined by first calculatingthe inductive reactance (XL) of the inductor at the resonant frequency (center frequency,fO), and then dividing the inductive reactance by total series resistance (RT). Therefore, Where
14. 14. The quality factor (QP) of a parallel resonant filter is determined by first calculatingthe inductive reactance (XL) of the inductor at the resonant frequency (center frequency,fO), and then dividing the total parallel resistance (RP) by the inductive reactance (XL).Therefore, Because the inductor wire resistance (RW) is in series with inductor L, the circuits inFigures 2-2 and 2-4 are not exactly parallel resonant circuits; the series combination ofinductance (L) and resistance (RW) must first be converted into an equivalent parallelnetwork with resistance REQ in parallel with inductance L. In Figure 2-2, the parallelequivalent resistance (REQ) will also be in parallel with resistor r and resistor RS making thetotal resistance of the parallel resonant circuit (RP), equal to the parallel equivalent ofresistors r, RS, and REQ. Therefore, RP can be solved from In Figure 2-4, the parallel equivalent resistance (REQ) will be in parallel with resistorR, making the total resistance of the parallel resonant circuit (RP) equal to the parallelequivalent of resistor R and REQ. Therefore, RP can be solved from The equation for converting resistance RW to the equivalent parallel resistance (REQ)is The parallel equivalent inductance LEQ is calculated from This equivalent inductance can be considered equal to the original inductance (L)for a high Q coil (QL 10)
15. 15. PROCEDUREBand-Pass FiltersStep 1 Open circuit file Fig 2-1. Make sure that the following Bode plotter settings are selected. Magnitude, Vertical (Log, F=0 dB, I=–20dB), Horizontal (Log, F=2 kHz, I=500Hz)Step 2 Run the simulation. Notice that the voltage gain in db has been plotted between the frequencies of 50Hz and 2Kz by the Bode plotter. Sketch the curve plot in the space provided.Question: Is the frequency response curve that of a band-pass filters? Explain why. Yes. A band-pass is a filtering device that permits only the frequencies within a certain band and rejects all other band. The plot curve shown above only allows the frequencies from 501.697 Hz to 1.993 kHz.Step 3 Move the cursor to the center of the curve at its peak point. Record the center frequency (f0) and the voltage gain in dB on the curve plot. fO= 996.84 Hz; AdB= – 1.637 dBStep 4 Based on the dB voltage gain, calculate the actual voltage gain (A) of the series resonant band-pass filter at the center frequency. A= 0.83
16. 16. Step 5 Move the cursor as close as possible to a point on the left side of the curve that is 3 dB down from dB gain measured at frequency fO. Record the approximate frequency (low-cutoff frequency, fC1) on the curve plot. Next, move the cursor as close as possible to a point on the right side of the curve that is 3 dB down from the dB gain measured at frequency fO. Record the approximate frequency (high-cutoff frequency, fC2) on the curve plot. fC1= 914.969 Hz fC2= 1.104 kHzStep 6 Based on the values of fC1 and fC2 measured on the curve plot, determine the bandwidth (BW) of the series resonant band-pass filter. BW= 189.031 HzStep 7 Based on the circuit component values in Figure 2-1, calculated the expected center frequency (fO) of the series resonant band-pass filter. fO(COMPONENT VALUE)= 1006.58 HzQuestion: How did the calculated value of the center frequency (fO) based on the circuitcomponent values compare with the measured value on the curve plot? The calculated fO based on the circuit component values and the measured value is almost equal. Their percentage difference is 0.97%.Step 8 Based on the values of fC1 and fC2, calculate the center frequency (fO). fO(fC1/fC2)= 1005.05 HzQuestion: How did the calculated value of the center frequency (fO) based on fC1 and fC2compare with the measured value on the curve plot? The percentage difference between the two is 0.82%.Step 9 Based on the circuit component values, calculate the quality factor (QS) of the series resonant band-pass filter. QS= 5.27
17. 17. Step 10 Based on the circuit quality factor (QS) and the center frequency (fO), calculate the expected bandwidth (BW) of the series resonant band-pass filter. BW(EXPECTED)= 191.00 HzQuestion: How did the expected bandwidth calculated from the value of QS and the centerfrequency compare with the bandwidth measured on the curve plot? The percentage difference between the two is 1.04%.Step 11 Change the resistance of R to 200 Ω. Run the simulation again. Measure the center frequency (fO) and the bandwidth (BW) from the curve plot and record the values. fO=996.84 Hz BW= 350.136 HzQuestions: What effect did changing the resistance R have on the center frequency of theseries resonant band-pass filter? What effect did changing the resistance of R have on thebandwidth of the series resonant band-pass filter? Explain. The center frequency remain constant or did not change its value while the bandwidth changed, it increased as the resistor increase. Therefore, the center frequency f O of a series resonant band-pass filter is constant with change in the resistance R while the bandwidth of the series resonant band-pass filter is directly proportional to the resistance R.Step 12 Change the capacitance of C to 0.1 µF. Run the simulation again. Measure the center frequency (fO) and the bandwidth (BW) on the curve plot and record the values Change the Bode plotter settings as needed. fO=1.583 kHz BW= 350 Hz
