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NATIONAL COLLEGE OF SCIENCE & TECHNOLOGY Amafel Bldg. Aguinaldo Highway Dasmariñas City, Cavite EXPERIMENT 5 Fourier Theory – Frequency Domain and Time DomainBani, Arviclyn C. September 01, 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 to simplifythe analysis. In the real world, electrical information signal are normally nonsinusoidal voltagewaveforms, such as audio signals, video signals, or computer data. Fourier theory provides apowerful means of analyzing communications systems by representing a nonsinusoidal signalas series of sinusoidal voltages added together. Fourier theory states that a complex voltagewaveform is essentially a composite 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 cosine wave equal to thefrequency of the periodic waveform, called the fundamental frequency, and a series of sine orcosine waves that are integer multiples of the fundamental frequency, called the 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 time domain,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 is represented on the horizontalaxis. Fourier theory provides a new way of expressing signals in the frequency domain,meaning that the voltage, current, or power is plotted as a function of frequency. Complexsignals containing many sine or cosine wave components are expressed as sine or cosinewave amplitudes at different frequencies, with amplitude represented on the vertical axis andfrequency represented 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 line along thehorizontal axis represents the sine or cosine wave frequencies. This is called a frequencyspectrum. In many cases the frequency domain is more useful than the time domain becauseit reveals the bandwidth requirements of the communications system in order to pass thesignal with minimal distortion. Test instruments displaying signals in both the time domain andthe frequency domain are available. The oscilloscope is used to display signals in the timedomain and the spectrum analyzer is used to display the frequency spectrum of signals in thefrequency domain.
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In the frequency domain, normally the harmonics decrease in amplitude as their frequencygets higher until the amplitude becomes negligible. The more harmonics added to make upthe composite waveshape, the more the composite waveshape will look like the originalwaveshape. Because it is impossible to design a communications system that will pass aninfinite number of frequencies (infinite bandwidth), a perfect reproduction of an original signalis impossible. In most cases, eliminate of the harmonics does not significantly alter the originalwaveform. The more information contained in a signal voltage waveform (after changingvoltages), the larger the number of high-frequency harmonics required to reproduce theoriginal waveform. Therefore, the more complex the signal waveform (the faster the voltagechanges), the wider the bandwidth required to pass it with minimal distortion. A formalrelationship between bandwidth and the amount of information communicated is calledHartley’s law, which states that the amount of information communicated is proportional to thebandwidth of the communications system and the transmission time.Because much of the information communicated today is digital, the accurate transmission ofbinary pulses through a communications system is important. Fourier analysis of binary pulsesis especially useful in communications because it provides a way to determine the bandwidthrequired for the accurate transmission of digital data. Although theoretically, thecommunications system must pass all the harmonics of a pulse waveshape, in reality,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 the pulse up time (t O) to thetime 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 a squarewave, the plot in the frequency domain will consist of a fundamental and all odd harmonics,with the even harmonics missing. The fundamental frequency will be equal to the frequency ofthe square wave. The amplitude of each odd harmonic will decrease in direct proportion to theodd harmonic frequency. Therefore,The circuit in Figure 5–1 will generate a square wave voltage by adding a series of sine wavevoltages as specified above. As the number of harmonics is decreased, the square wave thatis produced will have more ripples. An infinite number of harmonics would be required toproduce 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 = EThe circuit in Figure 5-2 will generate a triangular voltage by adding a series of cosine wavevoltages. In order to generate a triangular wave, each harmonic frequency must be an odd
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multiple of the fundamental with no even harmonics. The fundamental frequency will be equalto the frequency of the triangular 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, the waveshape will beshifted up by the amount of the dc voltage.For a series of periodic pulses with other than a 50% duty cycle, the plot in the frequencydomain will consist of a fundamental and even and odd harmonics. The fundamentalfrequency will be equal to the frequency of the periodic pulse train. The amplitude (A) of eachharmonic will depend on the value of the duty cycle. A general frequency domain plot of aperiodic pulse train with a duty cycle other than 50% is shown in the figure on page 57. Theoutline of peaks if the individual frequency components is called envelope of the frequencyspectrum. The first zero-amplitude frequency crossing point is labelled f o = 1/to, there to is theup time of the pulse train. The first zero-amplitude frequency crossing point fo) determines theminimum bandwidth (BW) 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 the pulse trainwith minimal distortion. Also note that the separation of the lines in the frequency spectrum isequal to the inverse of the time period (1/T) of the pulse train. Therefore a higher frequencypulse train requires a wider bandwidth (BW) because f = 1/TThe circuit in Figure 5-3 will demonstrate the difference between the time domain and thefrequency domain. It will also determine how filtering out some of the harmonics effects theoutput waveshape compared to the original input waveshape. The frequency generator(XFG1) will generate a periodic pulse waveform applied to the input of the filter (5). At theoutput of the filter (7), the oscilloscope will display the periodic pulse waveform in the timedomain, and the spectrum analyzer will display the frequency spectrum of the periodic pulsewaveform in the frequency domain. The Bode plotter will display the Bode plot of the filter so
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that the filter bandwidth can be measured. The filter is a 2-pole low-pass Butterworth activefilter using a 741 op-amp.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 (red0 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 have the same value of frequency.
