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# Exp amplitude modulation (8)

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### Exp amplitude modulation (8)

1. 1. NATIONAL COLLEGE OF SCIENCE AND TECHNOLOGY Amafel Building, Aguinaldo Highway Dasmariñas City, Cavite Experiment No. 5 AMPLITUDE MODULATIONLopera, Jericho James August 09. 2011Signal Spectra and Signal Processing/BSECE 41A1 Score: Engr. Grace Ramones Instructor
2. 2. OBJECTIVES: 1. Demonstrate an amplitude-modulated carrier in the time domain for different modulation indexes and modulating frequencies. 2. Determine the modulation index and percent modulation of an amplitude-modulated carrier from the time domain curve plot. 3. Demonstrate an amplitude-modulated carrier in the frequency domain for different modulation indexes and modulating frequencies. 4. Compare the side-frequency voltage levels to the carrier voltage level in an amplitude-modulated carrier for different modulation indexes. 5. Determine the signal bandwidth of an amplitude-modulated carrier for different modulating signal frequencies. 6. Demonstrate how a complex modulating signal generates many side frequencies to form the upper and lower sidebands.
3. 3. SAMPLE COMPUTATIONS:Step 3Step 4Step 7Step 8Step 9Step 9 QStep 10Step 10 QStep 12Step 13Step 13 QStep 16Step 17
4. 4. Step 19Step 20Step 21 QStep 26Step 28Step 30Step 35
5. 5. DATA SHEET:MaterialsTwo function generatorsOne dual-trace oscilloscopeOne spectrum analyzerOne 1N4001 diodeOne dc voltage supplyOne 2.35 nF capacitorOne 1 mH inductorResistors: 100 Ω, 2 kΩ, 10 kΩ, 20 kΩTheory:The primary purpose of a communications system is to transmit and receive information such as audio,video, or binary data over a communications medium or channel. The basic components in acommunications s ystem are the transmitter communications medium or channel, and the receiver.Possible communications media are wire cable, fiber optic cable, and free space. Before the informationcan be transmitted, it must be converted into an electrical signal compatible with the communicationsmedium, this is the purpose of the transmitter while the purpose of the receiver is to receive thetransmitted signal from the channel and convert it into its original information signal form. Is the originalelectrical information signal is transmitted directly over the communications channel, it is called basebandtransmission. An example of a communications system that uses baseband transmission is the telephonesystem.Noise is defined as undesirable electrical energy that enters the communications system and interfereswith the transmitted message. All communication systems are subject to noise in both the communicationchannel and the receiver. Channel noise comes from the atmosphere (lightning), outer space (radiationemitted by the sun and stars), and electrical equipment (electric motors and fluorescent lights). Receivernoise comes from electronic components such as resistors and transistors, which generate noise due tothermal agitation of the atoms during electrical current flow. In some cases obliterates the message andin other cases it results in only partial interference. Although noise cannot be completely eliminated, it canbe reduced considerably.Often the original electrical information (baseband) signal is not compatible with the communicationsmedium. In that case, this baseband signal is used to modulate a higher-frequency sine wave signal thatis in a frequency spectrum that is compatible with the communications medium. This higher-frequencysine wave signal is called a carrier. W hen the carrier frequency is in the electromagnetic spectrum it iscalled a radio frequency (RF) wave, and it radiates into space more efficiently and propagates a longerdistance than a baseband signal. W hen information is transmitted over a fiber optic cable, the carrierfrequency is in the optical spectrum. The process of using a baseband signal to modulate a carrier iscalled broadband transmission.There are basically three ways to make a baseband signal modulate a sine wave carrier: amplitudemodulation (AM), frequency modulation (FM), and phase modulation (PM). In amplitude modulation (AM),the baseband information signal varies the amplitude of the higher-frequency carrier. In frequencymodulation (FM), the baseband information