Frequency Modulation Questions and Answers

Answer: C

Medium wave channels are nine or ten kilohertz wide, while wideband FM needs something like 180, so the band could hold only a handful of stations. That is why FM broadcasting waited for the very high frequency bands to become usable. Receivers work perfectly well at medium wave. Propagation does not destroy frequency modulation, though it does cause selective fading. And transmitter power is not the constraint. Spectrum availability is what determines where each scheme is deployed.

Answer: D

The constant envelope lets the transmitter run its output device close to saturation, which matters enormously where power is scarce and efficiency is everything. Its bandwidth is wide rather than narrow, which is affordable at microwave frequencies. It tolerates frequency error better than SSB but that is not the driver. And the carrier is present throughout. Efficiency and noise performance together are why angle modulation dominated analogue microwave telephony.

Answer: A

Power goes with the square of amplitude:

\[ J_0^{2} = 0.77^{2} = 0.59 \]

so about 59 per cent remains in the carrier, the rest having moved into the sidebands. 77 per cent is the amplitude ratio quoted as a power. 44 per cent is the first sideband amplitude. And 19 per cent is that amplitude squared, which is one sideband's power fraction. Remembering to square before comparing powers is essential when reading Bessel tables.

Answer: B

Total power is constant, so whatever appears in the sidebands must come out of the carrier, and at some indices the carrier vanishes entirely. Output power does not fall; that is the whole point of the constant-envelope property. The limiter is a receiver stage. And deviation does not cancel anything; it redistributes energy according to the Bessel functions. This redistribution is the clearest way to see why FM power arithmetic differs so sharply from AM's.

Answer: C

The improvement rises roughly with the square of the modulation index, so doubling the deviation is worth about six decibels at the output for the same transmitted power, provided the receiver bandwidth is widened to match and the signal stays above threshold. Constant power does not mean constant performance, which is the counter-intuitive heart of the trade. The wider bandwidth admits more noise but the demodulation gain more than compensates. Below threshold the trade collapses, which is a real and separate caveat.

Answer: D

The cost is bandwidth, and therefore fewer channels in a given allocation, which is exactly what the noise improvement is bought with. Efficiency is unaffected, the envelope remaining constant. The modulator is no more complex for larger deviation. And the capture effect improves rather than disappears. Naming the currency of the trade correctly - spectrum for noise - is what makes FM design decisions comprehensible.

Answer: A

A discriminator turns frequency deviation into a voltage, which is exactly the demodulation FM requires. The reverse operation, voltage to frequency, is what a modulator or a voltage-controlled oscillator does. The two amplitude-and-phase conversions describe intermediate mechanisms inside particular detectors rather than the function of the stage. Naming the input and output quantities of each block is the fastest way to reason about an unfamiliar receiver diagram.

Answer: B

The limiter is unique to FM, deliberately destroying the amplitude information that an AM receiver depends on. The mixer, intermediate frequency amplifier and gain control all appear in both architectures. Spotting the limiter on a block diagram is therefore a reliable way to identify an FM receiver, and understanding why it is there - constant envelope means amplitude carries only noise - explains the scheme's noise advantage in one step.

Answer: C

Applying the rule:

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

so 16 kHz, which fits comfortably in a 25 kHz land mobile channel. 10 kHz is twice the deviation, omitting the audio term that matters most at low index. 6 kHz is twice the audio frequency, the narrowband limit that applies only when the index is much less than one. And 25 kHz is the channel spacing rather than the occupied bandwidth.

Answer: D

Regulators set spacing wider than the occupied bandwidth so that the Bessel tail beyond Carson's rule, frequency errors and receiver imperfections do not cause interference. FM broadcasting's 180 kHz in a 200 kHz channel is the standard example. Equality would leave no margin. Occupied bandwidth exceeding the spacing is precisely the over-deviation fault. And audio bandwidth alone would ignore the deviation entirely.