• Author / Uploaded
• soo chi

Citation preview

AMPLITUDE MODULATION

1-38

Section

2

= Amplitude

Modulation

Read it till it Hertz!

DEFINITION. Modulation is the process of impressing or imparting a lowfrequency source information (voice, audio, video, or data) onto a highfrequency bandpass signal with a carrier frequency fc by the introduction of amplitude, frequency or phase perturbation. DEFINITION. Demodulation is the reverse process where the received signals are transformed back to their original form. Amplitude Modulation is an analog modulation scheme in which the amplitude of a relatively high-frequency carrier signal is varied in accordance with the instantaneous amplitude of an information signal.

DEFINITION.

A. .THE AM WAVEFORM.

1.

Instantaneous Amplitude of the Modulated Wave

ν AM ( t ) = ( Vc + em ) sin ωc t

The modulating signal is more often an arbitrary waveform, such as audio signal. However, an analysis of sine wave modulation is very useful, since Fourier analysis often allows complex signals to be express as a series of sinusoids.

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO

Hence,

ν AM (t) = ( Vc + Vm sin ωmt ) sin ωc t

By applying little trigonometry,

ν AM (t) = Vc sin ωc t −

mVc mVc cos ( ωc + ωm ) t + cos ( ωc − ωm ) t 2 2

where : ν AM (t) = AM modulated wave Vc = peak amplitude of the carrier in V em = instantaneous amplitude of the modulating signal in V ωc = angular frequency of the carrier wave in rad/s ωm = angular frequency of the modulating signal in rad/s m = modulation index

Read it till it Hertz…jma ª

First, note that the amplitude of the carrier after modulation is the same as it was before modulation

(In short, AM is a bit of a misnomer). ª

Second, the amplitude of the upper and lower side frequencies is a function of both the carrier amplitude and the modulation index.

ª

Third, the carrier component is a +sine function, the upper side frequency a –cosine function, and the lower side frequency a +cosine function or the carrier is 90° out of phase with both the upper and lower

side frequencies, and the upper and lower side frequency are 180° out of phase with each other.

Loading ECE SUPERBook

1-39

AMPLITUDE MODULATION

1-40

2.

Coefficient of Modulation (m) A term used to describe the amount (modulation) present in an AM waveform.

of

amplitude

changed

Mathematically, where : Vm = peak amplitude of the

V m= m Vc

modulating signal in V Vc = peak amplitude of the carrier signal in V

In terms of AM Envelope,

m=

Vmax − Vmin Vmax + Vmin

Vmax = Vc + Vm Vmin = Vc − Vm

ECE Board Exam: APRIL 2004

An AM signal has the following characteristics: carrier frequency = 150 MHz; modulating frequency = 3 MHz; peak carrier voltage = 40 volts; and peak modulating voltage is 30 volts. Calculate the peak voltage of the lower sideband frequency. 40

Solution: VSB

mV c V 30 = ⇒m= m = = 0 .75 2 Vc 40 =

0 .75 (40 ) = 15 V 2

15

fc-fm

Sample Problem:

15

fc

fc+fm

Calculate the modulation index for a standard AM transmission, if the maximum peak voltage of the modulated wave is 150 V and the modulating signal voltage is 50 V peak.

Solution: m=

Vm ⇒ Vc = Vmax − Vm Vc

50 150 − 50 = 0 .5 =

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO Several Cases of m: a.

Undermodulation (mVc)

0

Loading ECE SUPERBook

1-41

AMPLITUDE MODULATION

1-42

Read it till it Hertz…jma OVERMODULATION creates side frequencies (harmonics) further from the carrier known as SPLATTER, BUCKSHOT, or SPURIOUS EMISSIONS which create interference to other radio services and since spectrum space is tightly controlled by law, overmodulation is illegal.

Sample Problem:

A modulating signal consists of a symmetrical triangular wave having zero dc component and peak-to-peak voltage of 11 V. Calculate the value of modulation index if it used to amplitude modulate a carrier of peak voltage 10 V.

