Frequency Modulation Questions and Answers

Q: Bandwidth of frequency modulation (FM) is given by :Where fm = frequency of modulating Signalfd = deviating frequency

The bandwidth is \(2(f_{m} + f_{d})\), Carson's rule.An FM signal has, in theory, infinitely many sidebands, but almost all of its power lies within a band of \(2(\Delta f + f_{m})\), where \(\Delta f = f_{d}\) is the frequency deviation and \(f_{m}\) the highest modulating frequency. This approximation is Carson's rule.For example, broadcast FM with \(f_{d} = 75\) kHz and \(f_{m} = 15\) kHz needs about \(2(75 + 15) = 180\) kHz. \(f_{m} + f_{d}\) is only half of that band, the average \(\frac{f_{m} + f_{d}}{2}\) has no physical meaning here, and \(2f_{m}\) is the bandwidth of AM or of very narrowband FM.

Q: Choose the correct statement :

The correct statement is that narrowband FM offers no improvement in SNR over AM.The noise advantage of FM comes from trading bandwidth for signal to noise ratio: the output SNR improvement grows with the square of the modulation index \(\beta\). In narrowband FM, \(\beta\) is small (well below 1) and the signal occupies about the same bandwidth as AM, \(2f_{m}\), so it offers essentially no SNR advantage over AM.The statements that NBFM is better than AM, or augments SNR, describe wideband FM. Saying it is worse than AM is also wrong; its performance is comparable.

Q: The main difference between Amplitude modulation and Frequency modulation is that :(a) Carrier frequency is higher than base band frequency(b) Frequency modulation has limited bandwidth(c) Frequency modulation has unlimited bandwidth(d) Intermediate frequency stage is found in frequency demodulator circuit

The main difference is (c) only: frequency modulation has unlimited bandwidth.Theoretically an FM wave has an infinite number of sidebands (described by Bessel functions), so its bandwidth is unlimited, although in practice Carson's rule gives the band that contains most of the power. An AM wave has just two sidebands and a bandwidth of twice the highest modulating frequency.Statement (a) is true of both AM and FM, so it is not a difference. Statement (b) is false. Statement (d) is not a difference either, since superheterodyne AM receivers also have an intermediate frequency stage. So any option including (a), (b) or (d) is wrong.

Q: Which is the true statement about frequency deviation in frequency modulation ?

In FM, frequency deviation is proportional to the amplitude of the modulating signal.Frequency modulation varies the instantaneous carrier frequency in step with the message:\[ f_{i}(t) = f_{c} + k_{f}\,m(t) \]The peak deviation is \(\Delta f = k_{f} A_{m}\), set by the modulating signal's amplitude. The modulating frequency sets how often the carrier swings, not how far.Deviation does not depend on the carrier frequency or on the carrier amplitude, which stays constant in FM. The modulating frequency enters the modulation index, \(\beta = \Delta f / f_{m}\), but not the deviation; deviation proportional to modulating frequency is a feature of phase modulation.

Q: Which of the following is disadvantage of FM over AM ?

The disadvantage of FM is that higher bandwidth is required.By Carson's rule, an FM signal needs a bandwidth of about \(2(\Delta f + f_{m})\). Broadcast FM, with a deviation of 75 kHz and audio up to 15 kHz, occupies about 180-200 kHz per channel, while AM needs only twice the highest audio frequency, about 10 kHz. FM therefore has to use VHF bands and fewer channels fit in a given spectrum.Needing less modulating power and giving better noise immunity are advantages of FM, not disadvantages. 'Carrier is of any shape' is not a property of either system.

Q: In FM, when frequency deviation doubled, then

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.

Bandwidth of frequency modulation (FM) is given by :
Where f_m = frequency of modulating Signal
f_d = deviating frequency

A.\(2(f_{m} + f_{d})\)
B.\(f_{m} + f_{d}\)
C.\(\frac{f_{m} + f_{d}}{2}\)
D.\(2f_{m}\)

Exam Year(s)

RSMSSB JE - 2025, RSMSSB - 2025,

Answer: A

The bandwidth is \(2(f_{m} + f_{d})\), Carson's rule.

An FM signal has, in theory, infinitely many sidebands, but almost all of its power lies within a band of \(2(\Delta f + f_{m})\), where \(\Delta f = f_{d}\) is the frequency deviation and \(f_{m}\) the highest modulating frequency. This approximation is Carson's rule.

For example, broadcast FM with \(f_{d} = 75\) kHz and \(f_{m} = 15\) kHz needs about \(2(75 + 15) = 180\) kHz. \(f_{m} + f_{d}\) is only half of that band, the average \(\frac{f_{m} + f_{d}}{2}\) has no physical meaning here, and \(2f_{m}\) is the bandwidth of AM or of very narrowband FM.

