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### Exp5 tagasa

1. 1. NATIONAL COLLEGE OF SCIENCE & TECHNOLOGY Amafel Bldg. Aguinaldo Highway Dasmariñas City, Cavite EXPERIMENT 5 Fourier Theory – Frequency Domain and Time DomainTagasa, Jerald A. September 06, 2011Signal Spectra and Signal Processing/BSECE 41A1 Score: Engr. Grace Ramones Instructor
2. 2. Objectives: 1. Learn how a square wave can be produced from a series of sine waves at different frequencies and amplitudes. 2. Learn how a triangular can be produced from a series of cosine waves at different frequencies and amplitudes. 3. Learn about the difference between curve plots in the time domain and the frequency domain. 4. Examine periodic pulses with different duty cycles in the time domain and in the frequency domain. 5. Examine what happens to periodic pulses with different duty cycles when passed through low-pass filter when the filter cutoff frequency is varied.
3. 3. Sample ComputationFrequencyDuty CycleFirst zero crossing point (f)BandwidthBW =
4. 4. Data Sheet:Materials:One function generatorOne oscilloscopeOne spectrum analyzerOne LM 741 op-ampTwo 5 nF variable capacitorsResistors: 5.86 kΩ, 10 kΩ, and 30 kΩTheory:Communications systems are normally studies using sinusoidal voltagewaveforms to simplify the analysis. In the real world, electricalinformation signal are normally nonsinusoidal voltage waveforms, such asaudio signals, video signals, or computer data. Fourier theory provides apowerful means of analyzing communications systems by representing anonsinusoidal signal as series of sinusoidal voltages added together.Fourier theory states that a complex voltage waveform is essentially acomposite of harmonically related sine or cosine waves at differentfrequencies and amplitudes determined by the particular signal waveshape.Any, nonsinusoidal periodic waveform can be broken down into sine or cosinewave equal to the frequency of the periodic waveform, called the fundamentalfrequency, and a series of sine or cosine waves that are integer multiplesof the fundamental frequency, called the harmonics. This series of sine orcosine wave is called a Fourier series.Most of the signals analyzed in a communications system are expressed in thetime domain, meaning that the voltage, current, or power is plotted as afunction of time. The voltage, current, or power is represented on thevertical axis and time is represented on the horizontal axis. Fourier theoryprovides a new way of expressing signals in the frequency domain, meaningthat the voltage, current, or power is plotted as a function of frequency.Complex signals containing many sine or cosine wave components are expressedas sine or cosine wave amplitudes at different frequencies, with amplituderepresented on the vertical axis and frequency represented on the horizontalaxis. The length of each of a series of vertical straight lines representsthe sine or cosine wave amplitudes, and the location of each line along thehorizontal axis represents the sine or cosine wave frequencies. This iscalled a frequency spectrum. In many cases the frequency domain is moreuseful than the time domain because it reveals the bandwidth requirements ofthe communications system in order to pass the signal with minimaldistortion. Test instruments displaying signals in both the time domain andthe frequency domain are available. The oscilloscope is used to displaysignals in the time domain and the spectrum analyzer is used to display thefrequency spectrum of signals in the frequency domain.In the frequency domain, normally the harmonics decrease in amplitude astheir frequency gets higher until the amplitude becomes negligible. The moreharmonics added to make up the composite waveshape, the more the composite
