Phase Modulation Questions and Answers

Answer: A

Phase modulation makes the phase departure proportional to the message amplitude, which is the exact analogue of frequency modulation's deviation proportional to amplitude. Proportionality to message frequency describes no useful scheme. Proportionality to the integral describes frequency modulation expressed as a phase, which is the dual case. And proportionality to the derivative would be FM written the other way round. Fixing which quantity is proportional to what is the whole basis of telling the two angle modulations apart.

Answer: B

For PM the index is simply the peak phase deviation, measured in radians, and it depends on the message amplitude alone. The second option is the FM index, which carries an extra division by the message frequency. Frequency deviation in hertz is an FM parameter with the wrong units for a phase. And the carrier's own phase has no fixed value to form a ratio with. That the PM index is dimensionless and frequency-independent is its most consequential difference from FM.

Answer: C

Multiplying sensitivity by peak amplitude:

\[ \beta_p = 2 \times 3 = 6 \ \mathrm{rad} \]

so 6 radians, a firmly wideband index. 1.5 divides the amplitude by the sensitivity, inverting the relationship. 0.67 inverts it the other way. And 5 corresponds to no combination of the figures. Because the result does not involve the message frequency at all, the same index applies to every tone the message contains, which is not true of FM. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.

Answer: D

The message frequency does not enter the PM index at all, which is the sharpest contrast with FM, where the index falls as the message frequency rises. Message amplitude and modulator sensitivity both scale the index directly, being its two factors. And the peak phase deviation is the index, so it cannot vary independently of it. This single asymmetry explains every difference in the two schemes' spectra and noise behaviour.

Answer: A

Substituting the equivalent deviation \( \beta_p f_m \) into Carson's rule gives \( B = 2f_m(1+\beta_p) \). Twice the index has the wrong units, an index being dimensionless. Twice the message frequency is the narrowband limit, valid only for a small index. And the product of index and message frequency is the equivalent frequency deviation rather than the bandwidth. Note that PM bandwidth is proportional to message frequency, so a higher note widens the spectrum in direct proportion.

Answer: B

Applying the rule:

\[ B = 2 \times 2(1+6) = 28 \ \mathrm{kHz} \]

giving 28 kHz. 24 kHz omits the unity term, which matters at any index. 14 kHz drops the factor of two. And 12 kHz is the equivalent frequency deviation, \( \beta_p f_m \), which is a component of the answer rather than the answer. Working the equivalent deviation out first and then applying Carson's rule keeps the two angle schemes on the same footing. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.

Answer: C

The index is unchanged at 6 because amplitude is unchanged, so

\[ B = 2 \times 4(1+6) = 56 \ \mathrm{kHz} \]

which is double the previous figure. That strict proportionality to message frequency is peculiar to PM: in FM the same change would lower the index and widen the bandwidth only modestly. Keeping the bandwidth unchanged ignores the frequency term. 34 kHz and 14 kHz follow from no correct substitution. The contrast is worth computing once for each scheme side by side.

Answer: D

Because frequency is the derivative of phase, integrating first cancels the differentiation implicit in the relationship and the result is frequency modulation. That equivalence is the basis of the Armstrong transmitter, which phase modulates a crystal oscillator and integrates the audio to obtain FM. Amplitude modulation needs a quite different modulator. Doubling an index requires a gain change. And single sideband is an amplitude scheme entirely.

Answer: A

The dual operation gives phase modulation: differentiating undoes the integration that the frequency-to-phase relationship performs. The pair of conversions is why the two schemes are studied together and why either modulator can produce either signal with the right pre-processing. Wideband FM would need more deviation rather than differentiation. And the two amplitude schemes named cannot be reached from an angle modulator at all.

Answer: B

Both schemes produce the same Bessel expansion, with infinitely many sideband pairs spaced at the message frequency; only the way the index is computed differs. That shared structure is why one set of Bessel tables serves for both. Neither has a finite number of pairs, so the second and third options both misstate the mathematics. And the spacing is the message frequency in both cases, not twice it.