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NATIONAL COLLEGE OF SCIENCE & TECHNOLOGY Amafel Bldg. Aguinaldo Highway Dasmariñas City, Cavite EXPERIMENT 5 Fourier Theory – Frequency Domain and Time DomainBoringot, Jeffrey B. September 1, 2011Signal Spectra and Signal Processing/BSECE 41A1 Score: Engr. Grace Ramones Instructor
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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.
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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 voltage waveforms tosimplify the analysis. In the real world, electrical information signal are normallynonsinusoidal voltage waveforms, such as audio signals, video signals, or computerdata. Fourier theory provides a powerful means of analyzing communicationssystems by representing a nonsinusoidal signal as series of sinusoidal voltages addedtogether. Fourier theory states that a complex voltage waveform is essentially acomposite of harmonically related sine or cosine waves at different frequencies andamplitudes determined by the particular signal waveshape. Any, nonsinusoidalperiodic waveform can be broken down into sine or cosine wave equal to thefrequency of the periodic waveform, called the fundamental frequency, and a series ofsine or cosine waves that are integer multiples of the fundamental frequency, calledthe harmonics. This series of sine or cosine wave is called a Fourier series.Most of the signals analyzed in a communications system are expressed in the timedomain, meaning that the voltage, current, or power is plotted as a function of time.The voltage, current, or power is represented on the vertical axis and time isrepresented on the horizontal axis. Fourier theory provides a new way of expressingsignals in the frequency domain, meaning that the voltage, current, or power isplotted as a function of frequency. Complex signals containing many sine or cosinewave components are expressed as sine or cosine wave amplitudes at differentfrequencies, with amplitude represented on the vertical axis and frequencyrepresented on the horizontal axis. The length of each of a series of vertical straightlines represents the sine or cosine wave amplitudes, and the location of each linealong the horizontal axis represents the sine or cosine wave frequencies. This is called
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a frequency spectrum. In many cases the frequency domain is more useful than thetime domain because it reveals the bandwidth requirements of the communicationssystem in order to pass the signal with minimal distortion. Test instrumentsdisplaying signals in both the time domain and the frequency domain are available.The oscilloscope is used to display signals in the time domain and the spectrumanalyzer is used to display the frequency spectrum of signals in the frequencydomain.In the frequency domain, normally the harmonics decrease in amplitude as theirfrequency gets higher until the amplitude becomes negligible. The more harmonicsadded to make up the composite waveshape, the more the composite waveshape willlook like the original waveshape. Because it is impossible to design a communicationssystem that will pass an infinite number of frequencies (infinite bandwidth), a perfectreproduction of an original signal is impossible. In most cases, eliminate of theharmonics does not significantly alter the original waveform. The more informationcontained in a signal voltage waveform (after changing voltages), the larger thenumber of high-frequency harmonics required to reproduce the original waveform.Therefore, the more complex the signal waveform (the faster the voltage changes), thewider the bandwidth required to pass it with minimal distortion. A formalrelationship between bandwidth and the amount of information communicated iscalled Hartley’s law, which states that the amount of information communicated isproportional to the bandwidth of the communications system and the transmissiontime.Because much of the information communicated today is digital, the accuratetransmission of binary pulses through a communications system is important. Fourieranalysis of binary pulses is especially useful in communications because it provides away to determine the bandwidth required for the accurate transmission of digitaldata. Although theoretically, the communications system must pass all the harmonicsof a pulse waveshape, in reality, relatively few of the harmonics are need to preservethe waveshape.The duty cycle of a series of periodic pulses is equal to the ratio of the pulse up time(tO) to the time period of one cycle (T) expressed as a percentage. Therefore,In the special case where a series of periodic pulses has a 50% duty cycle, called asquare wave, the plot in the frequency domain will consist of a fundamental and all
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odd harmonics, with the even harmonics missing. The fundamental frequency will beequal to the frequency of the square wave. The amplitude of each odd harmonic willdecrease in direct proportion to the odd harmonic frequency. Therefore,The circuit in Figure 5–1 will generate a square wave voltage by adding a series of sinewave voltages as specified above. As the number of harmonics is decreased, thesquare wave that is produced will have more ripples. An infinite number ofharmonics would be required to produce a perfectly flat square wave.