18. 18. Questions: What effect did changing the capacitance of C have on the center frequency ofthe series resonant band-pass filter? What effect did changing the capacitance of C have onthe bandwidth of the series resonant band-pass filter? Explain. The bandwidth remains constant or did not change its value. The center frequency however changed, it increased as the resistor increase. Therefore, the bandwidth of a series resonant band-pass filter is affected by the change of capacitance C while the center frequency of the series resonant band-pass filter is inversely proportional to the capacitance C.Step 13 Change the inductance of L to 50 mH. Run the simulation again. Measure the center frequency (fO) and the bandwidth (BW) on the curve plot and record the values. Change the Bode plotter settings as needed. fO=1.423 kHz BW= 386 HzQuestions: What effect did the changing the inductance of L has on the center frequency ofthe series resonant band-pass filter? What effect did changing the inductance of L have onthe bandwidth of the series resonant band-pass filter? Explain. The center frequency and bandwidth was compared to the original circuit. By changing the inductance, bandwidth BW and center frequency f O increased. Therefore, bandwidth and the center frequency of the series resonant band-pass filter are inversely proportional to the inductance of L
19. 19. Step 14 Open circuit file Fig 2-2. Make sure that the following Bode plotter settings are selected: Magnitude, Vertical (Log, F=0 dB, I=–20dB), Horizontal (Log, F=2 kHz, I=500Hz)Step 15 Run the simulation. Notice that the voltage gain in dB has been plotted between the frequencies of 500 Hz and 2 kHz by the Bode plotter. Sketch the curve plot in the space provided.Question: Is the frequency response curve that of a band-pass filters? Explain why. Yes. The curve plot above is a band-pass because it permits only the frequencies within bandpass from 501.697 Hz to 1.993 kHz and rejects all other band.Step 16 Move the cursor to the center of the curve at its peak. Record the center frequency (fO) and the voltage gain in dB on the curve plot. fO= 1.003 kHz AdB= – 2.189 dBStep 17 Based on the dB voltage gain, calculate the actual voltage gain (A) of the parallel resonant band-pass filter at the center frequency. A= 0.78Step 18 Move the cursor at its close as possible to a point on the left side of the curve that is 3dB down from the dB gain measured at frequency fO. Record the approximate frequency (low-cutoff frequency, fC1) on the curve plot. Next, move the cursor as close as possible to a point on the right side of the curve that is 3 dB down from the
20. 20. dB gain measured at frequency fO. Record the approximate frequency (high-cutoff frequency fC2) on the curve plot. fC1= 929.014 Hz fC2= 1.093 kHzStep 19 Based on the values of fC1 and fC2, determine the bandwidth (BW) of the parallel resonant band-pass filter. BW= 163.186 HzStep 20 Based on the circuit component values in Fig 2-2, calculate the expected center frequency (fO) of the parallel resonant band-pass filter. fO(COMPONENT VALUE)= 1006.58 HzQuestion: How did the calculated values of the center frequency (fO) base on the circuitcomponent values compare with measured values recorded on the curve plot? The percentage difference between the calculated fO based on the circuit component values and the measured value is only 0.37%.Step 21 Based on the values of fC1 and fC2, calculate the center frequency (fO). fO(fC1/fC2)= 1007.68 HzQuestion: How did the calculated value of the center frequency (fO) base on the fC1 and fC2compare with the measured value on the curve plot? The percentage difference between the two is 0.47%.Step 22 Based on the value of L and RW, calculate the quality factor (QL) of the inductor. QL=31.62Step 23 Based on the quality factor (QL) of the inductor, calculate the equivalent parallel inductor resistance (REQ) across the tank circuit. REQ= 20016.488Ω
21. 21. Step 24 Based on the value of REQ, RS, and R, calculate the total parallel resistance (RP) across the tank circuit. RP= 3922.20ΩStep 25 Based on the value of RP, calculate the quality factor (QP) of the parallel resonant band-pass filter. QP= 6.2015Step 26 Based on the filter quality (QP) and the center frequency (fO), calculate the expected bandwidth (BW) of the parallel resonant band-pass filter. BW= 162.31 HzQuestion: How did the expected bandwidth calculated from the value of QP and the centerfrequency compare with the bandwidth measured on the curve plot? The percentage difference between the two is 0.54%.Step 27 Change the resistance of R to 5kΩ. Run the simulation again. Measure the center frequency (fO) and the bandwidth (BW) from the curve plot and record the values. fO=1.003 kHz BW= 289.456 HzQuestions: What effect did changing the resistance of R have on the center frequency of theparallel resonant band-pass filter? What effect did changing the resistance R have on thebandwidth of the parallel resonant band-pass filter? Explain why. The center frequency remain constant or did not change its value while the bandwidth changed, it increased as the resistor decrease. Therefore, the center frequency f O of a series resonant band-pass filter is constant with change in the resistance R while the bandwidth of the series resonant band-pass filter is inversely proportional to the resistance R.