signal varies the frequency of the higher-frequency carrierand the carrier amplitude remains constant. In phase modulation (PM), the baseband information signalvaries the phase of the high-frequency carrier. Phase modulation (PM) is different fork of frequencymodulation and the carrier is similar in appearance to a frequenc y-modulated signal carrier. Therefore,both FM and PM are often referred to as an angle modulation. In this experiment, you will examine thecharacteristics of amplitude modulation (AM).In amplitude modulation, the carrier frequenc y remains constant, but the instantaneous value of thecarrier amplitude varies in accordance with the amplitude variations of the modulating signal. Animaginary line joining the peaks of the modulated carrier waveform, called the envelope, is the sameshape as the modulating signal with the zero reference line coinciding with the peak value of theunmodulated carrier. The relationship between the peak voltage of the modulating signal (Vm) and thepeak voltage of the unmodulated carrier (Vc) is the modulation index (m), therefore,
6. 6. Multiplying the modulation index (m) by 100 gives the percent modulation. W hen the peak voltage of themodulating signal is equal to the peak voltage of the unmodulated carrier, the percent modulation is100%. An unmodulated carrier has a percent modulation of 0%. W hen the peak voltage of the modulatingsignal (Vm) exceeds the peak voltage of the unmodulated carrier (Vc) overmodulation will occur, resultingin distortion of the modulating (baseband) signal when it is recovered from the modulated carrier.Therefore, if it is important that the peak voltage of the modulating signal be equal to or less than thepeak voltage of the unmodulated signal carrier (equal to or less than 100% modulation) with amplitudemodulation.Often the percent modulation must be measured from the modulated carrier displayed on an oscilloscope.W hen the AM signal is displayed on an oscilloscope, the modulation index can be computed fromwhere Vmax is the maximum peak-to-peak voltage of the modulated carrier and Vmin is the minimum peak-to-peak voltage of the modulated carrier. Notice that when Vmax = 0, the modulation index (m) is equal to 1(100% modulation), and when Vmin = Vmax, the modulation index is equal to 0 (0% modulation).when a single-frequency sine wave amplitude modulates a carrier, the modulating process causes twoside frequencies to be generated above and below the carrier frequency be an amount equal to themodulating frequency (fm). The upper side frequency (fuse) and the lower side frequency (fluff) can bedetermine fromA complex modulating signal, such as a square wave, consists of a fundamental sine wave frequenc y andmany harmonics, causing many side frequencies to be generated. The highest upper side frequenc y andthe lowest lower side frequency are determined be the highest harmonic frequency (fm (max), and thehighest lower side frequency and the lower upper side frequency are determined by the lowest harmonicfrequency (fm (min)). The band of frequencies between (fC + fm (min)) and (fC + fm (max)) is called the uppersideband. The band of frequencies between (fC – fm (min)) and (fC – fm (max)) is called the lowers sideband.The difference between the highest upper side frequency (fC + fm (max)) and the lowest lower sidefrequency (fC – fm (max)) is called the bandwidth occupied by the modulated carrier. Therefore, thebandwidth (BW ) can be calculated fromThis bandwidth occupied by the modulated carrier is the reason a modulated carrier is referred to asbroadband transmission.The higher the modulating signal frequencies (meaning more information is being transmitted) the widerthe modulated carrier bandwidth. This is the reason a video signal occupies more bandwidth than anaudio signal. Because signals transmitted on the same frequency interfere with one another, this carrierbandwidth places a limit on the number of modulated carriers that can occupy a given communicationschannel. Also, when a carrier is overmodulated, the resulting distorted waveshape generates a harmonicthat generate additional side frequencies. This causes the transmitted bandwidth to increase and interferewith other signals. This harmonic-generated sideband interference is called splatter.W hen an AM signal on an oscilloscope, it is observed in the time domain (voltage as a function time). Thetime domain display gives no indication of sidebands. In order to observe the sidebands generated by themodulated carrier, the frequency spectrum of the modulated carrier must be displayed in the frequenc ydomain (sine wave voltage levels as a function of frequency) on a spectrum analyzer.A sine wave modulated AM signal is a composite of a carrier and two side frequencies, and each of thesesignals transmits power. The total power transmitted (PT) is the sum of each carrier power ((PC) and thepower in the two side frequencies (PUSF and PUSB). Therefore,
