NATIONAL COLLEGE OF SCIENCE AND TECHNOLOGYAmafel Building, Aguinaldo Highway Dasmariñas City, Cavite

Experiment No. 5AMPLITUDE MODULATION

Agdon, Berverlyn B. August 09. 2011Signal Spectra and Signal Processing/BSECE 41A1 Score:

Engr. Grace RamonesInstructor

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.

SAMPLE COMPUTATIONS:Step 3Step 3

m=V mV c

=1V1V

=1

Step 4 Step 4

m=V max−V minV max+V min

=1.940V−4.680µV1.940V +4.680µV

=1

Step 7 Step 7

f m=f USF−f c=104.959kHz−100kHz=4.959kHz

f m=f c−f LSF=100kHz−95.041kHz=4.959kHz

% difference=|5kHz−4.959kHz5 kHz |×100%=0.82%

Step 8 Step 8

BW=2 f m=2 (5kHz )=10kHz

Step 9 Step 9

BW=f USF−f LSF=104.959 kHz−95.041kHz=9.918kHz

Step 9 QStep 9 Q

% difference=|10kHz−9.918kHz10 kHz |×100%=0.82%

Step 10 Step 10

V USF=V LSF=V C(m2 )=1V ( 12 )=0.5VStep 10 QStep 10 Q

% difference=|458.267mV−500mV458.267mV |×100%=9.11%

Step 12 Step 12

m=V mV c

=0.5V1V

=0.5

Step 13 Step 13

m=V max−V minV max+V min

=1.477V−480.8547mV1.477V +480.8547mV

=0.51

Step 13 QStep 13 Q

% difference=|0.5−0..510.5 |×100%=2%

Step 16 Step 16

BW=f USF−f LSF=104.959 kHz−95.041kHz=9.918kHz

Step 17 Step 17

V USF=V LSF=V C(m2 )=1V ( 0.52 )=0.25VStep 19 Step 19

m=V mV c

=0.1µV1V

=0.1×10−6

Step 20 Step 20

m=V max−V minV max+V min

=995.220mV−995.220mV995.220mV+995.220mV

=0

Step 21 QStep 21 Q

% difference=|1V−998.441mV1V |×100%=0.16%

Step 26 Step 26

BW=f USF−f LSF=109.917 kHz – 90.083kHz=19.834kHz

Step 28 Step 28

BW=f USF−f LSF=119.835 kHz – 80.165kHz=39.67kHz

Step 30Step 30

m=V max−V minV max+V min

=1.926V−0V1.926V +0V

=1

Step 35 Step 35

m=V max−V minV max+V min

=1.444V−479.211mV1.444V +479.211mV

=0.50

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 a communications system are the transmitter communications medium 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 be converted into an electrical signal compatible with the communications medium, this is the purpose of the transmitter while the purpose of the receiver is to receive the transmitted signal from the channel and convert it into its original information signal form. Is the original electrical information signal is transmitted directly over the communications channel, it is called baseband transmission. An example of a communications system that uses baseband transmission is the telephone system.Noise is defined as undesirable electrical energy that enters the communications system and interferes with the transmitted message. All communication systems are subject to noise in both the communication 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 and fluorescent lights). Receiver noise comes from electronic components such as resistors and transistors, 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 the communications medium. In that case, this baseband signal is used to modulate a higher-frequency sine wave signal that is in a frequency spectrum that is compatible with the communications medium. This higher-frequency sine wave signal is called a carrier. When the carrier frequency is in the electromagnetic spectrum it is called a radio frequency (RF) wave, and it radiates into space more efficiently and propagates a longer distance than a baseband signal. When information is transmitted over a fiber optic cable, the carrier frequency is in the optical spectrum. The process of using 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: amplitude modulation (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 frequency modulation (FM), the baseband information signal varies the frequency of the higher-frequency carrier and the carrier amplitude remains constant. In phase modulation (PM), the baseband 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. In this experiment, you will examine the characteristics of amplitude modulation (AM).In amplitude modulation, the carrier frequency remains constant, but the instantaneous value of the carrier amplitude varies in accordance with the amplitude variations of the modulating signal. An imaginary line joining the peaks of the modulated carrier waveform, called the envelope, is the same shape as the modulating signal with the zero reference line coinciding with the peak value of the unmodulated carrier. The relationship between the peak voltage of the modulating signal (Vm) and the peak voltage of the unmodulated carrier (Vc) is the modulation index (m), therefore,m=

