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Amplitude Modulation Questions and Answers

81.

A carrier of \( 100 \ \mathrm{V} \) peak is modulated to an index of \( 0.4 \). The envelope maximum is

Answer: A

The envelope reaches the carrier plus the message amplitude:

\[ V_{max} = 100(1+0.4) = 140 \ \mathrm{V} \]

so 140 volts, with a minimum of 60 volts. 104 V would follow from adding four per cent rather than forty. 40 V is the message amplitude alone. 160 V corresponds to an index of 0.6. The peak envelope voltage is what sizes the output devices and the antenna insulation, so it is a design figure rather than a merely academic one, and it grows to twice the carrier at full modulation.

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Answer: B

At full modulation the message amplitude equals the carrier, so the envelope swings from twice the carrier down to zero, just touching the axis without crossing it. Any deeper modulation would require a negative envelope, which is what folds and distorts. Half the carrier corresponds to an index of a half. Equality with the carrier is the unmodulated case. And twice the carrier is the maximum rather than the minimum. Recognising the just-touching envelope on an oscilloscope is the quickest way to identify a hundred per cent modulation.

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Answer: C

The envelope follows the message, so a square wave gives a square envelope stepping between the two amplitudes the two message levels imply. That is the basis of on-off keying when the message is a switched carrier. A sinusoidal envelope would require a sinusoidal message. Integration would occur only if the modulator included it. And a constant envelope with phase reversals describes a phase-keyed signal, which is an angle modulation rather than an amplitude one.

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Answer: D

Hard clipping generates harmonics of the audio, and each harmonic produces its own sideband pair further from the carrier, so the transmitted spectrum spreads well beyond the nominal channel. That splatter is exactly what a peak limiter is fitted to prevent, and why limiting is preferred to clipping. Properly limited AM and DSB-SC both occupy twice the audio bandwidth, and single sideband occupies only the audio bandwidth, so all three of the alternatives are narrower.

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Answer: A

A limiter reduces gain smoothly as the peak approaches the limit, so it holds the index below unity without the harmonic generation a hard clipper produces and therefore without splatter into adjacent channels. A clipper achieves the same peak control but at the cost of distortion. Both raise the average depth; that is not the distinguishing point. A monitor is still needed. And nothing makes an index above unity safe in full-carrier AM.

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Answer: B

The flattened trough on an envelope display is the direct evidence: the envelope has hit zero and folded, which is exactly what over-modulation means. Antenna current does rise with modulation depth, but a rise alone cannot distinguish deep legal modulation from illegal over-modulation. Carrier frequency is unaffected. And a falling audio input would reduce the depth rather than reveal an excess of it. Confirming the fault on the waveform rather than inferring it from a meter is what makes the envelope display indispensable.

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87.

Full-carrier amplitude modulation is also written as _____ in the standard abbreviations.

Answer: C

It is DSB-FC, double sideband full carrier, with DSB-SC being the suppressed-carrier form, SSB-SC single sideband suppressed carrier, and VSB vestigial sideband. The abbreviations are worth knowing because data sheets and standards use them without expansion, and mistaking DSB-FC for DSB-SC changes both the receiver required and the power budget by a large factor. All four describe amplitude modulations, differing only in what has been removed from the spectrum before transmission.

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88.

Which component of an AM receiver sets its adjacent-channel selectivity?

Answer: D

The intermediate frequency filter does, since it operates at a fixed frequency where a sharp response can be built once and used across the tuning range. That concentration of selectivity is the central advantage of the superheterodyne architecture. The antenna and its preselector provide image rejection rather than channel selectivity. De-emphasis belongs to FM. And the gain-control detector measures level rather than discriminating between channels.

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89.

What is the consequence of an intermediate frequency filter that is too narrow?

Answer: A

Narrowing the filter cuts the outer parts of the sidebands, which carry the highest audio frequencies, so the recovered sound loses its treble - the familiar muffled quality of a receiver switched to its narrow setting for interference rejection. Image rejection depends on filtering before the mixer. Impulse noise is if anything reduced by a narrower bandwidth. And the transmitted bandwidth is a property of the transmitter, entirely unaffected by the receiver's choices.

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Answer: B

Well below the diode's forward voltage the characteristic is curved, so the detector behaves as a square-law device with the distortion that implies, and its output falls faster than the input. That is why substantial intermediate frequency gain precedes the detector and why the automatic gain control keeps the level high. The diode is less linear near zero, not more. Level independence would be desirable but is not what a junction gives. And exact linearity below the forward drop is the opposite of the truth.

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