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SIGNAL SPECTRA EXPERIMENT AMPLITUDE MODULATION
 

SIGNAL SPECTRA EXPERIMENT AMPLITUDE MODULATION

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    SIGNAL SPECTRA EXPERIMENT AMPLITUDE MODULATION SIGNAL SPECTRA EXPERIMENT AMPLITUDE MODULATION Document Transcript

    • NATIONAL COLLEGE OF SCIENCE AND TECHNOLOGY Amafel Building, Aguinaldo Highway Dasmariñas City, Cavite Experiment No. 5 AMPLITUDE MODULATIONCauan, Sarah Krystelle P. August 09. 2011Signal Spectra and Signal Processing/BSECE 41A1 Score: Engr. Grace Ramones Instructor
    • OBJECTIVES1. 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.
    • SAMPLE COMPUTATIONSStep 3 (Expected Modulating index)Step 4 (Measured Modulating index)Step 7 Question (Modulating Frequency)Step 8 (Calculated Bandwidth)Step 9 (Measured Bandwidth)Step 9 Question (Percent Difference)Step 10 (Voltage Amplitude of Side Frequencies)Step 10 Question (Percent Difference)Step 12 (Expected Modulating index)Step 13 (Calculated Modulating index)Step 13 Question (Percent Difference)
    • Step 16 (Calculated Bandwidth)Step 17 (Voltage Amplitude of Side Frequencies)Step 19 (Expected Modulating index)Step 20 (Measured Modulating index)Step 21 Question (Percent Difference)Step 26 (Bandwidth)Step 28 (Bandwidth)Step 20 (Measured Modulating index)Step 20 (Measured Modulating index)
    • DATA SHEETMaterialsTwo 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 asaudio, video, or binary data over a communications medium or channel. The basic components in acommunications system are the transmitter communications medium or channel, and the receiver.Possible communications media are wire cable, fiber optic cable, and free space. Before theinformation can be transmitted, it must be converted into an electrical signal compatible with thecommunications medium, this is the purpose of the transmitter while the purpose of the receiver isto receive the transmitted signal from the channel and convert it into its original information signalform. Is the original electrical information signal is transmitted directly over the communicationschannel, it is called baseband transmission. An example of a communications system that usesbaseband transmission is the telephone system.Noise is defined as undesirable electrical energy that enters the communications system andinterferes with the transmitted message. All communication systems are subject to noise in both thecommunication channel and the receiver. Channel noise comes from the atmosphere (lightning),outer space (radiation emitted by the sun and stars), and electrical equipment (electric motors andfluorescent lights). Receiver noise comes from electronic components such as resistors andtransistors, which generate noise due to thermal agitation of the atoms during electrical current flow.In some cases obliterates 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 a higher-frequencysine wave signal that is in a frequency spectrum that is compatible with the communicationsmedium. This higher-frequency sine wave signal is called a carrier. When the carrier frequency is inthe electromagnetic spectrum it is called a radio frequency (RF) wave, and it radiates into spacemore efficiently and propagates a longer distance than a baseband signal. When information istransmitted over a fiber optic cable, the carrier frequency is in the optical spectrum. The process ofusing a baseband signal to modulate a carrier is called 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. Infrequency modulation (FM), the baseband information signal varies the frequency of the higher-frequency carrier and the carrier amplitude remains constant. In phase modulation (PM), thebaseband information 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 to a frequency-modulated signal carrier. Therefore, both FM and PM are often referred to as an angle modulation. Inthis experiment, you will examine the characteristics of amplitude modulation (AM).
