Frequency Modulation Questions and Answers

Answer: B

The statement that is not an advantage of FM over AM is that a lower bandwidth is required, option (B). Frequency modulation spreads the signal over a much wider band than amplitude modulation, which is why commercial FM uses a channel width of 200 kHz, against 10 kHz for AM. Better noise immunity, option (A), is the chief advantage of FM, since noise adds mainly to amplitude, which an FM receiver removes with a limiter. In FM the transmitted power is more useful, option (C), because all the power goes into the information-carrying sidebands and carrier at constant amplitude. Less modulating power, option (D), is required, since the modulator works at a low level.

Answer: C

The key takes statements I and III as correct, so option (C) is right, which makes statements II and IV the false ones. The Armstrong system generates frequency modulation indirectly: a crystal-controlled carrier is phase modulated by a signal that has been integrated, which is the phase and frequency modulation of statement I, and the small frequency deviation is then raised by frequency multiplication, statement III. Statement II is false, since the system is not amplitude modulation, although an amplitude component is removed with a limiter. Statement IV is false, because the integrator in the modulating path, a form of equalisation, is necessary. Options (A), (B) and (D) include a false statement as a correct one.

Answer: C

The approximate bandwidth is 3.2 kHz. The message is \(5\sin(2\pi\,100\,t)\), so \(f_{m} = 100\) Hz, and the modulated signal carries a phase deviation of 15 rad, so the modulation index is \(\beta = 15\). Carson's rule gives

\[ BW \approx 2(\beta + 1) f_{m} = 2(15 + 1)(100) \]

\[ BW = 3200\ \text{Hz} = 3.2\ \text{kHz} \]

Options (A) and (B) are of the order of the message frequency alone and ignore the many significant sidebands that a modulation index of 15 produces. Option (D), 4 kHz, does not follow from Carson's rule for these values. The 100 kHz carrier frequency does not enter the bandwidth.