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Correct Answer : A

Both are true and the second is the reason: identical short pulses mean the average transmitted power is low and independent of the modulation, whereas a width-modulated train's power rises with its average duty cycle. The efficiency is bought with a stricter timing requirement, which is the standard shape of these trades. Where a scheme's average power does not depend on the message, the transmitter can be sized for the average rather than the worst case.

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Correct Answer : B

The figure of merit is defined at equal transmitted power and equal message bandwidth, with the channel bandwidth left free to differ - which is what makes the comparison meaningful, since the extra channel bandwidth is precisely what FM spends to buy its advantage. Holding channel bandwidth equal would forbid FM the resource it trades in. Equalising the output ratio asks a different question, namely what power each scheme needs. And receiver noise figure is common to both and irrelevant to the comparison. Stating the conditions of a comparison is as important as the numbers it yields.

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Correct Answer : A

A passive loss ahead of the gain adds its decibels straight to the system noise figure, which is why front-end filters are made as low-loss as their selectivity allows and why an amplifier is sometimes placed before them. Being passive does not exempt a component from the cascade formula; on the contrary, a lossy passive stage is the worst place to put attenuation. And the effect is the full loss rather than half of it. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted.

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Correct Answer : D

It is the width of an ideal rectangular filter passing the same noise power as the real one, and it differs a little from the 3 dB bandwidth because real filters have sloping skirts. Using the 3 dB figure in a noise calculation introduces a small error, usually a few per cent. The transmitted bandwidth is a property of the signal. And noise is measurable everywhere. Defining an equivalent rectangle is the standard way of reducing a shaped response to one number.

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Correct Answer : C

Channel separation is the figure, measured by driving one channel and reading the level appearing in the other, and it exercises the matrix, the subcarrier regeneration and the phase accuracy together. Capture ratio concerns two competing signals, image rejection one interfering frequency, and amplitude suppression the limiter. Each specification tests one mechanism, and separation is the one that tests the stereo system end to end.

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Correct Answer : A

The decoder needs the composite recovered with its amplitudes and phases intact, since the difference signal is detected coherently against the regenerated subcarrier. Deviation and carrier frequency describe the transmitted signal rather than what the decoder consumes. Pilot level and intermediate frequency are conditions rather than the signal. And audio bandwidth and de-emphasis act after decoding. Coherent detection always imposes requirements on phase as well as amplitude, which is why the pilot exists.

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Correct Answer : C

The first two are correct and the third is exactly what the multiplex was designed to avoid: a mono receiver uses the sum signal at the bottom of the baseband and discards everything above it without modification. That backward compatibility shaped the whole arrangement, from the choice of sum and difference over left and right to the placing of the pilot above the audio band. Designing a new service around the receivers already in use is a recurring broadcasting constraint.

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Correct Answer : B

The image mixes to the same intermediate frequency as the wanted signal, so no filter after the mixer can separate them and the rejection must happen before. Impedance matching is a secondary benefit of the input circuit. Oscillator isolation matters but is not the preselector's purpose. And audio overload is prevented by gain control. This is the same reasoning as anti-alias filtering, and it recurs wherever an operation is irreversible.

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Correct Answer : B

Because the amplitude carries nothing, distorting or clipping it destroys no information, so a saturated amplifier is entirely acceptable and the efficiency gain is free. Wide bandwidth is a cost rather than an enabler. Carrier strength varies with index and is irrelevant. And a large index is common but not necessary; narrowband FM tolerates nonlinearity just as well. This single property is why angle modulation dominates power-limited links.

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Correct Answer : B

It is the peak phase deviation, and in PM that quantity is also the modulation index, which is why the two terms are used interchangeably here in a way they never are for FM. Peak frequency deviation is the FM parameter. Modulation depth is the AM index expressed as a percentage. And the deviation ratio is an FM system parameter defined at the highest message frequency. Each family has its own vocabulary and mixing them is the most reliable way to reach a wrong answer.