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Figure 5 – 1 Square Wave Fourier Series XSC1 V6 15 R1 1 J1 Ext T rig + _ 0 10.0kΩ A B 10 V _ _ + + Key = A V1 R2 J2 9 2 10 Vpk 10.0kΩ 6 1kHz Key = B 0° V2 10 R3 3 J3 R7 100Ω 3.33 Vpk 10.0kΩ 3kHz 0 V3 Key = C 0° 12 R4 4 J4 2 Vpk 10.0kΩ 5kHz Key = D 0° V4 14 R5 5 J5 1.43 Vpk 10.0kΩ 7kHz 0° Key = E V5 J6 R6 8 0 13 1.11 Vpk 10.0kΩ 9kHz Key = F 0°Figure 5 – 2 Triangular Wave Fourier Series XSC1 V6 12 R1 1 J1 Ext T rig + _ 0 10.0kΩ A B 10 V _ _ + + Key = A V1 R2 J2 13 2 10 Vpk 10.0kΩ 1kHz Key = B 90° V2 8 R3 3 J3 R7 1.11 Vpk 100Ω 10.0kΩ 3kHz 0 90° V3 Key = C R4 J4 6 9 4 0.4 Vpk 10.0kΩ 5kHz 90° V4 Key = D 0 11 R5 5 J5 0.2 Vpk 10.0kΩ 7kHz 90° Key = E
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The circuit in Figure 5-2 will generate a triangular voltage by adding a series of cosinewave voltages. In order to generate a triangular wave, each harmonic frequency mustbe an odd multiple of the fundamental with no even harmonics. The fundamentalfrequency will be equal to the frequency of the triangular wave, the amplitude of eachharmonic will decrease in direct proportion to the square of the odd harmonicfrequency. Therefore,Whenever a dc voltage is added to a periodic time varying voltage, the waveshapewill be shifted up by the amount of the dc voltage.For a series of periodic pulses with other than a 50% duty cycle, the plot in thefrequency domain will consist of a fundamental and even and odd harmonics. Thefundamental frequency will be equal to the frequency of the periodic pulse train. Theamplitude (A) of each harmonic will depend on the value of the duty cycle. A generalfrequency domain plot of a periodic pulse train with a duty cycle other than 50% isshown in the figure on page 57. The outline of peaks if the individual frequencycomponents is called envelope of the frequency spectrum. The first zero-amplitudefrequency crossing point is labelled fo = 1/to, there to is the up time of the pulse train.The first zero-amplitude frequency crossing point fo) determines the minimumbandwidth (BW0 required for passing the pulse train with minimal distortion.Therefore,Notice than the lower the value of to the wider the bandwidth required to pass thepulse train with minimal distortion. Also note that the separation of the lines in thefrequency 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
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The circuit in Figure 5-3 will demonstrate the difference between the time domain andthe frequency domain. It will also determine how filtering out some of the harmonicseffects the output waveshape compared to the original3 input waveshape. Thefrequency generator (XFG1) will generate a periodic pulse waveform applied to theinput of the filter (5). At the output of the filter (70, the oscilloscope will display theperiodic pulse waveform in the time domain, and the spectrum analyzer will displaythe frequency spectrum of the periodic pulse waveform in the frequency domain. TheBode plotter will display the Bode plot of the filter so that the filter bandwidth can bemeasured. The filter is a 2-pole low-pass Butterworth active filter using a 741 op-amp.
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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 and the square wave frequency (f)? They are both 1 kHz. They have the same value.
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What is the relationship between the sine wave harmonic frequencies (frequencies of sine wave generators f3, f5, f7, and f9 in figure 5-1) and the sine wave fundamental frequency (f1)? The sine wave harmonic frequency is different with the sine wave fundamental. The harmonics frequency has lot of ripples.What is the relationship between the amplitude of the harmonic sine wave generators and the amplitude of the fundamental sine wave generator? The amplitude of the odd harmonics will decrease in direct proportion to odd 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.Question: What happened to the square wave curve plot? Explain why.The amplitude increased. This is due to the dc voltage applied to the previous circuit.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.
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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 because the harmonic frequency generators had been eliminated.Step 9 Open circuit file FIG 5-2. 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 = 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.
