Amplitude Modulation Questions and Answers

Answer: C

The first two are correct and the third is the distinguishing point of full-carrier AM: because the carrier is present the envelope follows the message, so a simple diode detector suffices and no coherent reference is needed. That cheapness is the whole commercial argument for the scheme, and it is what suppressed-carrier systems trade away in exchange for power efficiency. Requiring coherent detection is a property of DSB-SC and SSB.

Answer: D

The first three are right and the fourth is not, so three match. Vestigial sideband occupies a little more than \( f_m \) but distinctly less than \( 2f_m \), which is the whole reason it exists: it captures most of the bandwidth saving of single sideband while keeping the low-frequency content that a sharp filter would destroy. Suppressing the carrier alone saves power without saving any bandwidth, which is the point of the second pair.

Answer: A

The relation is \( P_t = P_c(1 + \mu^{2}/2) \). Doubling the index term would overstate the sideband power fourfold. Omitting the half overstates it twofold and would put the efficiency ceiling at fifty per cent rather than a third. And a linear term in the index has the wrong dependence altogether, predicting sideband power at small index far above the truth. Each sideband holds \( \mu^{2}/4 \) of the carrier power, which is where the half comes from.

Answer: B

The indices combine in quadrature first:

\[ \begin{aligned} \mu_{eff}^{2} &= 0.36 + 0.64 = 1.00 \\ \frac{P_t}{P_c} &= 1 + \frac{1.00}{2} = 1.50 \end{aligned} \]

so 1.50 times, exactly the full-modulation figure, and the transmitter is right at its limit. 1.70 adds the indices before squaring. 1.98 squares their sum. 1.25 uses only the larger tone. Note the practical warning: two tones each individually modest can together reach a hundred per cent modulation.

Answer: C

The monitor exists to hold the working point in a narrow band: deep enough for efficiency and signal-to-noise ratio, shallow enough that peaks never exceed unity index and splatter into neighbouring channels. Frequency measurement, antenna matching and programme logging are all done by other equipment for other reasons, though a modern transmitter may well include all of them in one rack.

Answer: D

Distortion appearing only on peaks and curable by reducing the transmitter's audio level points to over-modulation: the envelope is being driven past its limit on the loudest passages. A gain control failure would distort at all levels or produce level-dependent noise. An image interferer would appear as a second programme or a whistle rather than as peak distortion. Diagonal clipping is a receiver fault and would not respond to the transmitter's audio level in this way.

Answer: A

Selective fading of one sideband unbalances the pair, so the envelope no longer reproduces the message faithfully and the recovered audio is distorted - a familiar problem on medium and short wave, and one reason single sideband is preferred for long-distance work. The carrier frequency is unaffected. Bandwidth is reduced only in the trivial sense that part of the signal is missing. And the message is degraded rather than lost entirely.

Answer: B

Receiver simplicity is the whole case for it: a diode, a capacitor and a resistor recover the message, needing no local reference and no phase tracking, which made mass broadcasting possible. AM is notably vulnerable to impulse noise rather than immune. Its transmitted power rises with modulation depth. And single sideband is narrower, so AM cannot claim the narrowest bandwidth.

Answer: C

An index of zero means no message is applied, so only the carrier is transmitted and the efficiency is zero: all the power and none of the information. It is the reference condition against which the antenna current and total power are compared when the index is measured. Suppressing the carrier is a quite different operation, achieved with a balanced modulator, and leaves the sidebands. Full modulation is an index of one, and folding requires an index above one. The zero case is worth naming because every power formula in the topic reduces to the carrier alone there.

Answer: D

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

\[ \begin{aligned} \frac{\mu^{2}}{4} &= \frac{50}{400} = 0.125 \\ \mu &= \sqrt{0.5} = 0.71 \end{aligned} \]

giving about 0.71, and a total power of 500 watts. 0.50 would put 25 watts in each sideband. 0.25 is the ratio of sideband to carrier power taken as the index. 1.00 would put 100 watts in each. Working from measured sideband power is how a spectrum analyser reading is converted into a modulation figure, and the factor of four is the step most often dropped.