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# Exp amplitude modulation (4)

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• 1. NATIONAL COLLEGE OF SCIENCE AND TECHNOLOGY Amafel Building, Aguinaldo Highway Dasmariñas City, Cavite Experiment No. 5 AMPLITUDE MODULATIONMaala, Michelle Anne C. August 09. 2011Signal Spectra and Signal Processing/BSECE 41A1 Score: Engr. Grace Ramones Instructor
• 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. SAMPLE COMPUTATIONS:Step 3Step 4Step 7Step 8Step 9Step 9 QStep 10Step 10 QStep 12Step 13Step 13 Q
• 4. Step 16Step 17Step 19Step 20Step 21 QStep 26Step 28Step 30Step 35
• 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 informationsuch as audio, video, or binary data over a communications medium or channel. Thebasic components in a communications system are the transmitter communicationsmedium or channel, and the receiver. Possible communications media are wire cable,fiber optic cable, and free space. Before the information can be transmitted, it must beconverted into an electrical signal compatible with the communications medium, this isthe 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 signalform. Is the original electrical information signal is transmitted directly over thecommunications channel, it is called baseband transmission. An example of acommunications system that uses baseband transmission is the telephone system.Noise is defined as undesirable electrical energy that enters the communicationssystem and interferes with the transmitted message. All communication systems aresubject to noise in both the communication channel and the receiver. Channel noisecomes from the atmosphere (lightning), outer space (radiation emitted by the sun andstars), and electrical equipment (electric motors and fluorescent lights). Receiver noisecomes from electronic components such as resistors and transistors, which generatenoise due to thermal agitation of the atoms during electrical current flow. In some casesobliterates the message and in other cases it results in only partial interference.Although noise cannot be completely eliminated, it can be reduced considerably.Often the original electrical information (baseband) signal is not compatible with thecommunications medium. In that case, this baseband signal is used to modulate ahigher-frequency sine wave signal that is in a frequency spectrum that is compatiblewith the communications medium. This higher-frequency sine wave signal is called acarrier. When the carrier frequency is in the electromagnetic spectrum it is called aradio frequency (RF) wave, and it radiates into space more efficiently and propagatesa longer distance than a baseband signal. When information is transmitted over a fiberoptic cable, the carrier frequency is in the optical spectrum. The process of using abaseband signal to modulate a carrier is called broadband transmission.There are basically three ways to make a baseband signal modulate a sine wavecarrier: amplitude modulation (AM), frequency modulation (FM), and phasemodulation (PM). In amplitude modulation (AM), the baseband information signalvaries the amplitude of the higher-frequency carrier. In frequency modulation (FM), thebaseband information signal varies the frequency of the higher-frequency carrier and
• 6. the carrier amplitude remains constant. In phase modulation (PM), the basebandinformation signal varies the phase of the high-frequency carrier. Phase modulation(PM) is different fork of frequency modulation and the carrier is similar in appearance toa frequency-modulated signal carrier. Therefore, both FM and PM are often referred toas an angle modulation. In this experiment, you will examine the characteristics ofamplitude modulation (AM).In amplitude modulation, the carrier frequency remains constant, but the instantaneousvalue of the carrier amplitude varies in accordance with the amplitude variations of themodulating signal. An imaginary line joining the peaks of the modulated carrierwaveform, called the envelope, is the same shape as the modulating signal with thezero reference line coinciding with the peak value of the unmodulated carrier. Therelationship between the peak voltage of the modulating signal (Vm) and the peakvoltage of the unmodulated carrier (Vc) is the modulation index (m), therefore,Multiplying the modulation index (m) by 100 gives the percent modulation. When thepeak voltage of the modulating signal is equal to the peak voltage of theunmodulated carrier, the percent modulation is 100%. An unmodulated carrier has apercent modulation of 0%. When the peak voltage of the modulating signal (Vm)exceeds the peak voltage of the unmodulated carrier (Vc) overmodulation will occur,resulting in distortion of the modulating (baseband) signal when it is recovered from themodulated carrier. Therefore, if it is important that the peak voltage of the modulatingsignal be equal to or less than the peak voltage of the unmodulated signal carrier(equal to or less than 100% modulation) with amplitude modulation.Often the percent modulation must be measured from the modulated carrier displayedon an oscilloscope. When the AM signal is displayed on an oscilloscope, the modulationindex can be computed fromwhere Vmax is the maximum peak-to-peak voltage of the modulated carrier and Vmin isthe 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, themodulation index is equal to 0 (0% modulation).when a single-frequency sine wave amplitude modulates a carrier, the modulatingprocess causes two side frequencies to be generated above and below the carrierfrequency be an amount equal to the modulating frequency (fm). The upper sidefrequency (fuse) and the lower side frequency (fluff) can be determine fromA complex modulating signal, such as a square wave, consists of a fundamental sinewave frequency and many harmonics, causing many side frequencies to begenerated. The highest upper side frequency and the lowest lower side frequency aredetermined be the highest harmonic frequency (fm (max), and the highest lower sidefrequency 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