V mV cMultiplying the modulation index (m) by 100 gives the percent modulation. When the peak voltage of the modulating signal is equal to the peak voltage of the unmodulated carrier, the percent modulation is 100%. An unmodulated carrier has a percent 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 the modulated carrier. Therefore, if it is important that the peak voltage of the modulating signal 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 displayed on an oscilloscope. When the AM signal is displayed on an oscilloscope, the modulation index can be computed from

m=V max−V minV max+V minwhere 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 two side frequencies to be generated above and below the carrier frequency be an amount equal to the modulating frequency (fm). The upper side frequency (fuse) and the lower side frequency (fluff) can be determine from

f USF=f c+ f m∧f LSF=f c−f mA complex modulating signal, such as a square wave, consists of a fundamental sine wave frequency and many harmonics, causing many side frequencies to be generated. The highest upper side frequency 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 by the 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 (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 side frequency (fC – fm (max)) is called the bandwidth occupied by the modulated carrier. Therefore, the bandwidth (BW) can be calculated fromBW=¿This bandwidth occupied by the modulated carrier is the reason a modulated carrier is referred to as broadband transmission.The higher the modulating signal frequencies (meaning more information is being transmitted) the wider 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 given communications channel. Also, when a carrier is overmodulated, the resulting distorted waveshape generates a harmonic that generate additional side frequencies. This causes the transmitted bandwidth to increase and interfere with other signals. This harmonic-generated sideband interference 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 sidebands generated by the modulated carrier, the frequency spectrum of the modulated carrier must be displayed in the frequency domain (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 these 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,

PT=PC+PUSF+PLSFThe total power transmitted (PT) can also be determined from the modulation index (m) using the equationPT=PC(1+m22 )=PC+PC(m22 )=PC+PC(m24 )+PC(m24 )

Therefore, the total power in the side frequencies (PSF) isPSF=PC(m22 )

and the power in each side frequency isPUSF=PLSF=PC(m24 )

Notice 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 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 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 an unmodulated 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 no information in the carrier. For 100% modulation, one-third of the total power transmitted is in the sidebands 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 the fundamental 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 the square root of the side-frequency power. Therefore the side-frequency voltage can be calculated fromPUSF=PLSF=PC(m

2

4 )=V C2

R (m2

4 )V USF

2

R=V LSF

2

R=V C

2

R (m2

4 )V USF

2=V LSF2=V C

2(m24 )V USF=V LSF=V C(m2 )

This 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 the fundamental and each harmonic.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 in the time domain. A spectrum analyzer has been attached to the output to display the frequency spectrum of the amplitude-modulated carrier in the frequency domain.

MULTIPLIER

1 V/V 0 V

Y

X

SUMMER

1 V/V 0 V

A

C

B

XFG1XSA1

TIN

2

30

XSC1

A B

Ext Trig+

+

_

_ + _

1

0

Carrier1 Vpk 100kHz 0° 0

4

Modulating Signal

ProcedureStep 1Open 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 2Run 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 3Based 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 4Determine 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 the curve plot compare with the expected value calculated in Step 3? The modulation index determined from the curve plot compare and the expected value calculated have no difference. They both have the 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 6Run 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 7Measure 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 mVfLSF = 95.041 kHz VLSF = 458.267 mVfUSF = 104.959 kHz VUSF = 458.249 mVQuestion: What was the frequency difference between the carrier frequency and each of the side frequencies? How did this compare with the modulating signal frequency?fc – fLSF = fm = 4.959 kHz fUSF – fc = fm = 4.959 kHzThese computed fm have 0.041 kHz or 0.82% difference compare with the modulating signal frequency given by the function generator. How did the frequency of the center spectral line compare with the carrier frequency?The frequency of the center spectral line and the carrier frequency are the same. They are both at 100 kHzStep 8Calculate 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 compare with the calculated bandwidth in Step 8?The bandwidth based from the frequency spectrum differs with 0.082 kHz or 0.82% compare with the expected bandwidth.

