Frequency Modulation Questions and Answers

Q: Which parameter would a regulator specify to protect neighbouring FM channels?

Regulators specify maximum deviation, since it is deviation that sets how far the spectrum extends and hence how much energy falls into the neighbours. A minimum power would help nobody's interference case. A minimum audio bandwidth is a quality matter. And the index at low audio frequencies is large but corresponds to a narrow spectrum in Carson terms, so constraining it would target the wrong quantity. Deviation is the single lever that controls occupied bandwidth.

Q: Why does an angle-modulated signal permit a saturated output amplifier?

Because the amplitude carries nothing, so distorting or clipping it costs no information and the amplifier can run at maximum efficiency. That single property separates the angle family from every amplitude scheme, where the envelope is the message. Bandwidth is wide rather than narrow. The carrier is present throughout. And the index is frequently far greater than one, as broadcasting's five demonstrates.

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.

101.

Which parameter would a regulator specify to protect neighbouring FM channels?

A.Maximum permitted deviation
B.Minimum transmitter power
C.Minimum audio bandwidth
D.Maximum modulation index at low frequencies

Answer: A

Regulators specify maximum deviation, since it is deviation that sets how far the spectrum extends and hence how much energy falls into the neighbours. A minimum power would help nobody's interference case. A minimum audio bandwidth is a quality matter. And the index at low audio frequencies is large but corresponds to a narrow spectrum in Carson terms, so constraining it would target the wrong quantity. Deviation is the single lever that controls occupied bandwidth.

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102.

A test tone of constant amplitude is swept upward in frequency. The FM index

A.Stays constant, both being set by the deviation
B.Falls while the occupied bandwidth rises
C.Rises while the occupied bandwidth falls
D.Falls and the occupied bandwidth falls with it

Answer: B

Constant amplitude fixes the deviation, so the index, being deviation over message frequency, falls as the tone rises, while Carson's rule adds the growing message frequency and the bandwidth increases. The two quantities move in opposite directions, which is the single most counter-intuitive result in the topic and the reason worst-case bandwidth is computed at the highest audio frequency rather than at the largest index.

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103.

Which observation on a spectrum analyser would confirm a deviation setting exactly?

A.Equal amplitudes of the first two sidebands
B.The total power remaining constant
C.A carrier null at a known index
D.The widest visible sideband

Answer: C

The Bessel function \( J_0 \) vanishes at particular indices, so a carrier null with a known modulating frequency fixes the index and hence the deviation exactly - the classical calibration method, needing no calibrated amplitude measurement at all. The outermost visible sideband depends on the analyser's noise floor. Equal first and second sidebands occur at one particular index and is a less sharp criterion. And constant total power is true of all FM and so distinguishes nothing.

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104.

Why does an angle-modulated signal permit a saturated output amplifier?

A.Its bandwidth is narrow enough to filter
B.Its carrier is suppressed
C.Its index is always less than one
D.Its amplitude carries no information

Answer: D

Because the amplitude carries nothing, so distorting or clipping it costs no information and the amplifier can run at maximum efficiency. That single property separates the angle family from every amplitude scheme, where the envelope is the message. Bandwidth is wide rather than narrow. The carrier is present throughout. And the index is frequently far greater than one, as broadcasting's five demonstrates.

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105.

Which comparison of pre-emphasis and a deviation limiter is correct?

A.Pre-emphasis raises high-frequency deviation, the limiter caps it
B.Both reduce the deviation on bright material
C.Pre-emphasis caps the deviation and the limiter raises it
D.Neither affects the deviation at all

Answer: A

They pull in opposite directions and that is why their order matters: pre-emphasis boosts treble and therefore increases deviation on bright material, and the limiter placed after it holds the peak deviation inside the channel. Putting the limiter first would let pre-emphasis push the signal back over the limit. Saying both reduce deviation misses the purpose of pre-emphasis, and the reversed description inverts both stages. The audio chain's order is a genuine design constraint rather than an arbitrary convention.

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106.

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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107.

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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108.

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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109.

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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110.

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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