• 7. called the upper sideband. The band of frequencies between (fC – fm (min)) and (fC – fm(max)) is called the lowers sideband. The difference between the highest upper sidefrequency (fC + fm (max)) and the lowest lower side frequency (fC – fm (max)) is called thebandwidth occupied by the modulated carrier. Therefore, the bandwidth (BW) can becalculated fromThis bandwidth occupied by the modulated carrier is the reason a modulated carrier isreferred to as broadband transmission.The higher the modulating signal frequencies (meaning more information is beingtransmitted) the wider the modulated carrier bandwidth. This is the reason a videosignal occupies more bandwidth than an audio signal. Because signals transmitted onthe same frequency interfere with one another, this carrier bandwidth places a limit onthe number of modulated carriers that can occupy a given communications channel.Also, when a carrier is overmodulated, the resulting distorted waveshape generates aharmonic that generate additional side frequencies. This causes the transmittedbandwidth to increase and interfere with other signals. This harmonic-generatedsideband interference is called splatter.When an AM signal on an oscilloscope, it is observed in the time domain (voltage as afunction time). The time domain display gives no indication of sidebands. In order toobserve the sidebands generated by the modulated carrier, the frequency spectrum ofthe modulated carrier must be displayed in the frequency domain (sine wave voltagelevels 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 these signals transmits power. The total power transmitted (PT) is the sum ofeach carrier power ((PC) and the power in the two side frequencies (PUSF and PUSB).Therefore,The total power transmitted (PT) can also be determined from the modulation index (m)using the equationTherefore, 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) and the carrier power does not depend on the modulationindex. When 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 side frequency (PUSF
• 8. and PLSF) is one-quarter of the carrier power (PC). When the percent modulation is 0%,the total side-frequency power (PSF) is zero because there are no side frequencies in anunmodulated carrier. Based on these results, it is easy to calculate that an amplitude-modulated carrier has all of the transmitted information in the sidebands and noinformation in the carrier. For 100% modulation, one-third of the total power transmittedis in the sidebands and two-thirds of the total power is wasted in the carrier, whichcontains no information. When a carrier is modulated by a complex waveform, thecombined modulation index of the fundamental and all of the harmonics determinesthe power in the sidebands. In a later experiment, you will see how we can remove andtransmit the same amount of information with less power.Because power is proportional to voltage squared, the voltage level of the frequenciesis equal to the square root of the side-frequency power. Therefore the side-frequencyvoltage can be calculated fromThis means that the voltage of each side frequency is equal to one-half the carriervoltage for 100% modulation of a sine-wave modulated carrier. When a carrier ismodulated by a complex waveform, the voltage of each side frequency cancalculated from the separate modulation indexes of the fundamental and eachharmonic.A circuit that mathematically multiplies a carrier and modulating (baseband) signal,and then adds the carrier to the result, will produce an amplitude-modulated carrier.Therefore, the circuit in Figure 6-1 will be used to demonstrate amplitude modulation.An oscilloscope has been attached to the output to display the modulated carrier inthe time domain. A spectrum analyzer has been attached to the output to display thefrequency spectrum of the amplitude-modulated carrier in the frequency domain.
• 9. 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.
• 10. 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 curveplot in Step 2.  m=1 or 100%Question: How did the value of the modulation index and percent modulationdetermined from the curve plot compare with the expected value calculated in Step3?  There is no difference between the values. They have same values.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 frequency (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: What was the frequency difference between the carrier frequency and eachof the side frequencies? How did this compare with the modulating signal frequency?  fc – fLSF = 4.959 kHz fUSF – fc = 4.959 kHz  There is a small difference between the values which is 0.82%.
• 11. How did the frequency of the center spectral line compare with the carrier frequency?  There is no difference between the values. They have same values.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 thefrequency spectral plot and record your answer on the spectral plot.  BW = 9.918 kHzQuestion: How did the bandwidth of the modulated carrier from the frequencyspectrum compare with the calculated bandwidth in Step 8?  There is a small difference between the values which is 0.82%.Step 10 Calculate the expected voltage amplitude of each side frequencyspectral line (VUSF and VLSF) based on the modulation index (m) and the carrier voltageamplitude  VUSF = VLSF = 0.5 VQuestions: How did the calculated voltage values compare with the measured valuesin on the spectral line?  There is a small difference between the values which is 9.11%.What was the relationship between the voltage levels of the side frequencies and thevoltage level of the carrier? Was 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.