7. 7. The total power transmitted (PT) can also be determined from the modulation index (m) using theequationTherefore, the total power in the side frequencies (PSF) isand the power in each side frequency isNotice that the power in the side frequencies depends on the modulation index (percent modulation) andthe carrier power does not depend on the modulation index. W hen the percent modulation is 100% (m =1), the total side-frequency power (PSF) is one-half of the carrier power (PC) and the power in each sidefrequency (PUSF and PLSF) is one-quarter of the carrier power (PC). W hen the percent modulation is 0%,the total side-frequency power (PSF) is zero because there are no side frequencies in an unmodulatedcarrier. Based on these results, it is easy to calculate that an amplitude-modulated carrier has all of thetransmitted information in the sidebands and no information in the carrier. For 100% modulation, one-thirdof the total power transmitted is in the sidebands and two-thirds of the total power is wasted in the carrier,which contains no information. W hen a carrier is modulated by a complex waveform, the combinedmodulation index of the fundamental and all of the harmonics determines the power in the sidebands. In alater experiment, you will see how we can remove and transmit the same amount of information with lesspower.Because power is proportional to voltage squared, the voltage level of the frequencies is equal to thesquare root of the side-frequency power. Therefore the side-frequenc y voltage can be calculated fromThis means that the voltage of each side frequency is equal to one-half the carrier voltage for 100%modulation of a sine-wave modulated carrier. W hen a carrier is modulated by a complex waveform, thevoltage of each side frequency can calculated from the separate modulation indexes of the fundamentaland each harmonic.A circuit that mathematically multiplies a carrier and modulating (baseband) signal, and then adds thecarrier to the result, will produce an amplitude-modulated carrier. Therefore, the circuit in Figure 6-1 willbe used to demonstrate amplitude modulation. An oscilloscope has been attached to the output to displaythe modulated carrier in the time domain. A spectrum analyzer has been attached to the output to displaythe frequency spectrum of the amplitude-modulated carrier in the frequency domain.
8. 8. XSA1 XSC1 Modulating Signal XFG1 Ext T rig + _ IN T A B + _ + _ 0 0 MULTIPLIER SUMMER 3 Y C X 2 A 1 1 V/V 4 1 V/V 0 V B 0V Carrier 1 Vpk 100kHz 0 0°ProcedureStep 1 Open circuit file Fig 6-1. This circuit will demonstrate how mathematical multiplication of a carrier and a modulating (baseband) signal, and then adding the carrier to the multiplication result, will produce an amplitude modulated carrier. Bring down the function generator enlargement and make sure that the following settings are selected: Sine Wave, Freq=5kHz, Ampl = 1V, Offset=0. Bring down the oscilloscope enlargement and make sure that the following settings are selected: Time base (Scale = 50 µs/Div, Xpos= 0, Y/T) Ch A (Scale = 1 V/Div, Ypos = 0, DC) Trigger (Pos edge, Level = 0, Auto).Step 2 Run the simulation to one full screen display, then pause the simulation. Notice that you have displayed an amplitude-modulated carrier curve plot on the oscilloscope screen. Draw the curve plot in the space provided and show the envelope on the drawing.Step 3 Based on the function generator amplitude (modulating sine wave voltage, Vm) and the voltage of the carrier sine wave (Vc), calculate the expected modulation index (m) and percent modulation.  m=1 or 100%Step 4 Determine the modulation index (m) and percent modulation from the curve plot in Step 2.