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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) of the triangular wave (blue) and the fundamental (red), and show the value of T on the curve plot. T = 1.00 msStep 12 Calculate the frequency (f) of the triangular wave from the time period (T). f = 1 kHzQuestions: What is the relationship between the fundamental frequency and the triangular wave frequency? They are the same.What is the relationship between the harmonic frequencies (frequencies of generators f3, f5, and f7 in figure 5-2) and the fundamental frequency (f1)? Each harmonic frequency is an odd multiple of the fundamental.What is the relationship between the amplitude of the harmonic generators and the amplitude of the fundamental generator?
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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.Question: What happened to the triangular wave curve plot? Explain. The waveshape shifted up. It is because dc voltage is added to a periodic time varying voltage; the waveshape will be shifted up by the amount of the 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.
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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 had already been eliminated.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 same shape disregarding any amplitude differences? Yes
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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 cycle setting onthe function generator? They have only diffence of 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
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f8 8 25.452 µV f9 9 552.582 mVQuestions: What conclusion can you draw about the difference between the even and odd harmonics for a square wave with the duty cycle (D) calculated in Step 19? The wave consists of odd harmonics while the even harmonics are almost zero.What conclusions can you draw about the amplitude of each odd harmonic compared 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 the duty cycle (D) calculated in Step 19? Yes.Based on the filter cutoff frequency (fC) measured in Step 20, how many of the square wave harmonics would you expect to be passed by this filter? Based on this answer, would you expect much distortion of the input square wave at the filter? Did your answer in Step 17 verify this conclusion? There are 21 square waves. Yes, because the more number of harmonics square wave the more 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 same shape, disregarding any amplitude differences? No, the input is square wave while the output is a sinusoidal wave.
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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 with the values in Table 5-1? The result is lower than the previous table.Based on the filter cutoff frequency (fc), how many of the square wave harmonics should be passed by this filter? Based on this answer, would you expect much distortion of the input square wave at the filter output? Did your answer in Step 22 verify this conclusion? There should be less than 5 square wave harmonics to be passed by this filter. Yes, there have much distortion in the input square wave at the filter 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
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enlargement and run the simulation to one full screen display, then 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 the same shape, disregarding any amplitude differences? Yes, they have the same shape.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 = 198.199 µsStep 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 cycle setting onthe function generator?The values of both D are the same.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.
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Question: Is this the frequency spectrum you expected for a square wave with duty cycle less than 50%? Yes, it is what I expected for 50% duty cycle.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.Step 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 on the curve plot? They have a difference of 117 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 = 4.719 kHzQuestion: Based on this answer and the cutoff frequency (fc) of the low-pass filter measure in Step 28, would you expect much distortion of the input square wave at the filter output? Did your answer in Step 25 verify this conclusion? No, because BW is inversely proportion to the distortion formed. Then, the higher the bandwidth, the lesser the distortion formed.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 the same shape, disregarding any amplitude differences? No, they do not have the same shape.
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Step 34 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 = 4.239 kHzQuestions: Was the cutoff frequency (fc) less than or greater than the minimumbandwidth (BW) required to pass the input waveshape with minimal distortion asdetermined in Step 32? fc is greater than BW required.Based on this answer, would you expect much distortion of the input pulsewaveshape at the filter output? Did your answer in Step 33 verify this conclusion? No, if the bandwidth is reduced, there will occur much distortion of theinput pulse waveshape at the filter output .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 the frequency plot in Step 29? It is inversely proportional. As the number of the harmonics increase, the amplitudedecrease.
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Conclusion: Based on the experiment, more harmonics added to the sine wave will result orgenerate a more complex waveshape. Square wave is used in the bode plot so harmonics areeasily observed. Triangular wave is used in the spectrum analyzer to see the difference of thefrequencies from each switch. Square wave consists of only fundamental frequency and theodd harmonics. In direct proportion to the odd harmonic frequency, the amplitude of eachodd harmonic will decrease. For triangular wave, the amplitude of each harmonic willdecrease in direct proportion to the square of the odd harmonic frequency. Moreover,decreasing the number of harmonic, the more ripple will appear to the generated square wavecurve plot.
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