• 12. 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 modulationdetermined from the curve plot compare with the expected value calculated in Step12?  There is a small difference between the values which is 2%.How did this percent modulation compare with the percent modulation in Step 3 and4? Explain any difference.  It decreased by almost 50%. The modulating signal and the percent modulation decreased by half because it is directly proportional to the Vm.Step 14 Bring down the spectrum analyzer enlargement. Run the simulation until the Resolution Frequency match, then pause the simulation. You have plotted the frequency spectrum for a modulated carrier. Draw the spectral plot in the space provided.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.
• 13.  BW = 9.918 kHzQuestion: How did the bandwidth of the modulated carrier from this frequencyspectrum compare with the bandwidth 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 valuesin on the spectral plot?  There is a small difference between the values which is 9.17%.What was the relationship between the voltage levels of the side frequencies and thevoltage level of the carrier? How did it compare with the results in Step 6?  It is one-fourth of the carrier. This is because the modulation index decreased by 50% It is half of the result in Step 6.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.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.  m = 0.1 × 10-6Step 20 Determine the modulation index (m) and percent modulation from thecurve plot in Step 18.  m=0Questions: How did the value of the modulation index and the percent modulationdetermined from the curve plot compare with the expected value calculated in Step19?  There is 0.1 × 10-6 difference.
• 14. How did this waveshape compare with the previous amplitude-modulatedwaveshapes? Explain any difference.  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 frequency of the spectral line compare with the carrierfrequency?  There is no difference between the values. They have same values.How did the voltage amplitude of the spectral line compare with the carrieramplitude?  There is a small difference between the values which is 0.16%.How did this frequency spectrum compare with the previous frequency spectrum?  The carrier frequency is only the signal present. There is no side frequencies.Step 23 Change the modulating frequency 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.
• 15. Question: How did this waveshape differ from the waveshape for a 5 kHz modulatingfrequency in Step 2?  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 frequency (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.  BW = 19.834 kHzQuestion: How did the bandwidth of the modulated carrier for a 10 kHz modulatingfrequency compare with the bandwidth for a 5 kHz modulating frequency in Step 6?
• 16.  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: Was there any difference between the amplitude of the side frequencies forthe 10 kHz plot in Step 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 frequency 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 kHzmodulating frequency? 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. Becausesquare wave consists of a fundamental sine wave frequency and many harmonics,there are lot of side frequencies generated.Step 30 Determine the modulation index (m) and percent modulation from thecurve plot in Step 29.  m=1
• 17. Step 31 Bring down the spectrum analyzer enlargement. Run the simulation until the Resolution Frequency match, the pause the simulation. You have plotted the frequency 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.Step 32 Measure the frequency of the spectral lines and record the answers onthe spectral plot. Neglect any side frequencies with amplitudes less than 10% of thecarrier 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 carrierdiffer from the spectrum for the sine wave modulated carrier in Step 6? Explain whythere were different.  it generates many side frequencies. It is because a square wave consists of a fundamental sine wave frequency and many harmonics.Step 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 carriercompare to the bandwidth of the 5-kHz sine wave modulated carrier in Step 6? Explainany difference.  There is a small difference between the values which is 0.002 kHz. The bandwidth is almost equal to Step 6.Step 34 Reduce the amplitude of the square wave to 0.5 V (500 mV) on thefunction generator. Bring down the oscilloscope enlargement and run the simulation to
• 18. one full screen display, then pause the simulation. Draw the curve plot in the spaceprovided.Step 35 Determine the modulation index (m) and percent modulation from thecurve plot in Step 34.  m = 0.5 = 50%Question: What is the difference between this curve plot and the one in Step 29?Explain.  In this curve plot, there is a modulating signal that is why the minimum amplitude is not at 0 V. The modulation index decreased by 50 % because of the amplitude of the modulating signal was reduced.
• 19. CONCLUSION: After conducting the experiment, I conclude that amplitude modulation orcommonly called AM, vary with carrier amplitude. The carrier frequency as remainconstant in AM. . The modulation index is the ratio of the amplitude of the modulating and carriersignal. It can also be measured through the maximum and the minimum voltage of themodulating signal In addition, the oscilloscope displays the time-domain of the amplitudemodulation on the other hand, the spectrum analyzer displays the frequency domain.The voltage of each frequency is half of the product of the carrier voltage and themodulation index. Then it is directly proportional to the carrier voltage and themodulation index. Because of a fundamental sine wave frequency and many harmonics of asquare wave, it is considered as a complex wave. This kind of wave generates manyside frequencies unlike the other wave which only consist of two side frequencies. The bandwidth of AM depends on the frequency of the modulating signal. It istwice the modulating frequency. Then, the bandwidth is directly proportional to themodulating frequency.