5. 5. waveshape will look like the original waveshape. Because it is impossible todesign a communications system that will pass an infinite number offrequencies (infinite bandwidth), a perfect reproduction of an originalsignal is impossible. In most cases, eliminate of the harmonics does notsignificantly alter the original waveform. The more information contained ina signal voltage waveform (after changing voltages), the larger the numberof high-frequency harmonics required to reproduce the original waveform.Therefore, the more complex the signal waveform (the faster the voltagechanges), the wider the bandwidth required to pass it with minimaldistortion. A formal relationship between bandwidth and the amount ofinformation communicated is called Hartley’s law, which states that theamount of information communicated is proportional to the bandwidth of thecommunications system and the transmission time.Because much of the information communicated today is digital, the accuratetransmission of binary pulses through a communications system is important.Fourier analysis of binary pulses is especially useful in communicationsbecause it provides a way to determine the bandwidth required for theaccurate transmission of digital data. Although theoretically, thecommunications system must pass all the harmonics of a pulse waveshape, inreality, relatively few of the harmonics are need to preserve the waveshape.The duty cycle of a series of periodic pulses is equal to the ratio of thepulse up time (tO) to the time period of one cycle (T) expressed as apercentage. Therefore,In the special case where a series of periodic pulses has a 50% duty cycle,called a square wave, the plot in the frequency domain will consist of afundamental and all odd harmonics, with the even harmonics missing. Thefundamental frequency will be equal to the frequency of the square wave. Theamplitude of each odd harmonic will decrease in direct proportion to the oddharmonic frequency. Therefore,The circuit in Figure 5–1 will generate a square wave voltage by adding aseries of sine wave voltages as specified above. As the number of harmonicsis decreased, the square wave that is produced will have more ripples. Aninfinite number of harmonics would be required to produce a perfectly flatsquare wave.
6. 6. Figure 5 – 1 Square Wave Fourier Series XSC1 Ext T rig V6 + R1 J1 _ A B 10.0kΩ + _ + _ 10 V Key = A V1 R2 J2 10 Vpk 10.0kΩ 1kHz Key = B 0° V2 R3 J3 4 155 0 8 160 14 13 12 R7 109 02 3 100Ω 3.33 Vpk 10.0kΩ 3kHz Key = C 0° V3 R4 J4 2 Vpk 10.0kΩ 5kHz Key = D 0° V4 R5 J5 1.43 Vpk 10.0kΩ 7kHz 0° Key = E V5 J6 R6 1.11 Vpk 10.0kΩ 9kHz Key = F 0° .The circuit in Figure 5-2 will generate a triangular voltage by adding aseries of cosine wave voltages. In order to generate a triangular wave, eachharmonic frequency must be an odd multiple of the fundamental with no evenharmonics. The fundamental frequency will be equal to the frequency of thetriangular wave, the amplitude of each harmonic will decrease in directproportion to the square of the odd harmonic frequency. Therefore,Whenever a dc voltage is added to a periodic time varying voltage, thewaveshape will be shifted up by the amount of the dc voltage.
7. 7. Figure 5 – 2 Triangular Wave Fourier Series XSC1 Ext T rig V6 + R1 J1 _ A B 10.0kΩ + _ + _ 15 V Key = A V1 R2 J2 10 Vpk 10.0kΩ 1kHz 90° V2 Key = B R3 J3 13 12 1 2 3 4 5 8 9 11 0 R7 6 0 1.11 Vpk 100Ω 10.0kΩ 3kHz 90° V3 Key = C R4 J4 0.4 Vpk 10.0kΩ 5kHz 90° V4 Key = D R5 J5 0.2 Vpk 10.0kΩ 7kHz 90° Key = EFor a series of periodic pulses with other than a 50% duty cycle, the plotin the frequency domain will consist of a fundamental and even and oddharmonics. The fundamental frequency will be equal to the frequency of theperiodic pulse train. The amplitude (A) of each harmonic will depend on thevalue of the duty cycle. A general frequency domain plot of a periodic pulsetrain with a duty cycle other than 50% is shown in the figure on page 57.The outline of peaks if the individual frequency components is calledenvelope of the frequency spectrum. The first zero-amplitude frequencycrossing point is labelled fo = 1/to, there to is the up time of the pulsetrain. The first zero-amplitude frequency crossing point fo) determines theminimum bandwidth (BW0 required for passing the pulse train with minimaldistortion.Therefore,