Solution: For Emax; Vm 11 = 10 + 2 2 = 15.5 V

Emax = Vc + For Emin;

Vm 11 = 10 − 2 2 = 4.5 V

Emin = Vc − For m; m=

Emax − Emin 15.5 − 4.5 = Emax + Emin 15.5 + 4.5

= 0.55 Answer : 0.55

1-43

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO B. .SIMULTANEOUS MODULATION.

mT = m12 + m22 + m32 … + mn2

where : m T = total modulation index m1 , m2 ...mn = modulation indices due to several modulating signals

ECE Board Exam: APRIL 2004

An AM transmitter is modulated by two sine waves at 1.5 kHz and 2.5 kHz, with modulation of 20 percent and 80 percent respectively. Calculate the effective modulation index.

Solution: mT =

m12 + m22 =

0 .22 + 0 .82 = 0 .824

Sample Problem:

Calculate the modulating voltage of an audio signal necessary to provide 100% modulation of a 100-V carrier that is simultaneously modulated by 2 audio waves with m1 and m2 equal to 75% and 45% respectively.

Solution: For m3 at 100 % mod ulation; m3 =

(

)

m2T − m12 + m22 =

(

12 − 0.752 + 0.452

)

= 48 .5 % For Modulating Voltage; Vm3 = m3 x Vc = 0.485 x 100 V = 48 .5 V Answer : Vm = 48 .5 V

Loading ECE SUPERBook

AMPLITUDE MODULATION

1-44

C. .SPECTRUM OF CONVENTIONAL AM SIGNAL. 1.

Frequency Spectrum

2.

Voltage Spectrum

3.

Power Spectrum

1-45

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO D. .CONVENTIONAL AM SYSTEM (DSBFC). 1.

AM Bandwidth

BW = 2fm

2.

Side Frequency (Sideband)

fSB = fc ± fm

Parameter

General Solution

Transmit Power

⎛ m2 ⎞ PT = Pc ⎜⎜ 1 + ⎟ 2 ⎟⎠ ⎝ m2Pc 4

Sideband Power

PUSB = PLSB =

Transmit Voltage

VT = Vc 1 +

m2 2

Transmit Current

I T = Ic 1 +

m2 2

where: BW = AM bandwidth fm = Modulating frequency in Hz Vc = Unmodulated carrier in V VT = Modulated carrier in V Pc = Unmodulated carrier power in W PT = Modulated power in W m = modulation index

Loading ECE SUPERBook

AMPLITUDE MODULATION

1-46

ECE Board Exam: NOV 2004

What will be the total sideband power of an AM transmitting station whose carrier power is 1200 W and a modulation of 95%?

Solution: PT = Pcarrier + PSB ⇒ where PSB = PSB =

0.95 2 (1200 ) = 541 .5 Watts 2

m2Pc 2

ECE Board Exam: APRIL 2004

Calculate the power in one sideband of an AM signal whose carrier power is 50 watts. The unmodulated current is 2 A while the modulated current is 2.4 A.

Solution: I T = Ic 1 +

m2 ⇒m= 2

Pusb = Plsb =

⎡⎛ I 2 ⎢⎜⎜ T ⎢⎝ Ic ⎣

2 ⎤ ⎞ ⎟ − 1⎥ = ⎟ ⎥ ⎠ ⎦

⎡ ⎛ 2 .4 ⎞ 2 ⎤ 2 ⎢⎜⎜ ⎟⎟ − 1⎥ = 0.938 ⎢⎝ 2 ⎠ ⎥ ⎣ ⎦

m2Pc 0 .938 2 (50 ) = = 11 watts 4 4

Sample Problem:

Calculate the total power and the power in each side frequency for a standard AM transmission that is sinusoidally modulated to a depth of 80% if the unmodulated carrier power is 50 kW.