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Correct Answer : A

The Bessel behaviour is shared with FM, so the carrier weakens as power moves into the sidebands and vanishes at particular indices before returning. Growth would violate the constancy of total power. Staying constant is the AM behaviour, where the carrier is fixed and the sidebands are added on top. And the nulls are isolated points rather than a permanent disappearance. The carrier nulls are as useful for calibrating a PM index as they are for an FM deviation.

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Correct Answer : A

Pre-emphasis exists to counter a rising noise spectrum, and PM does not have one: its detector output noise is flat, so lifting the treble at the transmitter would buy nothing and cost bandwidth. Bandwidth comparisons depend on the index and the message frequency rather than favouring PM generally. The PM index is constant with frequency rather than larger at the top. And detector noise figures are comparable; it is the spectral shape that differs.

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Correct Answer : D

Vestigial sideband keeps one full sideband and a small part of the other, retaining the low-frequency content that a sharp filter would destroy while saving most of the bandwidth SSB would save. It is emphatically an amplitude scheme rather than an angle one. Its carrier is retained rather than suppressed, so envelope detection works. And it is not single sideband with a pilot, which would be a different compromise addressing the carrier rather than the filter problem.

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Correct Answer : C

Unbalancing lets a carrier component through, and with enough of it the signal becomes ordinary full-carrier AM - which is one practical way to generate AM from a balanced modulator, and also the fault mode that shows up as poor carrier suppression. Sideband suppression needs a filter or a quadrature network. The output does not vanish; the sidebands are unaffected. And message polarity is not what balance controls.

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Correct Answer : A

Its role is entirely to serve as a reference large enough that the envelope follows the message, which is what lets a diode recover the audio without any local oscillator. It carries no message energy at all. Propagation is indifferent to whether a carrier is present. And bandwidth is set by the message frequency, since the sidebands sit at the message frequency either side of the carrier regardless of the carrier's amplitude. Understanding the carrier as a receiver aid rather than a transmission necessity is the key to the whole family.

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Correct Answer : A

DSB-SC is, by name and by construction. Full-carrier AM retains its carrier, as does vestigial sideband television, deliberately, so that envelope detection can be used. On-off keying switches a carrier on and off, so the carrier is present whenever anything is being sent. The distinction matters because it determines whether the receiver needs a coherent reference, which is the single largest difference in receiver cost.

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Correct Answer : A

The detector follows the magnitude, so it delivers a full-wave rectified version of the message, rich in harmonics and quite unlike the original. Something is certainly recovered, which is what makes the fault confusing in practice: the audio is present but wrong. Perfect recovery needs a coherent detector. And a simple polarity reversal would be harmless by comparison with what actually happens.

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Correct Answer : A

The information is carried in a varying envelope, so a nonlinear stage would distort it and generate intermodulation products that spread outside the channel. That is the practical cost of the scheme, since linear amplifiers are less efficient than the class C stages an AM or FM transmitter can use. There is no carrier to preserve. Bandwidth grows only as a symptom of the distortion. And phase linearity matters but is not the reason amplitude linearity is required.

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Correct Answer : A

The carrier is the mean of the envelope extremes:

\[ A_c = \frac{12+4}{2} = 8 \ \mathrm{V} \]

so 8 volts, with a message amplitude of 4 volts and therefore an index of 0.5. Reading the maximum as the carrier ignores the modulation entirely. The minimum is the carrier less the message. And 16 volts is the sum of the extremes, which is twice the carrier. Both the carrier and the index can be read from one oscilloscope trace this way, which is why the envelope display remains a standard bench tool.

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Correct Answer : B

Two: one above and one below the carrier, at plus and minus the message frequency. A single sideband appears only after one has been deliberately filtered out. Four would arise from two modulating tones. And the message does not ride on the carrier without producing sidebands; the sidebands are precisely how the information is carried, since a pure carrier conveys nothing.