Frequency Modulation Questions and Answers

Answer: C

It uses the phase relationship that develops between primary and secondary of a tuned transformer as the frequency departs from resonance, converting that into a differential output. It is linear over a useful range but responds to amplitude as well, so it needs a limiter ahead of it. Amplitude alone carries no FM information. Envelope rate of change is not what it senses. And two rectified slopes describes the balanced slope detector, a distinct and cruder arrangement.

Answer: D

The ratio detector's diode arrangement makes its output depend on the ratio of two voltages rather than their difference, so amplitude variation largely cancels and a separate limiter may be unnecessary. It still uses tuned circuits. Both detectors work through phase to sense frequency. And the last option misstates the reason: FM has constant amplitude at the transmitter, but noise and fading make it vary by the time it reaches the receiver, which is exactly why limiting matters.

Answer: A

The loop forces its oscillator to follow the input frequency, so the control voltage is a faithful copy of the message. It is highly linear, needs no critical tuned circuits and is easily integrated, which is why it dominates modern receivers. Rectification recovers an envelope, not a frequency. Multiplying by a fixed carrier is a coherent AM detector. And zero-crossing counting is a different technique, workable but noisier at low signal levels.

Answer: B

The phase-locked loop needs no tuned transformers and few external components, so it integrates readily and holds its alignment without adjustment. All three alternatives depend on carefully tuned inductive components that resist integration and drift with age and temperature. That practical difference, rather than any theoretical advantage, is what displaced the discriminator family from consumer equipment.

Answer: C

The signal is multiplied by a copy of itself passed through a network giving ninety degrees at the centre frequency, so the product varies with frequency deviation. It uses one tuned circuit and integrates well, which made it common in television sound stages. Rectifying sidebands is not a detection method. A crystal comparison would give frequency error rather than the message. And the envelope of a limited signal is constant by construction.

Answer: D

Index first, then Carson's rule:

\[ \begin{aligned} \beta &= \frac{10}{2} = 5 \\ B &= 2(10+2) = 24 \ \mathrm{kHz} \end{aligned} \]

so 5 and 24 kHz. The 20 kHz option omits the message term, which matters most when the index is small. 0.2 inverts the index. And 10 is the deviation quoted as an index. Doing the two steps in that order, index then bandwidth, avoids the commonest confusion in FM numerical work.

Answer: A

Doubling the amplitude doubles the deviation to 20 kHz:

\[ \begin{aligned} \beta &= \frac{20}{2} = 10 \\ B &= 2(20+2) = 44 \ \mathrm{kHz} \end{aligned} \]

giving 10 and 44 kHz. The bandwidth grows but not in proportion, because the message term is unchanged. Keeping the old bandwidth would ignore the deviation change entirely. And 20 as an index confuses deviation with index. Noticing that amplitude affects deviation while frequency affects index is the heart of FM behaviour.

Answer: B

Deviation is unchanged at 10 kHz while the message frequency rises to 4 kHz:

\[ \begin{aligned} \beta &= \frac{10}{4} = 2.5 \\ B &= 2(10+4) = 28 \ \mathrm{kHz} \end{aligned} \]

so 2.5 and 28 kHz. The index falls while the bandwidth rises, which is the counter-intuitive result worth remembering: a higher note occupies more spectrum but with a smaller index. Options keeping the old index or bandwidth miss one half of the change.

Answer: C

Carson's rule gives 180 kHz, so

\[ 200 - 180 = 20 \ \mathrm{kHz} \]

leaving about 20 kHz of guard band, which accommodates the Bessel tail beyond Carson's ninety-eight per cent and any frequency error. 125 kHz would follow from comparing the channel with twice the deviation alone. 50 kHz and 5 kHz do not follow from the figures. The allocation is a good example of an engineering rule and a regulatory channel plan being designed together.

Answer: D

The regime is decided by the index, so raising the deviation until it greatly exceeds the highest message frequency is what moves a signal from narrowband to wideband. Carrier frequency plays no part. Power affects range rather than spectrum. And narrowing the audio bandwidth raises the index at the top of the band but reduces the information carried, which is a different trade. Index is the only quantity that defines the regime.