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Choose the correct statement :

A.Narrowband FM offers better SNR than AM
B.Narrowband FM offers poor SNR than AM
C.Narrowband FM offers no improvement in SNR over AM
D.Narrowband FM augments SNR as compared to AM

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The correct statement is that narrowband FM offers no improvement in SNR over AM.

The noise advantage of FM comes from trading bandwidth for signal to noise ratio: the output SNR improvement grows with the square of the modulation index \(\beta\). In narrowband FM, \(\beta\) is small (well below 1) and the signal occupies about the same bandwidth as AM, \(2f_{m}\), so it offers essentially no SNR advantage over AM.

The statements that NBFM is better than AM, or augments SNR, describe wideband FM. Saying it is worse than AM is also wrong; its performance is comparable.

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The main difference between Amplitude modulation and Frequency modulation is that :

(a) Carrier frequency is higher than base band frequency
(b) Frequency modulation has limited bandwidth
(c) Frequency modulation has unlimited bandwidth
(d) Intermediate frequency stage is found in frequency demodulator circuit

A.(a) and (c)
B.(c) and (d)
C.(b) and (d)
D.(c) only

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: D

The main difference is (c) only: frequency modulation has unlimited bandwidth.

Theoretically an FM wave has an infinite number of sidebands (described by Bessel functions), so its bandwidth is unlimited, although in practice Carson's rule gives the band that contains most of the power. An AM wave has just two sidebands and a bandwidth of twice the highest modulating frequency.

Statement (a) is true of both AM and FM, so it is not a difference. Statement (b) is false. Statement (d) is not a difference either, since superheterodyne AM receivers also have an intermediate frequency stage. So any option including (a), (b) or (d) is wrong.

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Which is the true statement about frequency deviation in frequency modulation ?

A.frequency deviation is proportional to carrier signal frequency
B.frequency deviation is proportional to amplitude of carrier signal
C.frequency deviation is proportional to modulating frequency
D.frequency deviation is proportional to amplitude of modulating signal

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: D

In FM, frequency deviation is proportional to the amplitude of the modulating signal.

Frequency modulation varies the instantaneous carrier frequency in step with the message:

\[ f_{i}(t) = f_{c} + k_{f}\,m(t) \]

The peak deviation is \(\Delta f = k_{f} A_{m}\), set by the modulating signal's amplitude. The modulating frequency sets how often the carrier swings, not how far.

Deviation does not depend on the carrier frequency or on the carrier amplitude, which stays constant in FM. The modulating frequency enters the modulation index, \(\beta = \Delta f / f_{m}\), but not the deviation; deviation proportional to modulating frequency is a feature of phase modulation.

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Which of the following is disadvantage of FM over AM ?

A.less modulating power is required
B.better noise immunity is provided
C.higher bandwidth is required
D.carrier is of any shape

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: C

The disadvantage of FM is that higher bandwidth is required.

By Carson's rule, an FM signal needs a bandwidth of about \(2(\Delta f + f_{m})\). Broadcast FM, with a deviation of 75 kHz and audio up to 15 kHz, occupies about 180-200 kHz per channel, while AM needs only twice the highest audio frequency, about 10 kHz. FM therefore has to use VHF bands and fewer channels fit in a given spectrum.

Needing less modulating power and giving better noise immunity are advantages of FM, not disadvantages. 'Carrier is of any shape' is not a property of either system.

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

A.Better noise immunity is provided
B.Lower bandwidth is required
C.The transmitted power is more useful
D.Less modulating power is required

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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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.

A.I and II are correct.
B.I, II and III are correct.
C.I and III are correct.
D.III and IV are correct.

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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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

A.0.1 kHz
B.1 kHz
C.3.2 kHz
D.4 kHz

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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In FM, when frequency deviation doubled, then

A.Modulation Index is decreased
B.Modulation Index is doubled
C.Modulation Index halved
D.No change occurs in modulation index values

Exam Year(s)

UGC NET - 2013,

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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10.

List - I	List - II
a. Frequency modulation	i. Envelope detection
b. Double sideband suppressed signal carrier	ii. Companding
c. PCM	iii. Balance modulator
d. Amplitude modulation	iv. Pre-emphasis and de-emphasis

Codes : a b c d

A.a-i, b-ii, c-iii, d-iv
B.a-ii, b-iii, c-i, d-iv
C.a-iii, b-i, c-ii, d-iv
D.a-iv, b-iii, c-ii, d-i

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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