9. 9.  m=1 or 100%Question: How did the value of the modulation index and percent modulation determined from the curveplot compare with the expected value calculated in Step 3?  The two values have no difference. They both have the same value.Step 5 Bring down the spectrum analyzer enlargement and make sure that the following settings are selected: Frequency (Center = 100 kHz, Span = 100 kHz), Amplitude (Lin, Range = 0.2 V/Div) Resolution = 500 Hz.Step 6 Run the simulation until the Resolution Frequency match, then pause the simulation. You have displayed the frequency spectrum for a modulated carrier. Draw the spectral plot in the space provided.Step 7 Measure the carrier frequency (fc), the upper side frequency (fUSF), the lowest side frequenc y (fLSF), and the voltage amplitude of each spectral line and record the answer on the spectral plot.  fc = 100 kHz Vc = 999.085 mV  fLSF = 95.041 kHz VLSF = 458.267 mV  fUSF = 104.959 kHz VUSF = 458.249 mVQuestion: W hat was the frequenc y difference between the carrier frequency and each of the sidefrequencies? How did this compare with the modulating signal frequency?  fc – fLSF = 4.959 kHz fUSF – fc = 4.959 kHz  They are almost equal. The difference is only 0.82%.How did the frequency of the center spectral line compare with the carrier frequency?  The two values have no difference. They both have the same value.Step 8 Calculate the bandwidth (BW ) of the modulated carrier based on the frequency of the modulating sine wave.  BW = 10 kHzStep 9 Determine the bandwidth (BW ) of the modulated carrier from the frequency spectral plot andrecord your answer on the spectral plot.  BW = 9.918 kHzQuestion: How did the bandwidth of the modulated carrier from the frequenc y spectrum compare with thecalculated bandwidth in Step 8?  They are almost equal. The difference is only 0.82%.
10. 10. Step 10 Calculate the expected voltage amplitude of each side frequency spectral line (VUSF and VLSF)based on the modulation index (m) and the carrier voltage amplitude  VUSF = VLSF = 0.5 VQuestions: How did the calculated voltage values compare with the measured values in on the spectralline?  They are almost equal. The difference is only 9.11%.W hat was the relationship between the voltage levels of the side frequencies and the voltage level of thecarrier? W as this what you expected for this modulation index?  It is one-half of the amplitude of the carrier. This is expected for a 1 modulation index.Step 11 Change the modulating signal amplitude (function generator amplitude) to 0.5 V (500 mV). Bring down the oscilloscope enlargement and run the simulation to one full-screen display, then pause the simulation. Notice that you have displayed an amplitude-modulated carrier curve plot on the oscilloscope screen. Draw the curve plot in the space provided and show the envelope on the drawing.Step 12 Based on the voltage of the modulating (baseband) sine wave (Vm) and the voltage of the carrier sine wave (Vc), calculate the expected modulation index (m) and the percent modulation.  m = 0.5 = 50%Step 13 Determine the modulation index (m) and the percent modulation from the curve plot in Step 11.  m = 0.51 = 51%Questions: How did the value of the modulation index and percent modulation determined from the curveplot compare with the expected value calculated in Step 12?  They are almost equal. The difference is only 2%.How did this percent modulation compare with the percent modulation in Step 3 and 4? Explain anydifference.  It decreased by almost 50%. The modulating signal decreased by half so percent modulation also decreased by 50%Step 14 Bring down the spectrum analyzer enlargement. Run the simulation until the Resolution Frequenc y match, then pause the simulation. You have plotted the frequency spectrum for a modulated carrier. Draw the spectral plot in the space provided.