Solution: For PT ; ⎛ ⎛ m2 ⎞⎟ 0 . 82 PT = Pc ⎜1 + = 50kW ⎜1 + ⎟ ⎜ ⎜ 2 ⎠ 2 ⎝ ⎝ = 66 kW For Each Side Frequency Power ; m2Pc (0 .82 )50 kW = 4 4 = 8 kW each

Pside =

Answer : PT = 66 kW , Pside = 8 kW

⎞ ⎟ ⎟ ⎠

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO

1-47

Sample Problem:

Calculate the amplitude and resulting side frequency if a carrier wave of frequency 10 MHz with a peak value of 10 V is amplitude modulated by a 5 kHz sine wave of amplitude 6 V.

Solution:

For m; m=

Vm 6 = Vc 10

= 0.6 For Side Frequency; fside = fc ± fm = 10 MHz ± 0.005 MHz = 10.005 MHz and 9.995 MHz For amplitude; mVc 0.6 (10 ) VSB = = = 3V 2 2 Answer : VSB = 3 V, fside = 10.005 MHz and 9.995 MHz

Sample Problem:

The output voltage of an AM transmitter is 40 V when sinusoidally modulated to a depth of 100%. Calculate the voltage at each side frequency when the modulation depth is reduced to 50%.

Solution: For Vc ; Vc =

VT 2

m 2 = 32 .65 V 1+

=

40 1+

12 2

For Each Side Frequency Voltage ; mVc (0.5)32 .65 V Vside = = 2 2 = 8 .16 V each Answer : Vc = 32 .65 V, Vside = 8 .16 V

Loading ECE SUPERBook

AMPLITUDE MODULATION

1-48

Read it till it Hertz…jma ª

The maximum bandwidth of an AM wave is twice of the maximum modulating frequency.

ª

The total power in an AM signal increases with modulation, reaching a value of 50% greater than that of the unmodulated carrier at 100% modulation.

ª

The extra power with modulation goes into the sidebands; the carrier power does not change with modulation.

ª

The useful power, that is, the power that carries the information, is rather small reaching a maximum of one-third of the total signal power at 100% modulation.

E. .SUPPRESSED-CARRIER AM SYSTEM. 1.

Double-Sideband Suppressed-Carrier (DSB-SC)

General Solution of Double Sideband Suppressed Carrier (DSB-SC)

ν(t) = −

mVc mVc cos 2 π(fc + fm )t + cos 2π(fc − fm )t 2 2

1-49

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO

Parameter

General Solution m2Pc 2

Transmit Power

PT =

Sideband Power

PUSB = PLSB =

Transmit Voltage

VT =

Transmit Current

IT =

m2Pc 4

mVc 2 mIc 2

ECE Board Exam: NOV 2004

A DSB-SC system must suppress the carrier by 50 dB from its original value of 10 W. To what value must the carrier be reduced?

Solution: SdB = 10 log Pred =

2.

Poriginal Preduced

10 50 dB 10 10

⇒ Pred =

Porig 10

S dB 10

= 0.1 x10 −3 W = 100 μW

Single-Sideband Full-Carrier (SSB-FC)

General Solution of Single Sideband Full Carrier (SSB-FC)

ν(t) = Vc sin2πfc t +

mVc cos 2π(fc − fm )t 2

Loading ECE SUPERBook

AMPLITUDE MODULATION

1-50

Parameter

General Solution m2Pc 4

Transmit Power

PT = Pc +

Sideband Power

PUSB or PLSB =

Transmit Voltage

VT = Vc 1 +

m2 4

Transmit Current

I T = Ic 1 +

m2 4

m2Pc 4

ECE Board Exam: NOV 2004

Assuming 100% modulation H3E system, what would be the transmitted power in the remaining sideband of an AM signal if the carrier power is 1000 watts?

Solution: For H3E ⇒ (SSBFC ) PSB =

3.

m2PC 12 (1000) = = 250 watts 4 4

Single-Sideband Suppressed-Carrier

General Solution of Single Sideband Suppressed Carrier (SSB-SC)

ν(t) =

mVc cos 2π(fc − fm )t 2

1-51

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO

Parameter

4.

