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DSB and SSB Modulation Questions and Answers

61.

Three schemes carrying the same 3 kHz message are compared.

SchemeBandwidthCarrier power
Full-carrier AM6 kHzTwo thirds of total
DSB-SC6 kHzNone
SSB3 kHzNone

Which conclusion follows from the table?

Answer: A

Reading across, SSB alone improves on both axes: half the bandwidth and no wasted carrier. DSB-SC saves the power but not the bandwidth, which is the reverse of the second option. Full-carrier AM saves neither and buys receiver simplicity instead. And the table itself refutes any claim of equivalence. Laying the three out this way is the quickest way to fix the comparison in memory.

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62.

Which scheme would you choose for a crowded high-frequency voice band?

Answer: B

SSB, because the band is crowded and fading: half the bandwidth doubles the number of channels and all the power in the message buys signal-to-noise ratio when propagation is marginal. Full-carrier AM wastes both. DSB-SC wastes the bandwidth. And wideband frequency modulation would need many times the bandwidth, which is why it is confined to the very high frequency bands and above where spectrum is plentiful.

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63.

Which scheme would you choose for a simple consumer broadcast receiver?

Answer: C

Full-carrier AM, because the receiver reduces to a diode and two components and needs no local reference at all. That is the historical answer and it remains the right one whenever receiver cost dominates the system. SSB and DSB-SC both demand a coherent or accurately reinserted carrier. And independent sideband is a point-to-point capacity technique with no place in broadcasting.

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Answer: D

Removing one of the two sidebands halves the occupied bandwidth, a factor of two, and it is the only bandwidth economy available within the amplitude family short of vestigial shaping. Carrier suppression, which the two schemes share, saves power but no bandwidth, which is why the third option is a common and instructive error. Neither a factor of three nor four arises from any spectral argument.

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Answer: A

DSB-SC needs only a balanced modulator and a coherent detector, with no sharp filter or wideband quadrature network, so it is markedly simpler to build than SSB. That is why it appears wherever bandwidth is not the binding constraint, as in the television colour subcarrier. It still needs linear amplification and still needs carrier recovery. And its noise performance per watt is comparable rather than better, with the penalty of twice the receiver bandwidth.

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Answer: B

Double sideband suffers badly when the carrier and sidebands fade unequally, distorting the envelope; SSB has no carrier to fade out of proportion, so the impairment is a level change rather than a distortion. It is not immune to fading, which still raises and lowers the received level. The lack of redundancy is real but costs less than the envelope distortion it avoids. And the reinserted carrier is generated locally, so propagation cannot touch it.

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Answer: C

Class C is nonlinear, and SSB carries its information in a varying envelope, so the output is badly distorted with intermodulation products spreading well outside the channel. That is why SSB transmitters use linear class AB stages and accept the efficiency penalty. The efficiency gain is real but unusable. The sideband is distorted rather than lost. And flattening the envelope destroys the message rather than preserving it.

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Answer: D

Full-carrier AM and DSB-SC both occupy twice the highest message frequency, differing only in whether a carrier line sits between the sidebands. SSB uses half that. Vestigial sideband uses a little more than half. The pairing is worth noting because it isolates the two independent economies: carrier suppression saves power at constant bandwidth, sideband removal saves bandwidth.

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Answer: A

An ordinary phase-locked loop needs a spectral line to lock to, and a suppressed-carrier signal has none; the Costas arrangement derives an error signal from the modulation itself and so locks without a carrier. Drift immunity is a property of any loop with adequate bandwidth. Capture range depends on design rather than on topology in the way implied. And multipliers are central to its operation, not absent from it.

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70.

A design team must fit twice as many voice channels into an existing high-frequency allocation without raising transmitter power.

Which change achieves it?

Answer: B

Halving each channel's bandwidth doubles the number that fit, and moving from double to single sideband does exactly that while also improving the power efficiency. Going the other way halves the capacity. Adding pilots consumes power and no bandwidth is saved. And widening the audio would reduce the channel count rather than increase it.

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