11. 11. Step 15 Measure the carrier frequency (fC), the upper side frequency (fUSF), the lower side frequency (fLSF), and the voltage amplitude of each spectral line and record the answers on the spectral plot.  fc = 100 kHz Vc = 998.442 mV  fLSF = 95.041 kHz VLSF = 228.996 mV  fUSF = 104.959 kHz VUSF = 228.968 mVStep 16 Determine the bandwidth (BW ) of the modulated carrier from the frequency spectral plot and record your answer on the spectral plot.  BW = 9.918 kHzQuestion: How did the bandwidth of the modulated carrier from this frequency spectrum compare with thebandwidth on the spectral plot in Step 6? Explain.  The difference is 0%. They have the same valuesStep 17 Calculate the expected voltage amplitude of each side frequency spectral line (VUSF) based on the modulation index (m) and the carrier voltage amplitude (VC).  VUSF = VLSF = 0.25 VQuestions: How did the calculated voltage values compare with the measured values in on the spectralplot?  They are almost equal. The difference is only 9.17%.W hat was the relationship between the voltage levels of the side frequencies and the voltage level of thecarrier? How did it compare with the results in Step 6?  It is one-fourth of the amplitude of the carrier. The percent modulation decreased by 50% so it is half of the result compare to other.Step 18 Change the modulating amplitude (function generator amplitude) to 0 V (1 µV). Bring down the oscilloscope enlargement and run the simulation to one full screen display, then pause the simulation. Draw the curve plot in the space provided.
12. 12. Step 19 Based on the voltage of the modulating (baseband) sine wave (Vm) and the voltage of the carrier sine wave (Vc), calculate the expected modulation index (m) and percent modulation. -6  m = 0.1 × 10Step 20 Determine the modulation index (m) and percent modulation from the curve plot in Step 18.  m=0Questions: How did the value of the modulation index and the percent modulation determined from thecurve plot compare with the expected value calculated in Step 19? -6  There is 0.1 × 10 difference.How did this waveshape compare with the previous amplitude-modulated waveshapes? Explain anydifference.  It is like the carrier signal waveshape alone, because the amplitude of the modulating signal is almost 0.Step 21 Bring down the spectrum analyzer. Run the Resolution Frequencies match, then pause the simulation. You have plotted the frequency spectrum for an unmodulated carrier. Draw the spectral plot in the space provided.Step 22 Measure the frequency and voltage amplitude of the spectral line and record the values on the spectral plot.  fc = 100 kHz Vc = 998.441 mVQuestions: How did the frequenc y of the spectral line compare with the carrier frequenc y?
13. 13.  The two values have no difference. They both have the same value.How did the voltage amplitude of the spectral line compare with the carrier amplitude?  They are almost equal. The difference is only 0.16%.How did this frequency spectrum compare with the previous frequenc y spectrum?  The carrier frequency is only the signal present. There is no side frequencies.Step 23 Change the modulating frequenc y to 10 kHz and the amplitude back to 1 V on the function generator. Bring down the oscilloscope and run the simulation to one full screen display, then pause the simulation. Notice that you have displayed an amplitude-modulated carrier curve plot on the oscilloscope screen. Draw the curve plot in the space provided and show the envelope on the drawing.Question: How did this waveshape differ from the waveshape for a 5 kHz modulating frequency in Step2?  Compare to waveshape in Step 2, the waveshape this time is thinner and has lot of cycle per second.Step 24 Bring down the spectrum analyzer enlargement. Run the simulation until the Resolution Frequencies match, then pause the simulation. You have plotted the frequency spectrum for a modulated carrier. Draw the spectral plot in the space provided.Step 25 Measure the carrier frequency (fC), the upper side frequenc y (fUSF), and the lower side frequency (fLSF) of the spectral lines and record the answers on the spectral plot.  fc = 100 kHz fLSF = 90.083 kHz fUSF = 109.917 kHzStep 26 Determine the bandwidth (BW ) of the modulated carrier from the frequency spectral plot and record your answer on the spectral plot.
