Phase Modulation Questions and Answers

Answer: C

Only when both the index and the message frequency are fixed do the two schemes give the same bandwidth, since the Bessel structure and Carson's rule are shared and the difference lies entirely in how each index arises. Power is constant in both anyway and so distinguishes nothing. Carrier frequency plays no part in bandwidth. And sensitivities are measured in different units and cannot be equated. Framing the comparison properly is half the work in answering questions about it.

Answer: D

It is phase modulation restricted to four phases, carrying two bits per symbol, and it is the direct digital descendant of analogue PM. Four tones would be frequency shift keying. Four amplitude levels would be amplitude shift keying. And two separate carriers would need twice the spectrum, whereas quadrature keying achieves its two bits on one carrier by using phase alone. The lineage from analogue PM to modern digital radio runs directly through this scheme. Most errors on this topic come from carrying a result across from a neighbouring scheme where it does not hold.

Answer: A

Phase decisions remain reliable at signal-to-noise ratios where amplitude decisions would fail, because a constant-envelope signal spends all its power on the information and the decision boundaries are angular. Carrier recovery is emphatically required, which is why differential variants exist. Its bandwidth is comparable with other schemes at the same symbol rate. And envelope detection is impossible, the envelope being constant by design.

Answer: B

It encodes the phase change from one symbol to the next, which removes the need for an absolute reference and so sidesteps the ambiguity inherent in phase. The price is a small penalty in noise performance, since two noisy symbols rather than one contribute to each decision. Absolute phase is what plain keying uses. Amplitude and frequency differences belong to other keying families. The trade of a reference against a noise penalty is a recurring pattern in coherent systems.

Answer: C

Phase modulation, of which keying is simply the case where the message takes discrete values. Frequency shift keying stands in the same relation to FM, and amplitude shift keying to AM. Single sideband has no direct keyed counterpart in common use. Tracing each digital scheme back to its analogue parent makes the digital family easier to learn, since the spectra and the detection principles carry straight over. Most errors on this topic come from carrying a result across from a neighbouring scheme where it does not hold.

Answer: D

Substituting into the rule:

\[ B = 2 \times 4(1+1) = 16 \ \mathrm{kHz} \]

so 16 kHz. 8 kHz is the narrowband limit, which an index of one is too large to justify. 12 kHz would follow from dropping a factor somewhere. And 4 kHz is the message frequency itself. An index of exactly one is a useful reference case, since the bandwidth is then precisely twice the narrowband figure and the arithmetic can be checked at a glance. A single measurement chosen to respond to the one asymmetry beats several that respond to shared behaviour.

Answer: A

The Bessel behaviour is shared with FM, so the carrier weakens as power moves into the sidebands and vanishes at particular indices before returning. Growth would violate the constancy of total power. Staying constant is the AM behaviour, where the carrier is fixed and the sidebands are added on top. And the nulls are isolated points rather than a permanent disappearance. The carrier nulls are as useful for calibrating a PM index as they are for an FM deviation.

Answer: B

Power is constant, because the amplitude is constant and modulation only redistributes energy between the carrier and the sidebands. That is shared with FM and stands in direct contrast to AM, whose total power grows with modulation depth. Neither the index nor the message frequency alters the total. Holding this fact makes every angle-modulation power question trivial, and it explains why a constant-envelope scheme suits a power-limited link.

Answer: C

PM's power is constant and its envelope carries nothing, whereas AM varies its power with modulation depth and puts the whole message in the envelope. Only AM among the two carries envelope information. Full-carrier AM famously needs no coherent detection, which is its chief advantage, so the last option is wrong for AM. Setting the angle family against the amplitude family on these two axes - power constancy and where the information lives - organises the whole subject.

Answer: D

A diode responds to amplitude, and in PM the amplitude is constant, so a diode detector recovers nothing but a direct-current level. The receiver must compare phase against a reference instead. Bandwidth does not trouble a diode. The carrier is present rather than suppressed. And the index affects the spectrum rather than the detector's suitability. This is the same reason an envelope detector fails on FM, and it follows directly from the constant envelope both schemes share.