8. 8. A f=1/to 2/to f Frequency Spectrum of a Pulse Train Notice than the lower the value of to the wider the bandwidth required to pass the pulse train with minimal distortion. Also note that the separation of the lines in the frequency spectrum is equal to the inverse of the time period (1/T) of the pulse train. Therefore a higher frequency pulse train requires a wider bandwidth (BW) because f = 1/T The circuit in Figure 5-3 will demonstrate the difference between the time domain and the frequency domain. It will also determine how filtering out some of the harmonics effects the output waveshape compared to the original3 input waveshape. The frequency generator (XFG1) will generate a periodic pulse waveform applied to the input of the filter (5). At the output of the filter (70, the oscilloscope will display the periodic pulse waveform in the time domain, and the spectrum analyzer will display the frequency spectrum of the periodic pulse waveform in the frequency domain. The Bode plotter will display the Bode plot of the filter so that the filter bandwidth can be measured. The filter is a 2-pole low-pass Butterworth active filter using a 741 op-amp. Figure 5-3 Time Domain and Frequency Domain XFG1 XSC1 C1 XSA1 Ext T rig + 2.5nF 50% _ Key=A A _ B _ IN T + + R1 R2 741 30kΩ 30kΩ 42 OPAMP_3T_VIRTUAL 0 6 0 31 R3 C2 R4 5.56kΩ 10kΩ XBP1 2.5nF 50% Key=A R5 IN OUT 10kΩ
9. 9. Procedure:Step 1 Open circuit file FIG 5-1. Make sure that the following oscilloscope settings are selected: Time base (Scale = 200 µs/Div, Xpos = 0, Y/t), Ch A (Scale = 5V/Div, Ypos = 0, DC), Ch B (Scale = 50 mV/Div, Ypos = 0, DC), Trigger (Pos edge, Level = 0, Auto). You will generate a square wave curve plot on the oscilloscope screen from a series of sine waves called a Fourier series.Step 2 Run the simulation. Notice that you have generated a square wave curve plot on the oscilloscope screen (blue curve) from a series of sine waves. Notice that you have also plotted the fundamental sine wave (red). Draw the square wave (blue) curve on the plot and the fundamental sine wave (red) curve plot in the space provided.Step 3 Use the cursors to measure the time periods for one cycle (T) of the square wave (blue) and the fundamental sine wave (red) and show the value of T on the curve plot. T1 = 1.00 ms T2 = 1.00 msStep 4 Calculate the frequency (f) of the square wave and the fundamental sine wave from the time period. f = 1 kHzQuestions: What is the relationship between the fundamental sine wave andthe square wave frequency (f)? They are the same.What is the relationship between the sine wave harmonic frequencies(frequencies of sine wave generators f3, f5, f7, and f9 in figure 5-1) andthe sine wave fundamental frequency (f1)? They are all odd multiples.What is the relationship between the amplitude of the harmonic sine wavegenerators and the amplitude of the fundamental sine wave generator? The amplitude of the odd harmonics decrease in direct proportion toodd harmonic frequency.Step 5 Press the A key to close switch A to add a dc voltage level to the square wave curve plot. (If the switch does not close, click the mouse arrow in the circuit window before pressing the A key). Run the simulation again. Change the oscilloscope settings as needed. Draw the new square wave (blue) curve plot on the space provided.
