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NATIONAL COLLEGE OF SCIENCE & TECHNOLOGY Amafel Bldg. Aguinaldo Highway Dasmariñas City, Cavite EXPERIMENT 5 Fourier Theory – Frequency Domain and Time DomainCauan, Sarah Krystelle P. September 06, 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 signal asseries 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, nonsinusoidalperiodic waveform can be broken down into sine or cosine wave equal to the frequency of theperiodic waveform, called the fundamental frequency, and a series of sine or cosine waves thatare integer multiples of 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 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 horizontal axis.Fourier theory provides a new way of expressing signals in the frequency domain, meaning thatthe voltage, current, or power is plotted as a function of frequency. Complex signals containingmany sine or cosine wave components are expressed as sine or cosine wave amplitudes atdifferent frequencies, with amplitude represented on the vertical axis and frequency representedon the horizontal axis. The length of each of a series of vertical straight lines represents the sineor cosine wave amplitudes, and the location of each line along the horizontal axis represents thesine or cosine wave frequencies. This is called a frequency spectrum. In many cases thefrequency domain is more useful than the time domain because it reveals the bandwidthrequirements of the communications system in order to pass the signal with minimal distortion.Test instruments displaying signals in both the time domain and the frequency domain areavailable. 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 frequency domain.
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In the frequency domain, normally the harmonics decrease in amplitude as their frequency getshigher until the amplitude becomes negligible. The more harmonics added to make up thecomposite waveshape, the more the composite waveshape will look like the original waveshape.Because it is impossible to design a communications system that will pass an infinite number offrequencies (infinite bandwidth), a perfect reproduction of an original signal is impossible. Inmost cases, eliminate of the harmonics does not significantly alter the original waveform. Themore information contained in a signal voltage waveform (after changing voltages), the largerthe number of high-frequency harmonics required to reproduce the original waveform.Therefore, the more complex the signal waveform (the faster the voltage changes), the wider thebandwidth required to pass it with minimal distortion. A formal relationship between bandwidthand the amount of information communicated is called Hartley’s law, which states that theamount of information communicated is proportional to the bandwidth of the communicationssystem 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 pulses isespecially 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, relativelyfew 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 square wave,the plot in the frequency domain will consist of a fundamental and all odd harmonics, with theeven harmonics missing. The fundamental frequency will be equal to the frequency of the squarewave. The amplitude 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 a series of sine wavevoltages as specified above. As the number of harmonics is decreased, the square wave that isproduced will have more ripples. An infinite number of harmonics would be required to producea perfectly flat square wave.
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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 a series of cosine wavevoltages. In order to generate a triangular wave, each harmonic frequency must be an oddmultiple of the fundamental with no even harmonics. The fundamental frequency will be equal tothe 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 be shiftedup by the amount of the dc voltage.
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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 plot in the frequency domainwill consist of a fundamental and even and odd harmonics. The fundamental frequency will beequal to the frequency of the periodic pulse train. The amplitude (A) of each harmonic willdepend on the value of the duty cycle. A general frequency domain plot of a periodic pulse trainwith a duty cycle other than 50% is shown in the figure on page 57. The outline of peaks if theindividual frequency components is called envelope of the frequency spectrum. The first zero-amplitude frequency 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 minimum bandwidth (BW0required for passing the pulse train with minimal distortion.Therefore,
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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.
