Noise in Analog Systems Questions and Answers

Q: Assertion (A) : From SNR point of view, FM is superior to PM by a factor of 3.Reason (R) : Narrowband FM offers no improvement in SNR over AM.

The answer is both (A) and (R) are true but (R) is not the correct explanation of (A).For sinusoidal modulation, the output SNR of FM is \(3\beta^{2}\) times a reference value, while that of PM with the same peak phase deviation is \(\beta^{2}\) times it. So with equal deviation, FM is better than PM by a factor of 3 (about 4.8 dB). Separately, narrowband FM (\(\beta\) well below 1) occupies the same bandwidth as AM and gives no SNR improvement over it; the FM advantage appears only in wideband operation.Both statements are standard results, but the FM-over-PM factor of 3 comes from the different way noise is weighted after the discriminator, not from the comparison of narrowband FM with AM.

Q: What causes thermal noise in a resistor?

Thermal noise arises from the random thermal motion of charge carriers, so it depends on temperature and bandwidth alone and is present in any resistance whatever the current through it. Discrete arrival across a barrier is shot noise, which needs a current to flow. Slow surface fluctuations give flicker noise, dominant at low frequency. And current division between electrodes gives partition noise. Each mechanism has its own dependence, which is how they are told apart in a measurement.

Q: The available thermal noise power in a bandwidth \( B \) is

The available power is \( kTB \), independent of the resistance value, which is why a noise floor can be quoted in dBm per hertz without naming an impedance. \( 4kTBR \) is the mean square open-circuit voltage, which does depend on resistance. \( 2qIB \) is shot noise power spectral density in current terms. And dividing by bandwidth would give a density rather than a power. Knowing which expression gives a power and which a voltage prevents most errors in noise arithmetic.

Q: At \( 290 \ \mathrm{K} \) the thermal noise floor per hertz is about

The standard figure is -174 dBm per hertz, and adding ten times the logarithm of the bandwidth gives the power in any band: -114 dBm in a megahertz, -144 dBm in a kilohertz. -204 dBm would correspond to a much lower temperature. Carrying the -174 figure and the decibel bandwidth correction together is the quickest way to estimate any receiver's noise floor, and it is worth committing to memory.

Q: The mean square thermal noise voltage across a resistance is

The open-circuit voltage is given by \( \overline{v^{2}} = 4kTBR \), so a larger resistance develops more noise voltage even though the available power is unchanged. \( kTB \) is that available power. \( 2qIB \) belongs to shot noise. And the last expression has the wrong dimensions. The factor of four appears because the expression is for the open-circuit voltage rather than the voltage across a matched load, a distinction worth checking whenever the formula is quoted.

Direction for [ Question No: 1 To 5 ] :

Read the paragraph and answer the questions :

Noise may be defined, in an electrical sense, as an unwanted form of energy tending to interfere with the proper and easy reception and reproduction of wanted signals. Many disturbances of the electrical nature produce noise in receiver, modifying the signal in an unwanted manner. Noise is divided into two groups. Noise whose sources are external to the receiver and other one is noise created within the receiver itself. Various forms of noise created outside the receiver i.e. external noise includes atmospheric and extra-terrestrial noise and industrial noise. Atmospheric noise becomes less severe at frequencies above 30 MHz. Extra-terrestrial noises are solar noise and cosmic noise. They radiate over a very broad frequency spectrum which includes the frequencies, we use for communication. These disturbances are at the eruption of corona flares and sunspots. The noise generated is due to the random motion of electrons, atoms and molecules. Kinetic theory states that the temperature of the system increases due to the motion of the particles.

In a communication system, noise is most likely to affect the signal

A.at a transmitter
B.in a channel
C.in the information source
D.at the destination

Exam Year(s)

UGC NET - 2012,

Answer: B

In a communication system noise is most likely to affect the signal in the channel, option B: the signal passes through the medium between the transmitter and the receiver, where atmospheric, industrial and extraterrestrial noise is picked up and the signal is weakest. The transmitter works at a high signal level, so noise is relatively small there, the information source is not where noise is added, and at the destination the signal has already been received and amplified. The answer is the channel.

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Indicate the false statement.

A.HF mixers are generally noiser than HF amplifiers.
B.Impulse noise voltage is independent of bandwidth.
C.Thermal noise is independent of the frequency at which it is measured.
D.Industrial noise is usually of the impulse type.

Exam Year(s)

UGC NET - 2012,

Answer: B

The false statement is that impulse noise voltage is independent of bandwidth. Impulse noise consists of short, strong pulses, and its voltage is directly proportional to the bandwidth of the receiver, so that a wider bandwidth passes more of it, which is why option B is false. The other statements are true: HF mixers are generally noisier than HF amplifiers, thermal noise power is independent of the frequency at which it is measured over the ordinary range, and industrial noise is typically impulsive.

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The value of a resistor creating thermal noise is doubled. The noise power generated is therefore

A.halved
B.quadrupled
C.doubled
D.unchanged

Exam Year(s)

UGC NET - 2012,

Answer: D

The available thermal noise power of a resistor is \( P_n = kTB \), where k is Boltzmann's constant, T the absolute temperature and B the bandwidth. The resistance does not appear in this expression, so doubling it leaves the noise power unchanged, option D. What does change is the noise voltage, \( v_n = \sqrt{4kTBR} \), which rises by the square root of two; but the question asks for power. Options A, B and C wrongly make the power depend on the resistance.