10. 10. Question: What happened to the square wave curve plot? Explain why. It shifted upward. It is because of by the applied dc voltage.Step 6 Press the F and E keys to open the switches F and E to eliminate the ninth and seventh harmonic sine waves. Run the simulation again. Draw the new curve plot (blue) in the space provided. Note any change on the graph.Step 7 Press the D key to open the switch D to eliminate the fifth harmonics sine wave. Run the simulation again. Draw the new curve plot (blue) in the space provided. Note any change on the graph.Step 8 Press the C key to open switch C and eliminate the third harmonic sine wave. Run the simulation again.Question: What happened to the square wave curve plot? Explain. It became sinusoidal wave. This is because all the harmonics are gone.Step 9 Open circuit file FIG 5-2. Make sure that the following oscilloscope settings are selected: Time base (Scale = 200 µs/Div,
11. 11. Xpos = 0, Y/t), Ch A (Scale = 5V/Div, Ypos = 0, DC), Ch B (Scale = 100 mV/Div, Ypos = 0, DC), Trigger (Pos edge, Level = 0, Auto). You will generate a triangular wave curve plot on the oscilloscope screen from a series of sine waves called a Fourier series.Step 10 Run the simulation. Notice that you have generated a triangular wave curve plot on the oscilloscope screen (blue curve) from the series of cosine waves. Notice that you have also plotted the fundamental cosine wave (red). Draw the triangular wave (blue) curve plot and the fundamental cosine wave (red) curve plot in the space provided.Step 11 Use the cursors to measure the time period for one cycle (T) ofthe triangular wave (blue) and the fundamental (red), and show the value ofT on the curve plot. T1 = 1.00 ms T2 = 1.00 msStep 12 Calculate the frequency (f) of the triangular wave from the timeperiod (T). f = 1 kHzQuestions: What is the relationship between the fundamental frequency andthe triangular wave frequency? They are the same.What is the relationship between the harmonic frequencies (frequencies ofgenerators f3, f5, and f7 in figure 5-2) and the fundamental frequency (f1)? They are all odd functions.What is the relationship between the amplitude of the harmonic generatorsand the amplitude of the fundamental generator? The amplitude of the harmonic generators decreases in direct proportion to the square of the odd harmonic frequencyStep 13 Press the A key to close switch A to add a dc voltage level to the triangular wave curve plot. Run the simulation again. Draw the new triangular wave (blue) curve plot on the space provided.
12. 12. Question: What happened to the triangular wave curve plot? Explain. It shifted upward. It is because of by the applied dc voltage.Step 14 Press the E and D keys to open switches E and D to eliminate the seventh and fifth harmonic sine waves. Run the simulation again. Draw the new curve plot (blue) in the space provided. Note any change on the graph.Step 15 Press the C key to open the switch C to eliminate the third harmonics sine wave. Run the simulation again.Question: What happened to the triangular wave curve plot? Explain. It became sine wave, because the harmonic sine waves are gone.Step 16 Open circuit FIG 5-3. Make sure that following function generator settings are selected: Square wave, Freq = 1 kHz, Duty cycle = 50%, Ampl – 2.5 V, Offset = 2.5 V. Make sure that the following oscilloscope settings are selected: Time base (Scale = 500 µs/Div, Xpos = 0, Y/T), Ch A (Scale = 5 V/Div, Ypos = 0, DC), Ch B (Scale = 5 V/Div, Ypos = 0, DC), Trigger (pos edge, Level = 0, Auto). You will plot a square wave in the time domain at the input and output of a two-pole low-pass Butterworth filter.Step 17 Bring down the oscilloscope enlargement and run the simulation to one full screen display, then pause the simulation. Notice that you are displaying square wave curve plot in the time domain (voltage as a function of time). The red curve plot is the filter input (5) and the blue curve plot is the filter output (7)Question: Are the filter input (red) and the output (blue) plots the sameshape disregarding any amplitude differences? Yes.
