DSB and SSB Modulation Questions and Answers

Answer: B

It saves bandwidth, halving the occupied channel, and that is the second and independent economy the suppressed-carrier family offers. Linearity is a requirement rather than a saving, and SSB needs more of it. Frequency stability requirements become tighter, not looser. And receiver complexity increases at every step away from full-carrier AM. Keeping the two savings separate - power from the carrier, bandwidth from the sideband - makes the whole comparison straightforward.

Answer: C

The output is scaled by the cosine of the phase error, and \( \cos 90^{\circ} = 0 \), so the recovered signal is zero. That quadrature null is the sharpest demonstration of why phase must be tracked and not merely frequency. Full amplitude requires zero error. Inversion follows a one-hundred-and-eighty-degree error. And the rectified message is what an envelope detector delivers, which is a different failure mode altogether.

Answer: D

A small phase error scales the output by a cosine close to unity, so the cost is only a slight loss of amplitude; the trouble comes when the error grows towards a quarter cycle. A beat note follows a frequency error rather than a fixed phase offset, which is an important distinction. Inversion needs half a cycle. And complete loss needs the full quarter cycle of the quadrature null.

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.

Answer: B

Every receiver would need a stable, accurately tuned local oscillator, which for decades meant an instrument beyond the reach of a mass-market set, and even now means a synthesiser rather than a diode. Bandwidth is its advantage, not its problem. Music is degraded by frequency error rather than impossible. And SSB transmitters of very high power exist; the barrier has always been at the receiving end.

Answer: C

The two residues appear at different frequencies: the carrier residue exactly at the suppressed carrier, the sideband residue mirrored on the other side of it, so a spectrum display separates them immediately. A single tone suffices for both. The units are a matter of presentation. And a power meter cannot resolve either, which is why the measurement is spectral.

Answer: D

Real modulators are imperfectly balanced, so a small carrier residue remains, typically forty or more decibels below the wanted sideband, and it is specified and trimmed. Perfect suppression is unattainable. A full-power pilot would defeat the purpose, though a small deliberate pilot is used in some systems. And the unwanted sideband appears mirrored about the carrier rather than at it.

Answer: A

The balanced modulator cancels the carrier and leaves the sum and difference:

\[ 100 \pm 1 = 99 \ \mathrm{and} \ 101 \ \mathrm{kHz} \]

so 99 and 101 kHz with nothing at 100 kHz. The second option is what a plain amplifier would pass. The third describes full-carrier AM, where the carrier survives. And the last lists the message and carrier frequencies rather than the products, which is the baseband view rather than the passband one.

Answer: B

Passing the higher product alone leaves upper sideband at 101 kHz, which is single sideband suppressed carrier. Passing the lower product would give lower sideband at 99 kHz. Keeping both is DSB-SC, which is what the modulator produced before filtering. And restoring the carrier would give full-carrier AM, which would require deliberately adding back what the balance removed.

Answer: C

It is simply which of the two products the filter admits, and in practice a transmitter selects the sideband either by switching between two filters or by moving the carrier oscillator to the other side of a single filter. Message polarity does not choose a sideband. Carrier phase affects neither. And filter order affects the sharpness of the transition rather than which side is passed.