    • In amplitude modulation, the carrier frequency 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 (V m) andthe peak voltage of the unmodulated carrier (Vc) is the modulation index (m), therefore,Multiplying the modulation index (m) by 100 gives the percent modulation. When the peak voltage ofthe modulating signal is equal to the peak voltage of the unmodulated carrier, the percentmodulation is 100%. An unmodulated carrier has a percent modulation of 0%. When the peakvoltage 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 isrecovered from the modulated carrier. Therefore, if it is important that the peak voltage of themodulating signal be equal to or less than the peak voltage of the unmodulated signal carrier (equalto or less than 100% modulation) with amplitude modulation.Often the percent modulation must be measured from the modulated carrier displayed on anoscilloscope. When the AM signal is displayed on an oscilloscope, the modulation index can becomputed fromwhere Vmax is the maximum peak-to-peak voltage of the modulated carrier and Vmin is the minimumpeak-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 causestwo side frequencies to be generated above and below the carrier frequency be an amount equal tothe modulating frequency (fm). The upper side frequency (fuse) and the lower side frequency (fluff) canbe determine fromA complex modulating signal, such as a square wave, consists of a fundamental sine wave frequencyand many harmonics, causing many side frequencies to be generated. The highest upper sidefrequency and the lowest lower side frequency are determined be the highest harmonic frequency(fm (max), and the highest lower side frequency and the lower upper side frequency are determined bythe lowest harmonic frequency (fm (min)). The band of frequencies between (fC + fm (min)) and (fC + fm(max)) is called the upper sideband. The band of frequencies between (f C – fm (min)) and (fC – fm (max)) iscalled the lowers sideband. The difference between the highest upper side frequency (fC + fm (max))and the lowest lower side frequency (fC – fm (max)) is called the bandwidth occupied by the modulatedcarrier. Therefore, the bandwidth (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) thewider the modulated carrier bandwidth. This is the reason a video signal occupies more bandwidth
    • than an audio signal. Because signals transmitted on the same frequency interfere with one another,this carrier bandwidth places a limit on the number of modulated carriers that can occupy a givencommunications channel. Also, when a carrier is overmodulated, the resulting distorted waveshapegenerates a harmonic that generate additional side frequencies. This causes the transmittedbandwidth to increase and interfere with other signals. This harmonic-generated sidebandinterference is called splatter.When an AM signal on an oscilloscope, it is observed in the time domain (voltage as a function time).The time domain display gives no indication of sidebands. In order to observe the sidebandsgenerated by the modulated carrier, the frequency spectrum of the modulated carrier must bedisplayed in the frequency domain (sine wave voltage levels as a function of frequency) on aspectrum analyzer.A sine wave modulated AM signal is a composite of a carrier and two side frequencies, and each ofthese signals transmits power. The total power transmitted (PT) is the sum of each 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 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)and the carrier power does not depend on the modulation index. When the percent modulation is100% (m = 1), the total side-frequency power (PSF) is one-half of the carrier power (PC) and thepower in each side frequency (PUSF and PLSF) is one-quarter of the carrier power (PC). When thepercent modulation is 0%, the total side-frequency power (PSF) is zero because there are no sidefrequencies in an unmodulated carrier. Based on these results, it is easy to calculate that anamplitude-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 transmitted is in thesidebands and two-thirds of the total power is wasted in the carrier, which contains no information.When a carrier is modulated by a complex waveform, the combined modulation index of thefundamental and all of the harmonics determines the power in the sidebands. In a later experiment,you will see how we can remove and transmit the same amount of information with less power.
    • 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-frequency voltage can be calculatedfromThis 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. When a carrier is modulated by a complex waveform,the voltage of each side frequency can calculated from the separate modulation indexes of thefundamental and each harmonic.A circuit that mathematically multiplies a carrier and modulating (baseband) signal, and then addsthe carrier to the result, will produce an amplitude-modulated carrier. Therefore, the circuit in Figure6-1 will be used to demonstrate amplitude modulation. An oscilloscope has been attached to theoutput to display the modulated carrier in the time domain. A spectrum analyzer has been attachedto the output to display the frequency spectrum of the amplitude-modulated carrier in the frequencydomain. 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. m=1 or 100%Question: How did the value of the modulation index and percent modulation determined from thecurve plot compare with the expected value calculated in Step 3? The modulation index measured from the curve plot and the expected value calculated are equal. They do not have any difference.
    • 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 each of the sidefrequencies? How did this compare with the modulating signal frequency? fc – fLSF = fm = 4.959 kHz fUSF – fc = fm = 4.959 kHz The result above and the modulating signal frequency from the function generator differs with 0.041 kHz or 0.82%How did the frequency of the center spectral line compare with the carrier frequency? Both the frequency of the center spectral line and the carrier frequency is 100 kHz. There is no difference between the values of the two.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 and record your answer on the spectral plot. BW = 9.918 kHzQuestion: How did the bandwidth of the modulated carrier from the frequency spectrum comparewith the calculated bandwidth in Step 8? The measured and the calculated bandwidths differ with just 0.082 kHz or 0.82%.
    • 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 thespectral line? There is only 0.0407 V difference between the calculated VUSF and the measured values in on the spectral line. It is only 9.11% difference.What was the relationship between the voltage levels of the side frequencies and the voltage level ofthe carrier? Was this what you expected for this modulation index? The voltage levels of the side is one-half of the carrier voltage. Yes, this is expected for a sine-wave modulated carrier with 100% modulation.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 thecurve plot compare with the expected value calculated in Step 12? The difference of the modulation indexes are only 0.01. The measured values is 2% different with the expected modulation index.How did this percent modulation compare with the percent modulation in Step 3 and 4? Explain anydifference.