13. 13. Step 18 Use the cursor to measure the time period (T) and the time (fo) of the input curve plot (red) and record the values. T= 1 ms to = 500.477µsStep 19 Calculate the pulse duty cycle (D) from the to and T D = 50.07%.Question: How did your calculated duty cycle compare with the duty cyclesetting on the function generator? The difference is 0.07%.Step 20 Bring down the Bode plotter enlargement to display the Bode plot of the filter. Make sure that the following Bode plotter settings are selected; Magnitude, Vertical (Log, F = 10 dB, I = -40 dB), Horizontal (Log, F = 200 kHz, I = 100 Hz). Run the simulation to completion. Use the cursor to measure the cutoff frequency (fC) of the low-pass filter and record the value. fC = 21.197Step 21 Bring down the analyzer enlargement. Make sure that the following spectrum analyzer settings are selected: Freq (Start = 0 kHz, Center = 5 kHz, End = 10 kHz), Ampl (Lin, Range = 1 V/Div), Res = 50 Hz. Run the simulation until the Resolution frequencies match, then pause the simulation. Notice that you have displayed the filter output square wave frequency spectrum in the frequency domain, use the cursor to measure the amplitude of the fundamental and each harmonic to the ninth and record your answers in table 5- 1. Table 5-1 Frequency (kHz) Amplitude f1 1 5.048 V f2 2 11.717 µV f3 3 1.683 V f4 4 15.533 µV f5 5 1.008 V f6 6 20.326 µV f7 7 713.390 mV f8 8 25.452 µV f9 9 552.582 mVQuestions: What conclusion can you draw about the difference between theeven and odd harmonics for a square wave with the duty cycle (D) calculatedin Step 19? The even harmonics is much lower compared with the odd harmonics. It is almost zero.What conclusions can you draw about the amplitude of each odd harmoniccompared to the fundamental for a square wave with the duty cycle (D)calculated in Step 19? The amplitude of odd harmonics decreases in direct proportion with the odd harmonic frequency.Was this frequency spectrum what you expected for a square wave with theduty cycle (D) calculated in Step 19? Yes.
14. 14. Based on the filter cutoff frequency (fC) measured in Step 20, how many ofthe square wave harmonics would you expect to be passed by this filter?Based on this answer, would you expect much distortion of the input squarewave at the filter? Did your answer in Step 17 verify this conclusion? There are square waves. Yes, there is much distortion in the input square wave.Step 22 Adjust both filter capacitors (C) to 50% (2.5 nF) each. (If the capacitors won’t change, click the mouse arrow in the circuit window). Bring down the oscilloscope enlargement and run the simulation to one full screen display, then pause the simulation. The red curve plot is the filter input and the blue curve plot is the filter output.Question: Are the filter input (red) and output (blue) curve plots the sameshape, disregarding any amplitude differences? No.Step 23 Bring down the Bode plotter enlargement to display the Bode plot of the filter. Use the cursor to measure the cutoff frequency (Fc of the low-pass filter and record the value. fc = 2.12 kHzStep 24 Bring down the spectrum analyzer enlargement to display the filter output frequency spectrum in the frequency domain, Run the simulation until the Resolution Frequencies match, then pause the simulation. Use cursor to measure the amplitude of the fundamental and each harmonic to the ninth and record your answers in Table 5- 2. Table 5-2 Frequency (kHz) Amplitude f1 1 4.4928 V f2 2 4.44397µV f3 3 792.585 mV f4 4 323.075 µV f5 5 178.663mV f6 6 224.681 µV f7 7 65.766 mV f8 8 172.430 µV f9 9 30.959 mVQuestions: How did the amplitude of each harmonic in Table 5-2 compare withthe values in Table 5-1? The amplitude of the harmonics is lower compare with the previous result.Based on the filter cutoff frequency (fc), how many of the square waveharmonics should be passed by this filter? Based on this answer, would youexpect much distortion of the input square wave at the filter output? Didyour answer in Step 22 verify this conclusion? There are less than 5 square wave harmonics. Yes, there is much distortion of the input square wave at output.Step 25 Change the both capacitor (C) back to 5% (0.25 nF). Change the duty cycle to 20% on the function generator. Bring down the oscilloscope enlargement and run the simulation to one full screen display, then