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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 fundamental frequency have odd values of the sine wave harmonic frequencies.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 frequency, the period and the other properties, except for the amplitude of the signal, are still the same The amplitude increased because of the additional dc voltage applied to the 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.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.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.
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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 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 have the same frequency.What is the relationship between the harmonic frequencies (frequencies of generators f 3, f5, and f7 in figure 5-2) and the fundamental frequency (f1)? The lowest frequency is the sine wave fundamental frequency. They are odd function with interval of 2 kHz.What is the relationship between the amplitude of the harmonic generators and the amplitude of the fundamental generator? The amplitude of each harmonic will decrease 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.
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Question: What happened to the triangular wave curve plot? Explain. The amplitude of the triangular wave curve increases. It is because of the additional dc voltage added to the circuit.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. The triangular wave curve became sine wave. It is because the harmonics sine waves are already 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
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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? YesStep 18 Use the cursor to measure the time period (T) and the time (f o) 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 on thefunction generator? They are almost equal.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.
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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 the even and odd harmonics for a square wave with the duty cycle (D) calculated in Step 19? There is only odd harmonics. All even harmonics are almost approaches 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 this is what I expectedBased 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 should be 21 square waves. Yes, I would you expect much distortion of the input square wave at the filter, 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?
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No they have different shape, the filter input (red) is square wave while the output curve is a sinusoidal wave.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 amplitude of the table is much lower compare with 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 5 square wave. Yes, there have much distortion in the input square wave at the filter output. Yes, answer in Step 22 verify this onclusion.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 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.
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Question: Are the filter input (red) and the output (blue) curve plots the same shape, disregarding any amplitude differences? Yes, the filter input and the output curve plots 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 on thefunction generator? They have the same value.Step 28 Bring down the Bode plotter enlargement to display the Bode plot of the filter. Use the cursor to measure the cutoff frequency (f C) 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.Question: Is this the frequency spectrum you expected for a square wave with duty cycle less than 50%? Yes, it is what I expected.Step 30 Use the cursor to measure the frequency of the first zero crossing point (f o) 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 Hz
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Step 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, 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, the filter input (red) and the output (blue) curve plots do not have the same shape.Step 34 Bring down the Bode plotter enlargement to display the Bode plot of the filter. Use the cursor to measure the cutoff frequency (f c) of the low-pass filter and record the value. fc = 4.239 kHzQuestions: Was the cutoff frequency (f c) less than or greater than the minimum bandwidth(BW) required to pass the input waveshape with minimal distortion as determined in Step 32? fc is greater than BW required.Based on this answer, would you expect much distortion of the input pulse waveshape at thefilter output? Did your answer in Step 33 verify this conclusion? No, if the bandwidth is reduced, there will occur much distortion of the inputpulse 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? As the number of harmonics increase, the amplitude of the harmonics decreases. That is why compare with the previous frequency plot, it is much lower.
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Conclusion I can therefore say that any sine wave can transform to any periodic function such astriangular and square wave through Fourier Series. Fourier Series is obtain through thesummation of the sine and cosine function. Adding more harmonics to the sine wave the morethe waveshape becomes a square wave or triangular wave. As the number of harmonics isdecreased, the square wave that is produced will have more ripples. An infinite number ofharmonics would be required to produce a perfectly flat square wave. In a triangular wave, theamplitude of each harmonic will decrease in direct proportion to the square of the oddharmonic frequency. For 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 odd harmonics.
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