    • It is decreased by half. The percent modulation is directly proportional to the modulating amplitude, that is why when the modulating amplitude reduced by 50% the percent modulation also decreased by half.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. BW = 9.918 kHzQuestion: How did the bandwidth of the modulated carrier from this frequency spectrum comparewith the bandwidth on the spectral plot in Step 6? Explain. The bandwidth of the modulated carrier from the frequency spectrum and the bandwidth on the spectral plot in Step 6 have both the same values. The bandwidth is not affected by the variation of the amplitude of the modulating signal but the frequency of the modulating signal.Step 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 thespectral plot?
    • The difference between the calculated voltage and the measured values based on the spectral plot is 9.17%.What was the relationship between the voltage levels of the side frequencies and the voltage level ofthe carrier? How did it compare with the results in Step 6? The voltage levels of the side frequencies are one-fourth of the voltage level of the carrier. Compare to Step 6, it decreases by half because of the reduction of the modulation index to 50%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 the curve plot in Step 18. m=0Questions: How did the value of the modulation index and the percent modulation determined fromthe curve plot compare with the expected value calculated in Step 19? The difference between the computed and the measured modulation indexes is only a 0.1 × 10-6 difference. It is almost equal to zero.How did this waveshape compare with the previous amplitude-modulated waveshapes? Explain anydifference. The waveshape of is like the waveshape of the carrier. The reason for this is that the modulating signal is almost zero, and appears like a carrier only.
    • 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 carrier frequency? The frequency of the spectral line and the carrier frequency are the same.How did the voltage amplitude of the spectral line compare with the carrier amplitude? The difference between the amplitude of the spectral line and the carrier amplitude is 0.16%.How did this frequency spectrum compare with the previous frequency spectrum? Because of the absence of the modulating frequency, the frequency spectrum only showed the carrier frequency. 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.
    • Question: How did this waveshape differ from the waveshape for a 5 kHz modulating frequency inStep 2? In this waveshape, there is greater number of cycles per second. This is because the frequency of the modulating signal increases.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 modulating frequencycompare with the bandwidth for a 5 kHz modulating frequency in Step 6? The bandwidth of the modulated carrier 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 for the 10 kHz plotin Step 24 and the 5 kHz in Step 6? Explain. Yes there is a difference of 41.458 mV. Meaning, the frequency of the modulating frequency affects the amplitude of the side frequency. It is inversely proportional to the amplitude.Step 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 kHz modulatingfrequency? Explain. The bandwidth increased by 20 kHz. The reason for this is that bandwidth is twice the modulating frequency.Step 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.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 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 on the spectral plot.Neglect any 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. There are lot of side bands generated in this frequency domain. Because square wave consists of a fundamental sine wave frequency and many harmonics, there are lot of side frequencies generated.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. 9.506 kHzQuestion: How did the bandwidth of the 5-kHz square-wave modulated carrier compare to thebandwidth of the 5-kHz sine wave modulated carrier in Step 6? Explain any difference. There is a 0.002 kHz difference between the waveshape. However, the bandwidth of the fUSF1 and fLSF1 is equal to the answer in Step 6.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.
    • Step 35 Determine the modulation index (m) and percent modulation from the curve plot in Step 34. m = 0.5 = 50%Question: What is the difference between this curve plot and the one in Step 29? Explain. The minimum amplitude modulated signal is not at 0 V unlike with Step 29. This is because the percent modulation is 50%.
    • CONCLUSION From the word itself, in amplitude-modulated signal the instantaneous value of the carrieramplitude varies while the carrier frequency is constant. When a circuit added a carrier to theproduct of a carrier and modulating signal will produce an amplitude-modulated carrier. The oscilloscope displays the time-domain which is the relationship of the time and theamplitude voltage. The percent modulation is the ratio of the peak amplitude of modulating signaland the peak amplitude of the carrier signal. It is directly proportional to the modulating signal andinversely proportional to the carrier signal. It can also be determined through the maximum and theminimum amplitude of the modulated signal. The amplitude variation of the carrier peaks has theshape of the modulating signal and is referred to as envelope. Moreover, the spectrum analyzer displays the frequency domain which is the relationship ofthe amplitude voltage and the frequency. The carrier frequency is at the center and has the highestvoltage amplitude. The two side bands is generated by a sine wave modulating signal. The signalsgenerated by the modulation process are called sidebands and occur at frequencies above and belowthe carrier frequency. The amplitude of the sidebands is half the modulation index and the carriervoltage. So the amplitude of the sidebands is directly proportional to the modulation index and thecarrier voltage. The bandwidth is twice the frequency of the modulating frequency. Lastly, a complex modulating signal generates lot of frequency. A square wave modulatingsignal is an example of complex signal because of its a fundamental sine wave frequency and manyharmonics.