Amplitude Modulation Questions and Answers

Q: In full-carrier amplitude modulation the carrier's

Amplitude modulation varies the carrier amplitude in step with the message while its frequency and phase stay fixed, so the message appears as the envelope of the transmitted waveform. Varying the frequency instead gives frequency modulation, and varying the phase gives phase modulation; both are angle modulations and behave quite differently in bandwidth and in noise. Varying two parameters at once is not a standard scheme and would make demodulation ambiguous.

Q: The envelope of an AM waveform carries

The envelope reproduces the message, which is exactly why a simple diode detector recovers it: rectify, smooth, and the message appears. That simplicity is AM's great advantage and the reason broadcast receivers could be made cheaply for a century. The carrier frequency is the rate at which the waveform oscillates inside the envelope. The sideband frequencies are properties of the spectrum. And no local oscillator is involved in envelope detection at all.

Q: For a tone-modulated AM signal the modulation index is

The index is \( \mu = A_m/A_c \), the ratio of message amplitude to carrier amplitude, and it fixes how deeply the envelope is driven. Inverting the ratio gives a number greater than one for normal modulation, which is not the convention. A ratio of frequencies is the deviation ratio in angle modulation, a different quantity entirely. And the power ratio is \( 1 + \mu^{2}/2 \), which contains the index but is not it.

Q: An AM envelope varies between \( 15 \ \mathrm{V} \) and \( 5 \ \mathrm{V} \). The modulation index is

From the envelope extremes:\[ \begin{aligned} \mu &= \frac{V_{max}-V_{min}}{V_{max}+V_{min}} \\ &= \frac{10}{20} = 0.5 \end{aligned} \]so 0.5, or fifty per cent modulation. 0.33 is the ratio of the minimum to the maximum. 3.0 is the ratio of maximum to minimum, which is greater than one and so cannot be an index for an undistorted signal. 0.67 comes from dividing the difference by the maximum alone. Reading the index off an oscilloscope trace this way is the standard bench measurement.

Q: Over-modulation occurs when the index exceeds

Beyond an index of one the envelope would have to go negative, so it folds instead: the carrier phase reverses and the recovered message is badly distorted, with splatter into adjacent channels. That is why broadcast transmitters are limited to just under a hundred per cent modulation and why a modulation monitor is a standard fitting. An index of a half is perfectly normal. Two would be heavily over-modulated. And ten per cent is merely shallow, wasting power but distorting nothing.

Q: The distortion produced by over-modulation appears as

The envelope cannot go below zero, so the troughs clip flat and the detector recovers a distorted waveform rich in harmonics that also spread the transmitted spectrum well beyond the nominal channel. The carrier frequency is unaffected. Bandwidth increases rather than reduces, which is precisely the interference problem. And the phase reversal that occurs is exactly what a diode envelope detector cannot follow, so the message certainly is changed.

Q: A carrier at \( 1000 \ \mathrm{kHz} \) is modulated by a \( 5 \ \mathrm{kHz} \) tone. The sidebands lie at

The sidebands sit at the sum and difference frequencies:\[ f_c \pm f_m = 1000 \pm 5 \ \mathrm{kHz} \]so 995 and 1005 kHz, with the carrier still present between them. The second option gives only the upper sideband and the carrier. 990 and 1010 kHz would follow from a ten kilohertz tone. And the last option lists the message and carrier frequencies rather than the sidebands, which is a confusion between the baseband and passband descriptions.

Q: A speech signal occupying \( 300 \ \mathrm{Hz} \) to \( 3.4 \ \mathrm{kHz} \) amplitude modulates a carrier. The transmitted bandwidth is about

Bandwidth is twice the highest message frequency:\[ B = 2 \times 3.4 = 6.8 \ \mathrm{kHz} \]so 6.8 kHz, which is why AM broadcast channels are spaced at nine or ten kilohertz. 3.4 kHz is the message's own upper limit. 3.1 kHz is the message bandwidth, from 300 Hz to 3.4 kHz, which is not what sets the transmitted width because the sidebands are measured from the carrier. 6.2 kHz doubles the message bandwidth instead of the highest frequency, a subtle and common slip.

Direction for [ Question No: 1 To 1 ] :

Directions : The following items consist of two statements, one labelled the "Assertion (A)" and the other labelled the "Reason (R)". You are to examine these two statements carefully and decide if the Assertion (A) and the Reason (R) are individually true and if so, whether the Reason is a correct explanation of the Assertion. Select your answers to these items using the codes given below and mark your answer accordingly.

Assertion (A) : In amplitude modulation technique the modulation index should be close to 1.
Reason (R) : The power carried by message signal in the side bands increases with increase in modulation index.

A.Both (A) and (R) are correct and (R) is correct explanation of (A).
B.Both (A) and (R) are correct, but (R) is not correct explanation of (A).
C.(A) is true, but (R) is false.
D.(A) is false, but (R) is true.

Exam Year(s)

UGC NET - 2014,

Answer: A

Both statements are true and the Reason explains the Assertion. In AM the message power sits in the sidebands, and with modulation index m the sideband power is a fraction \(\frac{m^{2}}{2+m^{2}}\) of the total, so it grows as m rises. A modulation index close to 1 puts the most power into the sidebands, the part that carries the information, without overmodulation.

That is the reason for keeping m near 1, so the Reason explains the Assertion.