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 spectral line?There is a small difference between the calculated and the measured values of the voltage amplitude of each side frequency spectral line which is 0.0407 V. The calculated value differs with 9.11% compare with the measure value.What was the relationship between the voltage levels of the side frequencies and the voltage level of the carrier? Was this what you expected for this modulation index?The voltage of each level is one-half of the carrier voltage. This is expected for 100% modulation of the sine-wave modulated carrier.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 curve plot compare with the expected value calculated in Step 12? There is 2% difference between the expected and measured modulation index

How did this percent modulation compare with the percent modulation in Step 3 and 4? Explain any difference.When the modulating signal amplitude decreased by half, the percent modulation and the modulation index also decreased by half. It is because the percent modulation is directly proportional to the amplitude of the modulating signalStep 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 mVfLSF = 95.041 kHz VLSF = 228.996 mVfUSF = 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 the bandwidth on the spectral plot in Step 6? Explain.They have the same bandwidth.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 V

Questions: How did the calculated voltage values compare with the measured values in on the spectral plot?There is 9.17% difference. What was the relationship between the voltage levels of the side frequencies and the voltage level of the carrier? How did it compare with the results in Step 6?It is one-forth of the carrier voltage. The voltage levels of each side frequency and the carrier are proportional to each other. The result of the side frequency voltage decreased by half because the modulation index also decreased by half.

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 from the curve plot compare with the expected value calculated in Step 19? They are almost equal. They only have a 0.1 × 10-6 difference.How did this waveshape compare with the previous amplitude-modulated waveshapes? Explain any difference.The waveshape is like the waveshape of a carrier signal. It is different with the amplitude-modulated waveshapes because there is no modulating signal added to the signal.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?They are the same.How did the voltage amplitude of the spectral line compare with the carrier amplitude?The difference is 0.16%How did this frequency spectrum compare with the previous frequency spectrum?It is different with other frequency spectrum. Because this time, 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 in Step 2?

There is a greater cycles per second in this waveshape. 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 kHzfLSF = 90.083 kHzfUSF = 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 frequency compare with the bandwidth for a 5 kHz modulating frequency in Step 6?The bandwidth increased by almost 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 plot in Step 24 and the 5 kHz in Step 6? Explain.There is a difference of 41.458 mV. As the frequency of the modulating signal increases, the voltage of the side frequencies decreases.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 modulating frequency? Explain.The bandwidth increased by 20 kHz. Doubling the difference of the modulating frequency is the bandwidth. In this case, the difference is 10 kHz, that is why the bandwidth increased by 20 kHz.

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 square wave. It differs from the first waveshape because it is like a digital signal and the other is sinusoidal.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 kHzfLSF1 = 95.041 kHz fLSF2 = 85.537 kHzfUSF1 = 104.959 kHzfUSF2 = 114.876 kHzQuestion: How did the frequency spectrum for the square-wave modulated carrier differ from the spectrum for the sine wave modulated carrier in Step 6? Explain why there were different.There are other side frequencies. The waveshape is complex compare with the previous waveshapes. 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 the bandwidth 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. 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 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 modulating index decreased by half compare with the curve in Step 29. The modulation index is directly proportional to the amplitude of the modulating signal, then, when the amplitude of the modulating signal was decreased by half, the modulating signal also reduces by half.

CONCLUSION:In amplitude-modulated signal the value of the carrier amplitude varies while the carrier frequency remains constant. The modulating signal depends on the voltages of the carrier and the modulating signal. It is directly proportional to the amplitude of modulating signal and inversely proportional to the amplitude of carrier signal. The envelope of the modulated signal is as of the waveshape of the modulating signal. When there is no modulating signal, the modulated signal is like the curve of carrier signal. Moreover, the voltages in the side frequencies depend on the modulation index or the percent modulation and the amplitude of the carrier signal. As the modulating frequency increases, the bandwidth also doubles. A complex modulating signal, such as the square wave modulating signal, generates many side frequencies to form the upper and lower sidebands.