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Frequency Modulation Questions and Answers

111.

Indicate which one of the following is not an advantage of FM over AM :

Exam Year(s)

UGC NET - 2012,

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.

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112.

Indicate the false statement in the case of Armstrong modulation system.

I. The system is basically phase and frequency modulation.
II. The system is also amplitude modulation.
III. Frequency multiplication must be used.
IV. Equalization is unnecessary.

Exam Year(s)

UGC NET - 2012,

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.

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113.

A signal m(f) = 5 sin (\(2\pi\) 100t) frequency modulates a carrier. The resulting FM signal is
{10 cos (\(2\pi\) \(10^{5}\)t) + 15 sin (\(2\pi\) 100 t)}.
The approximate bandwidth of FM signal is

Exam Year(s)

UGC NET - 2013,

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.

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Answer: B

When the frequency deviation in FM is doubled, the modulation index is doubled.

The modulation index is the frequency deviation divided by the modulating frequency. If the modulating frequency is unchanged, doubling the deviation doubles the index.

Decreasing it, halving it or leaving it unchanged would each require the deviation to fall, to be halved or to stay the same, which is not what is stated.

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115.

List - IList - II
a. Frequency modulationi. Envelope detection
b. Double sideband suppressed signal carrierii. Companding
c. PCMiii. Balance modulator
d. Amplitude modulationiv. Pre-emphasis and de-emphasis

Codes : a b c d

Exam Year(s)

UGC NET - 2013,

Answer: D

Frequency modulation uses pre-emphasis and de-emphasis to improve the noise performance (iv); a double sideband suppressed carrier signal is produced by a balanced modulator (iii); PCM uses companding to improve the quantization of weak signals (ii); and amplitude modulation is demodulated by an envelope detector (i). So the code is a-iv, b-iii, c-ii, d-i. The answer is a-iv, b-iii, c-ii, d-i.

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