DSB and SSB Modulation Questions and Answers

Answer: C

The gap between the two sidebands is only twice the lowest message frequency, a few hundred hertz, so at a high carrier the filter would need a fractional bandwidth far beyond any practical resonator. At a few hundred kilohertz or a few megahertz the same absolute transition is a manageable fraction. Filters certainly exist far above a few megahertz, but not with this selectivity. Balanced modulators work at any frequency. And suppression is needed regardless.

Answer: D

Two balanced modulators are fed with carrier and message each split into quadrature components, and summing their outputs cancels one sideband while reinforcing the other. It avoids the sharp filter but demands an accurate ninety-degree network across the whole message band, which is the hard part. Filtering at the operating frequency is the problem the method sidesteps. Squaring and limiting are unrelated operations that would destroy the signal.

Answer: A

Weaver's third method uses audio-frequency quadrature at a single frequency in the middle of the band rather than a wideband phase shifter, which removes the hardest component of the phasing method at the cost of more mixers. It still needs balanced modulators, still needs translation to the operating frequency, and still needs a stable carrier oscillator. Its advantage is specific and worth knowing precisely because it is narrow.

Answer: B

Single sideband occupies just the message band:

\[ B = 3400 - 300 = 3100 \ \mathrm{Hz} \]

so 3100 Hz, and the channel spacing on the high-frequency bands is set accordingly. 6800 Hz is the double-sideband figure, twice the highest message frequency. 3400 Hz is the upper message limit rather than the width. And 6200 Hz doubles the message bandwidth, which is the width DSB would occupy if the message truly began at 300 Hz and the sidebands were measured from the carrier.

Answer: C

Convention puts upper sideband above ten megahertz and lower sideband below it, a purely arbitrary agreement that nonetheless matters: two stations on opposite sidebands cannot understand one another at all, since the recovered audio is spectrally inverted. Choosing at random would make contacts impossible. And alternating would be worse still. The convention is a good example of an interoperability rule with no technical basis beyond the need for agreement.

Answer: D

The receiver reinserts a local carrier at the right frequency and mixes, translating the sideband back to baseband. That oscillator is the beat frequency oscillator of traditional sets. Envelope detection cannot work, since one sideband alone has an envelope unrelated to the message. Limiting and discriminating is FM demodulation. And squaring produces the message multiplied by itself along with unwanted products. Which of the two economies applies is worth naming before any arithmetic is attempted.

Answer: A

A carrier error translates the whole recovered spectrum by that amount:

\[ f_{out} = 1000 + 50 = 1050 \ \mathrm{Hz} \]

so 1050 Hz, and every other component shifts by the same fifty hertz rather than in proportion. That is why speech remains intelligible with a modest error but music is ruined: the harmonic relationships are destroyed. The sign of the shift depends on which sideband is in use, so 950 Hz would follow for the other sideband. 1000 Hz would need perfect reinsertion.

Answer: B

Every component shifts by the same absolute amount, so the harmonic series is no longer harmonic and music sounds discordant, while speech intelligibility depends more on formant patterns and survives a shift of tens of hertz. Bandwidth is not the issue. Speech certainly has harmonics; they are simply less critical to recognition. And no SSB receiver uses envelope detection, whatever the programme material. The pattern repeats across the family: what the transmitter sheds, the receiver must supply.

Answer: C

It supplies the carrier that the transmitter suppressed, mixing with the received sideband to bring it down to baseband. Tuning it slightly is how an operator corrects the pitch of received speech. Gain control is a separate loop driven by the detected level. Image rejection is a front-end filtering matter. And de-emphasis belongs to frequency modulation, having no place in an SSB receiver at all. Placing the scheme on the power-and-bandwidth grid settles most questions of this kind at once.

Answer: D

At an index of unity each sideband carries a quarter of the carrier power:

\[ P_{sb} = \frac{400}{4} = 100 \ \mathrm{W} \]

so 100 watts, with 600 watts radiated in total. 200 W is the pair together. 400 W is the carrier itself. 50 W would correspond to an index of about 0.71. The figure matters because an SSB transmitter conveying the same intelligence needs only that hundred watts, which is the headline efficiency argument for the scheme. It is the same reasoning that decides whether a service keeps its carrier or discards it.