FM Transmitters and Receivers Questions and Answers

Answer: C

The pilot is at 19 kHz, half the subcarrier frequency, and the receiver doubles it to regenerate the suppressed 38 kHz subcarrier with the correct phase. 38 kHz is that subcarrier itself, transmitted with its carrier suppressed. 57 kHz carries the radio data subcarrier, three times the pilot. And 15 kHz is the top of the audio band. Each frequency in the multiplex is a simple multiple of the pilot, which is what makes coherent regeneration possible.

Answer: D

Suppressing it frees deviation that would otherwise be spent on a component carrying no information, and the small pilot at half the frequency is a far cheaper way to give the receiver its phase reference. Bandwidth is unchanged by carrier suppression, as in any double-sideband scheme. Envelope detection is impossible for a suppressed-carrier signal. And the pilot is placed deliberately above the audio band precisely so it does not interfere. Most errors on this topic come from carrying a result across from a neighbouring scheme where it does not hold.

Answer: A

The difference signal is double-sideband suppressed carrier about 38 kHz, so it spans 23 to 53 kHz for audio to 15 kHz. The sum signal occupies 0 to 15 kHz and is what a mono receiver uses. 57 kHz and above carries radio data. And 15 to 19 kHz is the guard space that keeps the pilot clear of the audio. Sketching the composite spectrum once makes the whole multiplex arrangement memorable.

Answer: B

It uses the 0 to 15 kHz sum signal and its audio filtering discards everything above, which is exactly the backward compatibility the multiplex was designed for. The difference signal alone would give an unlistenable result. Decoding both channels requires a stereo decoder the mono set does not have. And the pilot is deliberately inaudible and harmless. Designing a new service so that existing receivers still work is a recurring constraint in broadcasting.

Answer: C

Adding and subtracting the recovered sum and difference signals gives the left and right channels directly, which is why the multiplex encodes them that way rather than as two separate audio bands. The channels are not separable by sideband. Envelope detection fails on a suppressed-carrier signal. And the pilot serves only to regenerate the subcarrier. The matrix at each end is the elegance of the scheme.

Answer: D

It sits at 57 kHz, the third harmonic of the pilot, which keeps it coherent with the rest of the multiplex and clear of the difference signal ending at 53 kHz. 38 kHz is the stereo subcarrier and 19 kHz the pilot. 76 kHz would be the fourth harmonic and lies outside the standard composite. The choice of a harmonic of the pilot is deliberate: it allows one reference to serve every subcarrier in the system.

Answer: A

With radio data included the composite reaches about 60 kHz, which is what the modulator must accept and the receiver's discriminator must reproduce before decoding. The 15 kHz audio limit applies to each channel rather than the composite. 38 kHz is the subcarrier in the middle of the difference signal. And 75 kHz is the peak deviation, a quite different quantity. Distinguishing the baseband from the deviation is essential to reading the system correctly.

Answer: B

10.7 MHz is the convention, chosen high enough to place the image well outside the band and to accommodate a passband of some 200 kHz. 455 kHz is the AM broadcast figure and would be far too low to pass an FM channel. 38 kHz is the stereo subcarrier. And 21.4 MHz is twice the intermediate frequency, which is the image offset rather than the intermediate frequency itself. The value follows from the channel width, which is the general rule for choosing one.

Answer: C

Adding the intermediate frequency:

\[ f_{LO} = 98.1 + 10.7 = 108.8 \ \mathrm{MHz} \]

giving 108.8 MHz, just above the top of the broadcast band. 87.4 MHz is low-side injection, which would place the oscillator inside the band and risk radiating into it. 98.1 MHz is the signal and 10.7 MHz the intermediate frequency. Keeping the oscillator out of the received band is one practical reason high-side injection is preferred here. Working one numerical case through both schemes side by side fixes the distinction better than any rule of thumb.

Answer: D

The image sits twice the intermediate frequency above the signal:

\[ f_{image} = 98.1 + 21.4 = 119.5 \ \mathrm{MHz} \]

so 119.5 MHz, comfortably above the broadcast band, which is exactly why a 10.7 MHz intermediate frequency works so well here: the image falls where no broadcast station transmits. 108.8 MHz is the oscillator. The other two are the signal and its low-side alternative. Choosing an intermediate frequency that places the image outside the band is the cleanest solution to image interference.