General Solution m2Pc 4

Transmit Power

PT =

Sideband Power

PUSB or PLSB =

Transmit Voltage

VT =

mVc 2

Transmit Current

IT =

mI c 2

m2Pc 4

Single-Sideband Reduced-Carrier (For 90% Suppression)

Transmit Power PT = 0.1Pc +

Sideband Power

m2Pc 4

PUSB = PLSB =

m2Pc 4

F. .PERCENTAGE POWER SAVING.

%PS =

PCAM − Ptx x100% PCAM

where : PCAM = Conventional or Standard AM Ptx = Transmitted AM system

Loading ECE SUPERBook

AMPLITUDE MODULATION

1-52

ECE Board Exam: NOV 2003

Determine the power saving in percent when the carrier is suppressed in an AM signal modulated to 80%.

Solution: For Conventional AM; ⎛ m2 PT = Pc ⎜1 + ⎜ 2 ⎝ = 1 .32 Pc

2 ⎞ ⎛ ⎟ = Pc ⎜1 + 0 .8 ⎟ ⎜ 2 ⎠ ⎝

⎞ ⎟ ⎟ ⎠

For Suppressed Carrier system ; m2Pc 0 .82 Pc = 2 2 = 0 .32 Pc

Ptx =

Perce ntage Power Saved P − Ptx %℘ = CAM x 100 % PCAM =

1 .32Pc − 0 .32Pc x 100 % = 75 .75 % 1 .32Pc

Sample Problem:

Calculate the percentage power saving for J3E system at 90% modulation.

Solution: For Conventional AM; ⎛ m2 PT = Pc ⎜1 + ⎜ 2 ⎝ = 1 .405 Pc

2 ⎛ ⎞ ⎟ = Pc ⎜1 + 0 .9 ⎟ ⎜ 2 ⎠ ⎝

⎞ ⎟ ⎟ ⎠

For J3E system ; 0 .92 m2 = Pc x 4 4 = 0 .2025 Pc

PJ3E = Pc x

Perce ntage Power Saved − Ptx P x 100 % %℘ = CAM PCAM =

1 .405Pc − 0 .2025 Pc x 100 % = 85 .6 % 1 .405Pc

1-53

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO Comparison between Different Systems (Applicable when m=1)

System

Carrier Power

Total Sideband Power

LSB or USB Power

Percentage Power Saving

DSB-FC (A3E)

66.67% of PT

33.33% of PT

16.67% of PT

0%

DSB-SC

0% of PT

100% of PT

50% of PT

66.67%

SSB-FC (H3E)

80% of PT

20% of PT

20% of PT

16.67%

SSB-SC (J3E)

0% of PT

100% of PT

100% of PT

83.33%

Read it till it Hertz…jma ª

The standard AM broadcast band starts at 535 kHz and ends at 1605 kHz.

ª

Carrier assignments start at 540 kHz and continue in a succession of 10-kilohertz increments until the upper limit of the broadcast band is reached (1610 kHz). This adds up to a total of 107 carrier assignments, or CHANNELS, over the entire broadcast band.

ª

Since interference between such closely spaced stations would be nearly impossible to prevent, the FCC (U.S.) avoids assigning adjacent channels to stations in the same area.

Loading ECE SUPERBook

AMPLITUDE MODULATION

1-54

I

H

1.

The A. B. C. D.

standard AM broadcast band ______. starts at 88 kHz and ends at 108 kHz starts at 535 kHz and ends at 1605 kHz starts at 535 MHz and ends at 1605 MHz starts at 88 MHz and ends at 108 MHz

2.

Two sinusoidal signals, V1 and V2, are fed into an ideal balanced mixer. V1 is a 20-MHz signal; V2 is a 5-MHz signal. What frequencies would you expect at the output of the mixer? A. 5 MHz and 15 MHz B. 20 MHz and 100 MHz C. 15 MHz and 25 MHz D. 5 MHz and 25 MHz

3.

An AM transmitter generates 100 watts with 0% modulation. How much power will it generate with 20% modulation? A. 50.12 watts B. 310.1 watts C. 102 watts D. 256 watts

4.