15. 15. pause the simulation. Notice that you have displayed a pulse curve plot on the oscilloscope in the time domain (voltage as a function of time). The red curve plot is the filter input and the blue curve plot is the filter output.Question: Are the filter input (red) and the output (blue) curve plots thesame shape, disregarding any amplitude differences? Yes.Step 26 Use the cursors to measure the time period (T) and the up time (to) of the input curve plot (red) and record the values. T= 1 ms to =Step 27 Calculate the pulse duty cycle (D) from the to and T. D = 19.82%Question: How did your calculated duty cycle compare with the duty cyclesetting on the function generator? Their difference is 0.18%Step 28 Bring down the Bode plotter enlargement to display the Bode plot of the filter. Use the cursor to measure the cutoff frequency (fC) of the low-pass filter and record the value. fC = 21.197 kHzStep 29 Bring down the spectrum analyzer enlargement to display the filter output frequency spectrum in the frequency domain. Run the simulation until the Resolution Frequencies match, then pause the simulation. Draw the frequency plot in the space provided. Also draw the envelope of the frequency spectrum. 5.041 kHzQuestion: Is this the frequency spectrum you expected for a square wave withduty cycle less than 50%? Yes.Step 30 Use the cursor to measure the frequency of the first zero crossing point (fo) of the spectrum envelope and record your answer on the graph. fo = 5.041 kHzStep 31 Based on the value of the to measured in Step 26, calculate the expected first zero crossing point (fo) of the spectrum envelope. fo = 5.045 kHzQuestion: How did your calculated value of fo compare the measured value onthe curve plot?
16. 16. The difference is 0.004 HzStep 32 Based on the value of fo, calculate the minimum bandwidth (BW) required for the filter to pass the input pulse waveshape with minimal distortion. BW = 5.045 kHzQuestion: Based on this answer and the cutoff frequency (fc) of the low-passfilter measure in Step 28, would you expect much distortion of the inputsquare wave at the filter output? Did your answer in Step 25 verify thisconclusion? No, there is less distortion.The higher the bandwidth, the lesser the distortion.Step 33 Adjust the filter capacitors (C) to 50% (2.5 nF) each. Bring down the oscilloscope enlargement and run the simulation to one full screen display, then pause the simulation. The red curve plot is the filter input and the blue curve plot is the filter output.Question: Are the filter input (red) and the output (blue) curve plots thesame shape, disregarding any amplitude differences? No.Step 34 Bring down the Bode plotter enlargement to display the Bode plotof the filter. Use the cursor to measure the cutoff frequency (fc) of thelow-pass filter and record the value. fc = 4.239 kHzQuestions: Was the cutoff frequency (fc) less than or greater than theminimum bandwidth (BW) required to pass the input waveshape with minimaldistortion as determined in Step 32? The fc is greater than the BW required to pass the input waveshape with minimal distortionBased on this answer, would you expect much distortion of the input pulsewaveshape at the filter output? Did your answer in Step 33 verify thisconclusion? No, there will have much distortion in the input waveshape at the output if the bandwidth is reduced.Step 35 Bring down the spectrum analyzer enlargement to display the filter output frequency spectrum in the frequency domain. Run the simulation until the Resolution Frequencies match, then pause the simulation.Question: What is the difference between this frequency plot and thefrequency plot in Step 29? In this frequency plot the amplitude is lower compared with the frequency plot in Step 29?
17. 17. Conclusion I conclude that a square wave and a triangular wave can be broken downinto a series of sine function (for square wave) and cosine function (fortriangular wave). Both the square wave and the triangular wave are oddfunction and the amplitude of the odd harmonics decrease in directproportion to odd harmonic frequency. Add to that, a nonsinusoidal functionis composed of fundamental sine wave and harmonics. The greater the value ofthe harmonics, the more it will become complex function. When a dc supply isadded to the circuit, the wave will shift upward but not increase its Vp. Ifa circuit has a duty cycle less than 20%, there will have even and oddharmonics. Lastly, bandwidth is inversely proportional to the up time of thepulse train. Bandwidth is also inversely proportional to the distortion ofthe input.