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### Pula

1. 1. NATIONAL COLLEGE OF SCIENCE AND TECHNOLOGY Amafel Bldg. Aguinaldo Highway Dasmariñas City, Cavite EXPERIMENT 5 “Fourier Theory – Frequency Domain and Time Domain”Pula, Rolando 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 COMPUTATIONDUTY CYCLEFREQUENCYFIRST ZERO CROSSING POINTBANDWIDTHBW =
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 voltage waveforms to simplify the analysis.In the real world, electrical information signal are normally nonsinusoidal voltage waveforms, such as audiosignals, video signals, or computer data. Fourier theory provides a powerful means of analyzingcommunications systems by representing a nonsinusoidal signal as series of sinusoidal voltages addedtogether. Fourier theory states that a complex voltage waveform is essentially a composite of harmonicallyrelated sine or cosine waves at different frequencies and amplitudes determined by the particular signalwaveshape. 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 or cosine wavesthat are integer multiples of the fundamental frequency, called the harmonics. This series of sine or cosinewave is called a Fourier series.Most of the signals analyzed in a communications system are expressed in the time domain, meaning thatthe voltage, current, or power is plotted as a function of time. The voltage, current, or power is representedon the vertical axis and time is represented on the horizontal axis. Fourier theory provides a new way ofexpressing signals in the frequency domain, meaning that the voltage, current, or power is plotted as afunction of frequency. Complex signals containing many sine or cosine wave components are expressed assine or cosine wave 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 straight linesrepresents the sine or cosine wave amplitudes, and the location of each line along the horizontal axisrepresents the sine 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 bandwidth requirements of thecommunications system in order to pass the signal with minimal distortion. Test instruments displayingsignals in both the time domain and the frequency domain are available. The oscilloscope is used to displaysignals in the time domain and the spectrum analyzer is used to display the frequency spectrum of signals inthe frequency domain.In the frequency domain, normally the harmonics decrease in amplitude as their frequency gets higher untilthe amplitude becomes negligible. The more harmonics added to make up the composite waveshape, themore the composite waveshape will look like the original waveshape. Because it is impossible to design acommunications system that will pass an infinite number of frequencies (infinite bandwidth), a perfectreproduction of an original signal is impossible. In most cases, eliminate of the harmonics does notsignificantly alter the original waveform. The more information contained in a signal voltage waveform (afterchanging voltages), the larger the number of high-frequency harmonics required to reproduce the originalwaveform. 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 bandwidth and theamount of information communicated is called Hartley’s law, which states that the amount of informationcommunicated is proportional to the bandwidth of the communications system and the transmission time.Because much of the information communicated today is digital, the accurate transmission of binary pulsesthrough a communications system is important. Fourier analysis of binary pulses is especially useful incommunications because it provides a way to determine the bandwidth required for the accurate
5. 5. transmission of digital data. Although theoretically, the communications system must pass all the harmonicsof 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 the time period ofone 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 inthe 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 of the square wave. The amplitude of each oddharmonic will decrease 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 sine wave voltages asspecified above. As the number of harmonics is decreased, the square wave that is produced will have moreripples. An infinite number of harmonics would be required to produce a perfectly flat square wave.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 10 14 13 12 6 R7 102 09 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 wave voltages. Inorder to generate a triangular wave, each harmonic frequency must be an odd multiple of the fundamentalwith no even harmonics. The fundamental frequency will be equal to the frequency of the triangular wave,the amplitude of each harmonic 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 waveshape will be shifted up by theamount of the dc voltage.
6. 6. 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 domain will consistof a fundamental and even and odd harmonics. The fundamental frequency will be equal to the frequency ofthe periodic pulse train. The amplitude (A) of each harmonic will depend on the value of the duty cycle. Ageneral frequency domain plot of a periodic pulse train with a duty cycle other than 50% is shown in thefigure on page 57. The outline of peaks if the individual frequency components is called envelope of thefrequency spectrum. The first zero-amplitude frequency crossing point is labelled f o = 1/to, there to is the uptime of the pulse train. The first zero-amplitude frequency crossing point f o) determines the minimumbandwidth (BW0 required for passing the pulse train with minimal distortion.Therefore, A f=1/to 2/to f Frequency Spectrum of a Pulse Train
7. 7. Notice than the lower the value of to the wider the bandwidth required to pass the pulse train with minimaldistortion. Also note that the separation of the lines in the frequency spectrum is equal to the inverse of thetime period (1/T) of the pulse train. Therefore a higher frequency pulse train requires a wider bandwidth (BW)because f = 1/TThe 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 theoriginal3 input waveshape. The frequency generator (XFG1) will generate a periodic pulse waveform appliedto the input of the filter (5). At the output of the filter (70, the oscilloscope will display the periodic pulsewaveform in the time domain, and the spectrum analyzer will display the frequency spectrum of the periodicpulse waveform in the frequency domain. The Bode plotter will display the Bode plot of the filter so that thefilter 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 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 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.
8. 8. 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 THE EQUAL.What is the relationship between the sine wave harmonic frequencies (frequencies of sine wave generatorsf3, f5, f7, and f9 in figure 5-1) and the sine wave fundamental frequency (f 1)? THE FREQUENCIES ARE ALL ODD MULTIPLES.What is the relationship between the amplitude of the harmonic sine wave generators and the amplitude ofthe fundamental sine wave generator? THE AMPLITUDE OF THE ODD HARMONICS 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. IT MOVED UPWARD. IT IS BECAUSE DC VOLTAGE ADDED.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.
9. 9. 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. ALL THE HARMONICS ARE ALL GONE SO IT APPEARS LIKE THE FUNDAMENTAL SINE WAVE.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.Step 11 Use the cursors to measure the time period for one cycle (T) of the triangular wave (blue) and thefundamental (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 wave frequency? THEY ARE THE EQUAL.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)? THEY ARE ALL ODD FUNCTIONS.What is the relationship between the amplitude of the harmonic generators and the amplitude of thefundamental generator? THE AMPLITUDE OF THE HARMONIC GENERATORS DECREASES IN DIRECT PROPORTION TO THE SQUARE OF THE ODD HARMONIC FREQUENCY
10. 10. Step 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. IT MOVED UPWARD. IT IS BECAUSE DC VOLTAGE ADDED.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, IT APPEARS AS A FUNDAMENTAL SINE WAVE.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 amplitudedifferences? YES.Step 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 = 500.477µS