FM Transmitters and Receivers Questions and Answers

Answer: D

It carries station identification, programme type, traffic messages and similar data, which receivers display or act upon. The phase reference is the pilot, at a third of its frequency. Duplicating the difference signal would waste deviation for no gain. And a calibration tone would be a test signal rather than a broadcast service. Adding a data channel above the existing multiplex, without disturbing either mono or stereo reception, is the same backward-compatible design pattern as stereo itself.

Answer: A

Power is constant, since the amplitude never varies and modulation merely redistributes energy among the carrier and sidebands. Bandwidth grows with deviation. The index falls as audio frequency rises. And deviation is proportional to audio level, which is what the limiter exists to cap. Identifying the one invariant answers every FM power question immediately and explains why class C amplification is available throughout the chain. A single measurement chosen to respond to the one asymmetry beats several that respond to shared behaviour.

Answer: B

Because amplitude carries nothing, clipping or compressing it destroys no information, so the stage can be driven hard for efficiency and the tuned output circuit restores a clean waveform. FM's bandwidth is wide rather than narrow. Its carrier is present throughout. And the deviation being small relative to the carrier is true but irrelevant to amplifier class. One property of the signal decides which amplifier classes are available, and it is worth checking that property first in any transmitter question.

Answer: C

An FM broadcast channel is about 180 kHz wide against AM's nine or ten, so the intermediate filter must be correspondingly wider, which is also why a higher intermediate frequency is used. Fading depends on propagation rather than modulation. Detectors do not set selectivity, filters do. And transmitter drift is controlled to a small fraction of the channel in both services. Filter width always follows the occupied bandwidth of the service.

Answer: D

Its high Q gives a flat passband with steep skirts, which is what adjacent-channel rejection requires and which a single tuned circuit cannot approach. It provides no gain; indeed it has insertion loss. Gain control is a separate loop. And it works at one fixed frequency only, which is precisely why it suits the intermediate stages of a superheterodyne rather than the tunable front end. Fixed-frequency components are always cheaper to make sharp.

Answer: A

AM suppression quantifies how much amplitude modulation on the input survives to the audio output, and it is a direct test of the limiter. Capture ratio concerns two competing FM signals. Image rejection concerns one particular offset frequency. And shape factor describes a filter's transition steepness. Every specification names one mechanism, and AM suppression names the one mechanism that gives angle modulation its advantage over the amplitude family.

Answer: B

Applying the rule:

\[ B = 2(75+15) = 180 \ \mathrm{kHz} \]

so 180 kHz, which is why the channel is 200 kHz wide. 150 kHz is twice the deviation with the audio term dropped. 200 kHz is the channel spacing, not the occupied bandwidth. And 90 kHz omits the factor of two. The twenty-kilohertz difference between the two figures is the guard band, and knowing which number is which is what distinguishes an occupied bandwidth from an allocation. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Answer: C

Carson's rule gives

\[ B = 2(5+3) = 16 \ \mathrm{kHz} \]

which fits comfortably in the 25 kHz channel such services use, so 16 kHz is the answer and 25 kHz is the allocation. 10 kHz omits the audio term, which matters most at low index. And 6 kHz is the narrowband limit valid only for an index well below one, which an index of 1.67 is not. Choosing between the full rule and its limit is decided by the index.

Answer: D

Dividing bandwidth by centre frequency:

\[ \frac{0.18}{10.7} = 0.017 \]

so about 1.7 per cent, which a ceramic filter achieves comfortably. The same 180 kHz at 98 MHz would be under 0.2 per cent, which is why the selectivity is provided after translation rather than at the signal frequency. That comparison is the numerical form of the superheterodyne's whole justification, and it is worth working once for each service.

Answer: A

Dividing again:

\[ \frac{0.18}{98} = 0.0018 \]

giving about 0.18 per cent, some ten times harder than the same job at 10.7 MHz. That single comparison explains why every broadcast receiver translates before filtering, and why a tuned radio-frequency receiver could never achieve consistent selectivity across a band. The arithmetic is trivial but the conclusion is the architectural one.