22. 22. Band-Stop (Notch) FiltersStep 28 Open the circuit file Fig 2-3. Make sure that the following Bode plotter settings are selected: Magnitude, Vertical (Log, F=0 dB, I=–20dB), Horizontal (Log, F=2 kHz, I=500Hz)Step 29 Run the simulation. Notice that the voltage gain in dB has been plotted between the frequencies of 500 Hz and 2 kHz by the Bode plotter. Sketch the curve plot in the space provided.Question: Is the frequency response curve that of a band-stop (notch) filters? Explain why. Yes. What was shown above is a band-stop filter because it rejects all frequencies that are within the bandpass, and giving easy passage only to frequencies outside the bandpass.Step 30 Move the cursor to the center of the curve at its lowest point. Record the center frequency (fO) on the curve plot. fO= 1.003 kHz;Step 31 Move the cursor to the highest point on the flat part of the curve and record the db gain on the curve plot. AdB= – 0.066 dBStep 32 Move the cursor as close as possible to a point on the left side of the curve that is 3 dB down from the highest dB gain on the flat part of the curve. Record the approximate frequency (low-cutoff frequency fC1) on the curve plot. Next, move the cursor as close as possible to a point on the right side of the curve that is 3 dB down from the highest dB gain on the flat part of the curve. Record the approximate frequency (high-cutoff frequency fC1) on the curve plot. fC1= 918.093 Hz fC2= 1.103 kHz
23. 23. Step 33 Based on the values of fC1 and fC2, determine the bandwidth (BW) of the series resonant band-stop (notch) filter. BW= 184.907 HzStep 34 Based on the circuit component values in Fig 2-3, calculate the expected center frequency (fO) of the series resonant band-stop (notch) filter. fO(COMPONENT VALUE)= 1006.58 HzQuestion: How did the calculated value of the center frequency (fO) base on circuitcomponent values compare with the measured value recorded on the curve plot? The percentage difference between the calculated fO based on the circuit component values and the measured value is only 0.36%.Step 35 Based on the values of fC1 and fC2, calculate the center frequency (fO) fO(fC1/fC2)= 1006.31 HzQuestion: How did the calculated value of the center frequency (fO) base on the fC1 and fC2compare with the measured value on the curve plot? The percentage difference between the two is 0.33%.Step 36 Based on the circuit component values calculate the quality factor (Qs) of the series resonant band-stop (notch) filter. QS= 5.27Step 37 Based on the circuit quality factor (QS) and the center frequency (fO), calculate the expected bandwidth (BW) of the series resonant band-stop (notch) filter. BW(EXPECTED)= 191.00 HzQuestion: How did the expected bandwidth calculated from the value of QS and the centerfrequency compare with the bandwidth measured on the curve plot? The percentage difference between the two is 3.30%.