If the carrier power is 1000 watts, what is the power in the USB at 70.7% modulation? A. 333.33 watts B. 125 watts C. 666.67 watts D. 70.7 watts

5.

A carrier is modulated by three audio tones are 0.3, 0.4, and 0.5, then what A. 0.636 C. 0.707

6.

You 100 A. C.

7.

A SSB transmitter is connected to a 50-ohm antenna. If the peak output voltage of the transmitter is 20 volts, what is the PEP? A. 6 watts B. 4 watts C. 8 watts D. 2 watts

8.

The total power in an AM signal increases with modulation, reaching a value of _____ greater than that of the unmodulated carrier at _____. A. 80%, 80% B. 32%, 80% modulation D. 64%, 80% modulation C. 50%, 75%

tones. If the modulation indexes for the is the total modulation index? B. 1.2 D. 0.9

look at an AM signal with an oscilloscope and see that the maximum Vpp is volts and the minimum Vpp is 25 volts. What is the modulation index? 0.25 B. 1.25 0.6 D. 0.75

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO 9.

1-55

The total power in an AM signal increases with modulation, reaching a value of _____ greater than that of the modulated wave at _____. A. 13%, 50% B. 22%, 75% modulation D. 33%, 90% modulation C. 50%, 50%

10. An AM transmitter supplies a 10 kW of carrier power to a 50 ohms load. It operates at a carrier frequency of 1.2 MHz and is 85% modulated by a 3 MHz sine wave. Calculate the rms voltage of the signal. APRIL 2004 A. 547 V B. 825 V C. 327 V D. 707.1 V 11. For H3E transmitter, the useful power, that is, the power that carries the information, is rather small reaching a maximum of _____. A. one-third of the total signal power at 100% modulation B. one-fifth of the total signal power at 100% modulation C. one-fourth of the total signal power at 100% modulation D. one-tent of the total signal power at 100% modulation 12. For A3E transmitter, the useful power, that is, the power that carries the information in one of the sideband, is rather small reaching a maximum of _____. A. one-third of the total signal power at 100% modulation B. one-fifth of the total signal power at 100% modulation C. one-fourth of the total signal power at 100% modulation D. one-sixth of the total signal power at 100% modulation 13. The power amplifier of an AM transmitter draws 100 watts from the power supply with no modulation. Assuming high-level modulation, how much power does the modulation amplifier deliver for 100% modulation? A. 16.67 watts B. 50 watts C. 33.33 watts D. 66.67 watts 14. If the final RF amplifier of an AM transmitter is powered by 100 volts DC, what is the maximum collector voltage at 100% modulation? A. 400 volts B. 50 volts C. 200 volts D. 100 volts 15. Determine the power saving in percent when the carrier is suppressed in an AM signal modulated to 80%. NOV 2003 A. 75.76% B. 82.82% C. 33.33% D. 16.67% 16. The most commonly used filter in SSB generation. A. Mechanical B. RC C. LC D. Low pass 17. Which of the following is not a technique of generating an SSB signal? A. phase shift method B. filter method C. weaver method D. Armstrong method