14. 14.  BW = 19.834 kHzQuestion: How did the bandwidth of the modulated carrier for a 10 kHz modulating frequency comparewith the bandwidth for a 5 kHz modulating frequency in Step 6?  Compare with the bandwidth in Step 6, the bandwidth this time increased by 10 kHz.Step 27 Measure the voltage amplitude of the side frequencies and record your answer on the spectral plot in Step 24.  Vc = 998.436 mV VLSF = 416.809 mV VUSF = 416.585 mVQuestion: W as there any difference between the amplitude of the side frequencies for the 10 kHz plot inStep 24 and the 5 kHz in Step 6? Explain.  There is 41.458 mV difference. As the frequency of the modulating signal increases, the amplitude of each sideband decreasesStep 28 Change the modulating frequenc y to 20 kHz on the function generator. Run the simulation until the Resolution Frequencies match, the pause the simulation. Measure the bandwidth (BW ) of the modulated carrier on the spectrum analyzer and record the value.  BW = 39.67 kHzQuestion: How did the bandwidth compare with the bandwidth for the 10 kHz modulating frequenc y?Explain.  The bandwidth of the waveshape increased twice of the modulating frequency that explains why it increases 20 kHzStep 29 Change the modulating signal frequency band to 5kHz and select square wave 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 carrier modulated by a square wave on the oscilloscope screen. Draw the curve plot in the space proved and show the envelope on the drawing.Question: How did this waveshape differ from the waveshape in step 2? The waveshape is complex wave. The modulating signal is square wave which consists of a fundamental sine wave frequency and many harmonics, so there are lot of side frequencies generated.Step 30 Determine the modulation index (m) and percent modulation from the curve plot in Step 29.  m=1Step 31 Bring down the spectrum analyzer enlargement. Run the simulation until the Resolution Frequenc y match, the pause the simulation. You have plotted the frequenc y spectrum for a square-wave modulated carrier. Draw the spectral plot in the space provided. Neglect any side frequencies with amplitudes less than 10% of the carrier amplitude.
15. 15. Step 32 Measure the frequency of the spectral lines and record the answers on the spectral plot. Neglectany side frequencies with amplitudes less than 10% of the carrier amplitude.  fc = 100 kHz  fLSF1 = 95.041 kHz fLSF2 = 85.537 kHz  fUSF1 = 104.959 kHz fUSF2 = 114.876 kHzQuestion: How did the frequency spectrum for the square-wave modulated carrier differ from thespectrum for the sine wave modulated carrier in Step 6? Explain why there were different.  it creates many side frequencies. It is because it is a complex waveStep 33 Determine the bandwidth (BW ) of the modulated carrier from the frequency spectral plot and record your answer on the spectral plot. Neglect any side frequencies with amplitudes less than 10% of the carrier amplitude.  BW = 9.506 kHzQuestion: How did the bandwidth of the 5-kHz square-wave modulated carrier compare to the bandwidthof the 5-kHz sine wave modulated carrier in Step 6? Explain any difference.  Difference is 0.002 kHz.Step 34 Reduce the amplitude of the square wave to 0.5 V (500 mV) on the function generator. Bringdown the oscilloscope enlargement and run the simulation to one full screen display, then pause thesimulation. Draw the curve plot in the space provided.
16. 16. Step 35 Determine the modulation index (m) and percent modulation from the curve plot in Step 34.  m = 0.5 = 50%Question: W hat is the difference between this curve plot and the one in Step 29? Explain.  The amplitude of the modulating signal is reduced by half. That explains why the minimum amplitude is not at 0 V. And also it is only 50 % modulation.
17. 17. CONCLUSION: An amplitude modulation is one of the type of modulation that can produced by a circuit which is amodulator that performs a mathematical multiplication of the carrier and the information signals. In thistype of modulation, the amplitude of the carrier is changed depending with the characteristics of themodulating signal while the carrier frequency remains constant. Oscilloscope shows the time domainwhile the spectrum analyzer is frequency domain. The envelope is like the shape of the modulatingsignal. Modulation signal is the ratio of the peak voltage of the modulating signal to the peak value of thecarrier. The carrier frequency is at the center. The amplitude of the sidebands is half the modulation indexand the carrier voltage. So the amplitude of the sidebands is directly proportional to the modulation indexand the carrier voltage Complex modulating signal such as square wave produces lot of side bandsbecause of the fundamental sine wave frequency and many harmonics.