24. 24. Step 38 Open circuit file Fig 2-4. Make sure that the following Bode plotter settings are selected: Magnitude, Vertical (Log, F=0 dB, I=–20dB), Horizontal (Log, F=2 kHz, I=500Hz)Step 39 Run the simulation. Notice that the voltage gain in dB has been plotted between the frequencies of 500 Hz and 2 kHz by the Bode plotter. Sketch the curve plot in the space provided.Question: Is the frequency response curve that of a band-stop (notch) filters? Explain why. Yes. The above curve plot is a band-stop filter because it rejects all frequencies that are within the bandpass, and giving easy passage only to frequencies outside the bandpass.Step 42 Move the cursor to the center of the curve at its lowest point. Record the center frequency (fO) on the curve plot. fO= 1.003 kHz;Step 41 Move the cursor to the highest point on the flat part of the curve and record the dB gain on the curve plot. AdB= – 0.055 dBStep 42 Move the cursor as close as possible to a point on the left side of the curve that is 3 dB down from the highest dB gain on the flat part of the curve. Record the approximate frequency (low-cutoff frequency fC1) on the curve plot. Next, move the cursor as close as possible to a point on the right side of the curve that is 3 dB down from the highest dB gain on the flat part of the curve. Record the approximate frequency (high-cutoff frequency fC1) on the curve plot. fC1= 914.969 Hz; fC2= 1.098 kHz
25. 25. Step 43 Based on the values of fC1 and fC2, determine the bandwidth (BW) of the parallel resonant band-stop (notch) filter. BW= 183.031 HzStep 44 Based on the circuit component values in Fig 2-4, calculate the expected center frequency (fO) of the parallel resonant band-stop (notch) filter. fO(COMPONENT VALUE)= 1006.58 HzQuestion: How did the calculated values of the center frequency (fO) base on the circuit component values compare with measured values recorded on the curve plot? The percentage difference between the calculated fO based on the circuit component values and the measured value is only 0.36%.Step 45 Based on the values of fC1 and fC2, calculate the center frequency (fO). fO(fC1/fC2)= 1002.32 HzQuestion: How did the calculated value of the center frequency (fO) base on the fC1 and fC2 compare with the measured value on the curve plot? The percentage difference between the two is 0.07%.Step 46 Based on the value of L and RW, calculate the quality factor (QL) of the inductor. QL=31.62Step 47 Based on the quality factor (QL) of the inductor, calculate the equivalent parallel inductor resistance (REQ) across the tank circuit. REQ= 20016.48ΩStep 48 Based on the value of REQ and R, calculate the total parallel resistance (RP) across the tank circuit. RP= 3333.79Ω
26. 26. Step 49 Based on the value of RP, calculate the quality factor (QP) of the parallel resonant band-stop (notch) filter. QP= 5.27Step 50 Based on the filter quality (QP) and the center frequency (fO), calculate the expected bandwidth (BW) of the parallel resonant band-stop (notch) filter. BW= 191 HzQuestion: How did the expected bandwidth calculated from the value of QP and the center frequency compare with the bandwidth measured on the curve plot? The percentage difference between the two is 4.35%.Step 51 Change the resistance of R to 2kΩ. Run the simulation again. Measure the center frequency (fO) and the bandwidth (BW) from the curve plot and record the values. fO=1.003 kHz BW= 337.136 HzQuestions: What effect did changing the resistance of R have on the center frequency of the parallel resonant band-stop (notch) filter? What effect did changing the resistance R have on the bandwidth of the parallel resonant band-stop (notch) filter? Explain. The center frequency remain constant or did not change its value while the bandwidth changed, it increases as the resistor decreases. Therefore, the center frequency f O of a series resonant band-pass filter is constant with the change of the value of the resistance of resistor R while the bandwidth of the series resonant band-pass filter is inversely proportional to the value of the resistance of resistor R.
27. 27. CONCLUSIONSBased on the output signal in the bode plotter the band-pass filter only allows frequencieswithin the band and blocks all frequencies outside the band. Its counterpart is the band-stop (notch) filter which blocks all frequencies within the band and allows all otherfrequencies outside the band.In addition to that, the voltage gain and the center frequency of the band-pass filter are atthe peak point of the curve plot. On the other hand, the voltage gain and the centerfrequency of the band-stop filter are at the lowest point of the curve plot. Bandwidth can bedetermine from subtracting the frequencies 3 dB down and up from dB gain measured atcenter frequency fO.Lastly, I concluded that the center frequency of the curve plot was not affected by theresistance and is inversely proportional to the capacitance and inductance. Bandwidth isnot affected by capacitance, but inversely proportional to the inductance. The bandwidth inL-C series resonant band-pass/stop filter is directly proportional to the resistance, whilebandwidth in L-C series resonant band-pass/stop filter is inversely proportional to theresistance.