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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 10kΩProcedure: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 asquare wave curve plot on the oscilloscope screen from a series of sine waves called a Fourierseries.Step 2 Run the simulation. Notice that you have generated a square wave curve plot on theoscilloscope screen (blue curve) from a series of sine waves. Notice that you have also plottedthe fundamental sine wave (red). Draw the square wave (blue) curve on the plot and thefundamental sine wave (red) curve plot in the space provided. T = 1.00 ms
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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 thetime period. f = 1 kHzQuestions: What is the relationship between the fundamental sine wave and the square wavefrequency (f)? Both the fundamental sine wave and the square wave frequency are 1 kHz. They are the sameWhat is the relationship between the sine wave harmonic frequencies (frequencies of sine wavegenerators f3, f5, f7, and f9 in figure 5-1) and the sine wave fundamental frequency (f1)? The sine wave harmonic frequencies and the sine wave fundamental frequency are all odd functions.What is the relationship between the amplitude of the harmonic sine wave generators and theamplitude of the fundamental sine wave generator? The amplitude of each harmonic sine wave generators is the quotient of amplitude of the fundamental sine wave generator to the sine wave harmonic frequencies. That is why it decreases as the sine wave harmonic frequencies increases.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 Akey). Run the simulation again. Change the oscilloscope settings as needed. Draw the new squarewave (blue) curve plot on the space provided.
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Question: What happened to the square wave curve plot? Explain why. The waveshape shifted upwards. This was caused by the additional dc voltage.Step 6 Press the F and E keys to open the switches F and E to eliminate the ninth and seventhharmonic sine waves. Run the simulation again. Draw the new curve plot (blue) in the spaceprovided. 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 thesimulation again. Draw the new curve plot (blue) in the space provided. Note any change on thegraph.
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Step 8 Press the C key to open switch C and eliminate the third harmonic sine wave. Run thesimulation again.Question: What happened to the square wave curve plot? Explain. It became sinusoidal. The square wave is consist of a fundamental and all odd harmonics but because of the absence of the sine wave harmonics, the waveshape became the sine wave fundamentalStep 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 atriangular wave curve plot on the oscilloscope screen from a series of sine waves called a Fourierseries.Step 10 Run the simulation. Notice that you have generated a triangular wave curve ploton the oscilloscope screen (blue curve) from the series of cosine waves. Notice that you havealso plotted the fundamental cosine wave (red). Draw the triangular wave (blue) curve plot andthe fundamental cosine wave (red) curve plot in the space provided. T = 1.00 ms
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Step 11 Use the cursors to measure the time period for one cycle (T) of the triangularwave (blue) and the fundamental (red), and show the value of T on the curve plot. T1 = 1.00 ms T2 = 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 wavefrequency? The triangular wave frequency and the sine wave fundamental frequency are the same.What is the relationship between the harmonic frequencies (frequencies of generators f3, f5, andf7 in figure 5-2) and the fundamental frequency (f1)? The sine wave harmonic frequencies and the sine wave fundamental frequency are all odd functions.What is the relationship between the amplitude of the harmonic generators and the amplitude ofthe fundamental generator? The amplitude of each harmonic sine wave generators is the quotient of amplitude of the fundamental sine wave generator to the square of the sine wave harmonic frequencies. That is why it decreases as the sine wave harmonic frequencies increases.Step 13 Press the A key to close switch A to add a dc voltage level to the triangular wavecurve plot. Run the simulation again. Draw the new triangular wave (blue) curve plot on thespace provided.