Loading ECE SUPERBook

1-56

AMPLITUDE MODULATION

18. An emission technique where the total current will be twice as much when the modulation index is doubled. A. R3E B. H3E C. A3E D. J3E 19. What is the maximum modulating signal frequency that can be used with an H3E system with 20-kHz bandwidth? A. 10 kHz B. 20 kHz C. 5 kHz D. 40 kHz 20. An AM broadcast transmitter is tested by feeding the RF output into a 50Ω (dummy) load. Tone modulation is applied. The carrier frequency is 850 kHz and the FCC licensed power output is 5 kW. The sinusoidal tone is set for 90% modulation. What is the average power that is being dissipated in the dummy load? A. 7.25 kW B. 7025 W C. 72.k5 W D. 725 W 21. How many percent of the total transmitted power is present in the sideband in an A3E system? A. 16.67% B. 80% C. 50% D. 33.33% 22. An intelligence signal is amplified by a 70% efficient amplifier before being combined with a 10-kW carrier to generate the AM signal. If it is desired to operate at 100% modulation, what is the dc input power to the final intelligence amplifier? A. 7.14 kW B. 5.14 kW C. 4.14 kW D. 6.14 kW 23. A transistor RF power amplifier operating class C is designed to produce 40 W output with a supply voltage of 60 V. If the efficiency is 70%, what is the average collector current? NOV 2003 A. 666.67 mA B. 952.4 mA C. 476.2 mA D. 238.1 mA 24. At 80% modulation H3E, what is the percentage power saving? A. 66.67% B. 33.33% C. 12.12% D. 16.67% 25. An AM transmitter has a 1-kW carrier and is modulated by three different sine waves having equal amplitudes. If the total modulation index is 80%, calculate the individual values of m in % and the total transmitted power. A. 56.2%, 1.32 kW B. 46.2%, 1.72 kW C. 46.2%, 1.32 kW D. 56.2%, 1.72 kW 26. The total bandwidth needed for an AM signal at 55.25 MHz with 0.5 MHz video modulation is ______. A. 0.5 MHz B. 1 MHz C. 101.5 MHz D. 10 MHz

Self-Sufficient Guide to ECE by JASON AMPOLOQUIO

1-57

27. A transmitter radiates 9 kW with the carrier unmodulated and 2.08 kW when the carrier is sinusoidally modulated and then suppressed. The modulation index is ______. A. 0.6 B. 0.8 C. 0.68 D. 0.58 28. The "envelope" of an AM signal is due to: A. the baseband signal B. C. the amplitude signal D. 29. The A. B. C. D.

the carrier signal none of the above

equation for full-carrier AM is: v(t) = (Ec + Em) x sin(ωct) v(t) = (Ec + Em) x sin(ωmt) + sin(ωct) v(t) = (Ec - Em) x sin(ωmt) x sin(ωct) v(t) = (Ec + Em sin(ωmt)) x sin(ωct)

30. An SSB transmitter has an average power ranging from 750-1000 W. What is the PEP? A. 9 kW B. 6 kW C. 3 kW D. 5 Kw 31. Calculate the average power of an SSB signal with 2-tone modulation if the peak voltage is 25 V, and assume that the load is 50Ω, resistive. A. 1.56 to 2.08 W B. 1.32 to 2.8 W C. 1.12 to 2.08 W D. 1.66 to 2.58 W 32. Determine the percentage power saving if the carrier and the USB is suppressed in an AM system modulated at 85%. A. 82.23% B. 66.67% C. 86.73% D. 89.71% 33. When a broadcast AM transmitter is 80% modulated, its antenna current is 15 A. What will be the new output current and the percentage increase when the modulation depth is increased by 95%? A. 15.78 A, 17% B. 13.1 A, 20.46% C. 13.1 A, 17% D. 15.78 A, 20.46% 34. In amplitude modulated wave equation, the carrier is _______ with both the upper and lower side frequencies, and the upper and lower side frequencies are _______ with each other. A. 90° out of phase, 180° out of phase B. 180° out of phase, 90° out of phase C. 90° out of phase, 270° out of phase D. 270° out of phase, 90° out of phase 35. A term used to describe the amount of amplitude changed present in an AM waveform. A. Deviation B. Coefficient of Modulation C. Shift D. Drift

Loading ECE SUPERBook

AMPLITUDE MODULATION

1-58

36. Calculate the modulation index for a standard AM transmission, if the maximum peak voltage of the modulated wave is 150 V and the modulating signal voltage is 50 V peak. A. 25% B. 75% C. 50% D. 100% 37. What is the value of modulation index and the relation between modulating signal amplitude and carrier amplitude if the transmitted AM wave is undermodulated. B. m>1, Vm>Vc A. m=1, Vm=Vc D. m= infinite, Vm=Vc=0 C. mVc A. m=1, Vm=Vc C. mVc A. m=1, Vm=Vc D. m= infinite, Vm=Vc=0 C. m