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In full-carrier amplitude modulation the carrier's

A.Amplitude varies with the message
B.Frequency varies with the message
C.Phase varies with the message while its amplitude is held constant
D.Amplitude and frequency both vary in proportion to the message

Answer: A

Amplitude modulation varies the carrier amplitude in step with the message while its frequency and phase stay fixed, so the message appears as the envelope of the transmitted waveform. Varying the frequency instead gives frequency modulation, and varying the phase gives phase modulation; both are angle modulations and behave quite differently in bandwidth and in noise. Varying two parameters at once is not a standard scheme and would make demodulation ambiguous.

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The envelope of an AM waveform carries

A.The phase of the local oscillator
B.The message waveform
C.The carrier frequency
D.The sum of the two sideband frequencies

Answer: B

The envelope reproduces the message, which is exactly why a simple diode detector recovers it: rectify, smooth, and the message appears. That simplicity is AM's great advantage and the reason broadcast receivers could be made cheaply for a century. The carrier frequency is the rate at which the waveform oscillates inside the envelope. The sideband frequencies are properties of the spectrum. And no local oscillator is involved in envelope detection at all.

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For a tone-modulated AM signal the modulation index is

A.Message frequency over carrier frequency
B.Total power over carrier power, measured at the antenna
C.Message amplitude over carrier amplitude
D.Carrier amplitude over message amplitude

Answer: C

The index is \( \mu = A_m/A_c \), the ratio of message amplitude to carrier amplitude, and it fixes how deeply the envelope is driven. Inverting the ratio gives a number greater than one for normal modulation, which is not the convention. A ratio of frequencies is the deviation ratio in angle modulation, a different quantity entirely. And the power ratio is \( 1 + \mu^{2}/2 \), which contains the index but is not it.

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An AM envelope varies between \( 15 \ \mathrm{V} \) and \( 5 \ \mathrm{V} \). The modulation index is

A.0.33
B.3.0
C.0.67
D.0.5

Answer: D

From the envelope extremes:

\[ \begin{aligned} \mu &= \frac{V_{max}-V_{min}}{V_{max}+V_{min}} \\ &= \frac{10}{20} = 0.5 \end{aligned} \]

so 0.5, or fifty per cent modulation. 0.33 is the ratio of the minimum to the maximum. 3.0 is the ratio of maximum to minimum, which is greater than one and so cannot be an index for an undistorted signal. 0.67 comes from dividing the difference by the maximum alone. Reading the index off an oscilloscope trace this way is the standard bench measurement.

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Over-modulation occurs when the index exceeds

A.Unity
B.One half
C.Two
D.Ten per cent

Answer: A

Beyond an index of one the envelope would have to go negative, so it folds instead: the carrier phase reverses and the recovered message is badly distorted, with splatter into adjacent channels. That is why broadcast transmitters are limited to just under a hundred per cent modulation and why a modulation monitor is a standard fitting. An index of a half is perfectly normal. Two would be heavily over-modulated. And ten per cent is merely shallow, wasting power but distorting nothing.

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The distortion produced by over-modulation appears as

A.A phase reversal that leaves the recovered message unchanged
B.Clipping of the envelope minima
C.A shift of the carrier frequency
D.A reduction of the transmitted bandwidth

Answer: B

The envelope cannot go below zero, so the troughs clip flat and the detector recovers a distorted waveform rich in harmonics that also spread the transmitted spectrum well beyond the nominal channel. The carrier frequency is unaffected. Bandwidth increases rather than reduces, which is precisely the interference problem. And the phase reversal that occurs is exactly what a diode envelope detector cannot follow, so the message certainly is changed.

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A carrier at \( 1000 \ \mathrm{kHz} \) is modulated by a \( 5 \ \mathrm{kHz} \) tone. The sidebands lie at

A.990 and 1010 kHz
B.5 and 1000 kHz
C.995 and 1005 kHz
D.1000 and 1005 kHz

Answer: C

The sidebands sit at the sum and difference frequencies:

\[ f_c \pm f_m = 1000 \pm 5 \ \mathrm{kHz} \]

so 995 and 1005 kHz, with the carrier still present between them. The second option gives only the upper sideband and the carrier. 990 and 1010 kHz would follow from a ten kilohertz tone. And the last option lists the message and carrier frequencies rather than the sidebands, which is a confusion between the baseband and passband descriptions.

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The bandwidth of a full-carrier AM signal modulated by a tone of frequency \( f_m \) is

A.f m
B.Half of f m
C.Four times f m
D.Twice f m

Answer: D

Both sidebands are present and each sits \( f_m \) from the carrier, so the occupied bandwidth is \( 2f_m \). That factor of two is what single sideband removes, and it is the first entry in every comparison of the modulation schemes. A bandwidth of \( f_m \) would describe single sideband. Half of \( f_m \) is less than the message needs and is impossible. Four times would follow from a wideband angle modulation rather than from AM.

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10.

A speech signal occupying \( 300 \ \mathrm{Hz} \) to \( 3.4 \ \mathrm{kHz} \) amplitude modulates a carrier. The transmitted bandwidth is about

A.6.8 kHz
B.3.4 kHz
C.3.1 kHz
D.6.2 kHz

Answer: A

Bandwidth is twice the highest message frequency:

\[ B = 2 \times 3.4 = 6.8 \ \mathrm{kHz} \]

so 6.8 kHz, which is why AM broadcast channels are spaced at nine or ten kilohertz. 3.4 kHz is the message's own upper limit. 3.1 kHz is the message bandwidth, from 300 Hz to 3.4 kHz, which is not what sets the transmitted width because the sidebands are measured from the carrier. 6.2 kHz doubles the message bandwidth instead of the highest frequency, a subtle and common slip.

Discuss in Forum

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Amplitude Modulation Questions and Answers

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