Amplitude Modulation Questions and Answers

Answer: C

Each sideband holds \( \mu^{2}/4 \) of the carrier power:

\[ \frac{0.25}{4} = 0.0625 = \frac{1}{16} \]

so a sixteenth each, an eighth in the two together. A quarter would be \( \mu^{2} \) itself. An eighth is the pair rather than one. And a half would need an index above unity. The small numbers are the point: at half modulation only about six per cent of the total power conveys the message.

Answer: D

Amplitude modulation has no mechanism for trading bandwidth against noise, which is what wideband FM does through its capture and deviation effects. Noise that adds to the amplitude adds directly to the recovered message, so AM is the most vulnerable of the schemes rather than immune. Performance improves with a larger index, since more of the power carries information. And it therefore certainly depends on the index.

Answer: A

The sum of the two signals is applied to a nonlinear device whose square term produces the product, and a band-pass filter then selects the modulated components around the carrier. It is the simplest low-level modulator. A linear mixer cannot generate new frequencies at all. Switching the carrier gives a chopper or switching modulator, a related but distinct method. And supply-voltage variation is high-level anode or collector modulation, used in transmitters rather than at low level.

Answer: B

In high-level, or collector, modulation the message varies the supply to the output stage, so the carrier amplitude follows it and the whole transmitter runs at high efficiency in class C. It is the classic broadcast arrangement. Modulating at the first stage is low-level modulation, which then requires a linear and less efficient chain afterwards. The matching network and the oscillator are not modulation points at all.

Answer: C

Amplifying a modulated signal requires linear operation, which rules out the efficient class C used for an unmodulated carrier, so the power stages waste more of their input. High-level modulation avoids that by modulating the supply of a class C stage instead. The message is amplified once. The carrier is not suppressed in full-carrier AM. And a properly designed linear stage does not attenuate the sidebands relative to the carrier.

Answer: D

The balance cancels the carrier at the output, leaving the two sidebands only, which is DSB-SC and the first stage in generating single sideband. Full carrier requires the carrier to be deliberately added. Frequency modulation needs a reactance or phase modulator instead. And vestigial sideband still requires a shaping filter after the modulator, so the last option is incomplete rather than merely wrong.

Answer: A

The order is 1, 2, 3, 4, 5. The balanced modulator must act before anything can be filtered, the filter must act while the signal is still at the low intermediate frequency where a sharp filter is realisable, and the translation to the operating frequency comes afterwards. Filtering before the modulator would have nothing to filter, and translating before filtering would demand an impossibly sharp filter at the final frequency.

Answer: B

The phasing method feeds two balanced modulators with carrier and message each split into quadrature components, and the outputs add so that one sideband cancels. It avoids the sharp filter but demands accurate ninety-degree networks over the whole message band. A crystal filter is the filter method. A phase-locked loop with a limiter is an FM detector. And a square-law device with a filter is a simple AM modulator.

Answer: C

The mixer translates every station to a single fixed intermediate frequency, so the selectivity and gain can be provided once, by fixed filters and amplifiers, instead of being retuned across the band. That single idea is why the architecture displaced all others. Direct conversion to baseband is a different and now common design, but it is not the superheterodyne. And an intermediate frequency that varied with tuning would defeat the purpose entirely.

Answer: D

High-side injection puts the oscillator above the signal:

\[ f_{LO} = 1200 + 455 = 1655 \ \mathrm{kHz} \]

so 1655 kHz. 745 kHz is low-side injection, which is also workable but requires a wider oscillator tuning ratio and is therefore less common in broadcast receivers. 455 kHz is the intermediate frequency itself. And 1200 kHz is the signal. The choice of side affects the image frequency, which is the next thing to check.