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One of the following is not a useful quantity for comparing the noise performance of receivers :

A.Input noise voltage
B.Equivalent noise resistance
C.Noise temperature
D.Noise figure

Exam Year(s)

UGC NET - 2012,

Answer: A

Input noise voltage is not a useful quantity for comparing the noise performance of receivers, because it depends on the bandwidth, the resistance and the circuit conditions, so it does not give a figure that applies in the same way to different receivers. Equivalent noise resistance, noise temperature and noise figure are all standard measures used to compare receivers. The answer is therefore option A.

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Indicate the noise whose source is in a category different from that of the other three.

A.Solar noise
B.Cosmic noise
C.Atmospheric noise
D.Galactic noise

Exam Year(s)

UGC NET - 2012,

Answer: C

Atmospheric noise comes from the earth's own atmosphere, lightning discharges in particular, while solar, cosmic and galactic noise all come from beyond the earth and are classed as extraterrestrial noise. The passage names the extra-terrestrial group as solar and cosmic noise, and galactic noise is the same cosmic type. So the one whose source is in a different category from the other three is atmospheric noise, option C.

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Direction for [ Question No: 6 To 6 ] :

Assertion-Reason type questions : The following items consist of two statements, one labelled as 'Assertion (A)', and the other labelled the 'Reason (R)'. You are to examine these two statements and decide if the Assertion (A) and the Reason (R) are individually true and if so, whether the reason is a correct explanation of the Assertion. Select your answer to these items using the codes given below and mark your answer sheet accordingly.

Codes :
(A) Both (A) and (R) are true and (R) is the correct explanation of (A)
(B) Both (A) and (R) are true but (R) is not the correct explanation of (A)
(C) (A) is true but (R) is false
(D) (A) is false and (R) is true

Assertion (A) : From SNR point of view, FM is superior to PM by a factor of 3.

Reason (R) : Narrowband FM offers no improvement in SNR over AM.

A.Both (A) and (R) are true and (R) is the correct explanation of (A)
B.Both (A) and (R) are true but (R) is not the correct explanation of (A)
C.(A) is true but (R) is false
D.(A) is false and (R) is true

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The answer is both (A) and (R) are true but (R) is not the correct explanation of (A).

For sinusoidal modulation, the output SNR of FM is \(3\beta^{2}\) times a reference value, while that of PM with the same peak phase deviation is \(\beta^{2}\) times it. So with equal deviation, FM is better than PM by a factor of 3 (about 4.8 dB). Separately, narrowband FM (\(\beta\) well below 1) occupies the same bandwidth as AM and gives no SNR improvement over it; the FM advantage appears only in wideband operation.

Both statements are standard results, but the FM-over-PM factor of 3 comes from the different way noise is weighted after the discriminator, not from the comparison of narrowband FM with AM.

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What causes thermal noise in a resistor?

A.Random motion of charge carriers
B.Discrete arrival of carriers across a junction barrier
C.Slow fluctuations in the surface state of the semiconductor
D.Division of current between two collecting electrodes

Answer: A

Thermal noise arises from the random thermal motion of charge carriers, so it depends on temperature and bandwidth alone and is present in any resistance whatever the current through it. Discrete arrival across a barrier is shot noise, which needs a current to flow. Slow surface fluctuations give flicker noise, dominant at low frequency. And current division between electrodes gives partition noise. Each mechanism has its own dependence, which is how they are told apart in a measurement.

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The available thermal noise power in a bandwidth \( B \) is

A.kT over B
B.kTB
C.4kTBR
D.2qIB

Answer: B

The available power is \( kTB \), independent of the resistance value, which is why a noise floor can be quoted in dBm per hertz without naming an impedance. \( 4kTBR \) is the mean square open-circuit voltage, which does depend on resistance. \( 2qIB \) is shot noise power spectral density in current terms. And dividing by bandwidth would give a density rather than a power. Knowing which expression gives a power and which a voltage prevents most errors in noise arithmetic.

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At \( 290 \ \mathrm{K} \) the thermal noise floor per hertz is about

A.-204 dBm
B.-144 dBm
C.-174 dBm
D.-114 dBm

Answer: C

The standard figure is -174 dBm per hertz, and adding ten times the logarithm of the bandwidth gives the power in any band: -114 dBm in a megahertz, -144 dBm in a kilohertz. -204 dBm would correspond to a much lower temperature. Carrying the -174 figure and the decibel bandwidth correction together is the quickest way to estimate any receiver's noise floor, and it is worth committing to memory.

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10.

The mean square thermal noise voltage across a resistance is

A.kTB
B.2qIB
C.kTBR squared
D.4kTBR

Answer: D

The open-circuit voltage is given by \( \overline{v^{2}} = 4kTBR \), so a larger resistance develops more noise voltage even though the available power is unchanged. \( kTB \) is that available power. \( 2qIB \) belongs to shot noise. And the last expression has the wrong dimensions. The factor of four appears because the expression is for the open-circuit voltage rather than the voltage across a matched load, a distinction worth checking whenever the formula is quoted.

Discuss in Forum

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Noise in Analog Systems Questions and Answers

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