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Question: What happened to the triangular wave curve plot? Explain. The waveshape shifted upwards. This was caused by the additional dc voltage.Step 14 Press the E and D keys to open switches E and D to eliminate the seventh andfifth harmonic sine waves. Run the simulation again. Draw the new curve plot (blue) in the spaceprovided. 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 plot became sinusoidal. The triangular wave is consist of a fundamental and all odd harmonics but because the sine wave harmonics are missing, the waveshape became the sine wave fundamental
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Step 16 Open circuit FIG 5-3. Make sure that following function generator settings areselected: Square wave, Freq = 1 kHz, Duty cycle = 50%, Ampl – 2.5 V, Offset = 2.5 V. Makesure 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 andoutput of a two-pole low-pass Butterworth filter.Step 17 Bring down the oscilloscope enlargement and run the simulation to one full screendisplay, then pause the simulation. Notice that you are displaying square wave curve plot in thetime domain (voltage as a function of time). The red curve plot is the filter input (5) and the bluecurve plot is the filter output (7)Question: Are the filter input (red) and the output (blue) plots the same shape disregarding anyamplitude differences? Yes, they both are square wave.Step 18 Use the cursor to measure the time period (T) and the time (fo) of the input curveplot (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? The calculated duty cycle and the duty cycle setting on the function generator have difference 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 = 10dB, 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 spectrumanalyzer 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, thenpause the simulation. Notice that you have displayed the filter output square wave frequencyspectrum in the frequency domain, use the cursor to measure the amplitude of the fundamentaland 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 oddharmonics for a square wave with the duty cycle (D) calculated in Step 19? For a series of periodic pulse with 50% duty cycle, the frequency domain consists of a fundamental and odd harmonics while the even harmonics are almost negligible.What conclusions can you draw about the amplitude of each odd harmonic compared to thefundamental for a square wave with the duty cycle (D) calculated in Step 19? The amplitude of each odd harmonic decreases as the fundamental frequency for a square wave. Also, the plot in the frequency domain consist of a fundamental and all odd harmonics, with the even harmonics missingWas 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 waveharmonics would you expect to be passed by this filter? Based on this answer, would you expectmuch distortion of the input square wave at the filter? Did your answer in Step 17 verify thisconclusion? There are 11 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’tchange, click the mouse arrow in the circuit window). Bring down the oscilloscope enlargementand run the simulation to one full screen display, then pause the simulation. The red curve plot isthe filter input and the blue curve plot is the filter output.
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Question: Are the filter input (red) and output (blue) curve plots the same shape, disregardingany amplitude differences? No. They do not have the same shape.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 outputfrequency spectrum in the frequency domain, Run the simulation until the ResolutionFrequencies match, then pause the simulation. Use cursor to measure the amplitude of thefundamental 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 inTable 5-1? Compare with the previous table, the amplitude of the harmonics is lower.Based on the filter cutoff frequency (fc), how many of the square wave harmonics should bepassed by this filter? Based on this answer, would you expect much distortion of the input squarewave 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, therehave 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 to20% on the function generator. Bring down the oscilloscope enlargement and run the simulationto one full screen display, then pause the simulation. Notice that you have displayed a pulsecurve plot on the oscilloscope in the time domain (voltage as a function of time). The red curveplot 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, disregardingany 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 inputcurve 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?The values of both D are the almost 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 outputfrequency spectrum in the frequency domain. Run the simulation until the ResolutionFrequencies 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 with duty cycle lessthan 50%? Yes.
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Step 30 Use the cursor to measure the frequency of the first zero crossing point (fo) of thespectrum 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 zerocrossing 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 0.004 HzStep 32 Based on the value of fo, calculate the minimum bandwidth (BW) required for thefilter 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 inStep 28, would you expect much distortion of the input square wave at the filter output? Did youranswer in Step 25 verify this conclusion? No, because BW is inversely proportion to the distortion formed. Then, the higher thebandwidth, the lesser the distortion formed.Step 33 Adjust the filter capacitors (C) to 50% (2.5 nF) each. Bring down the oscilloscopeenlargement and run the simulation to one full screen display, then pause the simulation. The redcurve 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, disregardingany amplitude differences? No, they 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 (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 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?
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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 outputfrequency spectrum in the frequency domain. Run the simulation until the ResolutionFrequencies match, then pause the simulation. Question: What is the difference between this frequency plot and the frequency plot inStep 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 or generate a morecomplex waveshape. Square wave is used in the bode plot so harmonics are easily observed.Triangular wave is used in the spectrum analyzer to see the difference of the frequencies fromeach switch. Square wave consists of only fundamental frequency and the odd harmonics. Indirect proportion to the odd harmonic frequency, the amplitude of each odd harmonic willdecrease. For triangular wave, the amplitude of each harmonic will decrease in direct proportionto the square of the odd harmonic frequency. Moreover, decreasing the number of harmonic, themore ripple will appear to the generated square wave curve plot.
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