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.

Q: For \( R = 1 \ \mathrm{k\Omega} \), \( B = 1 \ \mathrm{MHz} \) at \( 290 \ \mathrm{K} \), the noise voltage is about

Substituting into the voltage expression:\[ \begin{aligned} \overline{v^{2}} &= 4kTBR \\ &= 1.6 \times 10^{-11} \ \mathrm{V^{2}} \\ v &= 4.0 \ \mathrm{\mu V} \end{aligned} \]so about 4 microvolts root mean square. The other options are decade errors either way. This is the figure that sets what a high-impedance input can usefully resolve, and it explains why sensitive front ends are built at low impedance.

Q: What causes shot noise?

Shot noise comes from the granularity of charge, so it requires a current and grows with it, unlike thermal noise which needs only a temperature. Thermal agitation is the resistor mechanism. Surface state fluctuation gives flicker noise. And mismatch reflections are a signal-path problem rather than a noise source. Because shot noise depends on current and thermal noise does not, the two can be separated experimentally by varying the bias. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Q: Which noise mechanism dominates at very low frequencies in semiconductors?

Flicker noise, with a spectral density falling roughly as one over frequency, dominates below a corner that may be anywhere from hertz to kilohertz depending on the device. Thermal and shot noise are essentially flat and take over above that corner. And partition noise occurs where current divides between electrodes. Knowing where each mechanism dominates tells a designer which device parameter to optimise for a given application. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted.

Q: Which of these is an external noise source?

Atmospheric static, along with galactic and man-made noise, arrives at the antenna from outside the receiver, whereas the other three are generated within it. The distinction matters because external noise can sometimes be reduced by siting, shielding or directional antennas, while internal noise can only be reduced by better devices. At medium wave the external contribution usually dominates, which is why a very low noise front end buys little there.

Q: Why does a very low noise front end buy little at medium wave?

At those frequencies atmospheric and man-made noise arriving at the antenna exceed anything the receiver adds, so improving the receiver's own noise figure changes the signal-to-noise ratio hardly at all. Bandwidth affects both contributions equally. Thermal noise density is flat with frequency. And antenna inefficiency attenuates signal and external noise together. Knowing which noise dominates decides whether an improvement is worth paying for.

Direction for [ Question No: 1 To 1 ] :

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.

Discuss in Forum

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.

Discuss in Forum

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

For \( R = 1 \ \mathrm{k\Omega} \), \( B = 1 \ \mathrm{MHz} \) at \( 290 \ \mathrm{K} \), the noise voltage is about

A.4 microvolts
B.40 microvolts
C.0.4 microvolts
D.4 millivolts

Answer: A

Substituting into the voltage expression:

\[ \begin{aligned} \overline{v^{2}} &= 4kTBR \\ &= 1.6 \times 10^{-11} \ \mathrm{V^{2}} \\ v &= 4.0 \ \mathrm{\mu V} \end{aligned} \]

so about 4 microvolts root mean square. The other options are decade errors either way. This is the figure that sets what a high-impedance input can usefully resolve, and it explains why sensitive front ends are built at low impedance.

Discuss in Forum

What causes shot noise?

A.Reflections from a mismatched transmission line
B.Discrete arrival of charge carriers
C.Random thermal agitation in a resistance
D.Fluctuating surface states at low frequency

Answer: B

Shot noise comes from the granularity of charge, so it requires a current and grows with it, unlike thermal noise which needs only a temperature. Thermal agitation is the resistor mechanism. Surface state fluctuation gives flicker noise. And mismatch reflections are a signal-path problem rather than a noise source. Because shot noise depends on current and thermal noise does not, the two can be separated experimentally by varying the bias. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Discuss in Forum

Which noise mechanism dominates at very low frequencies in semiconductors?

A.Shot noise
B.Partition noise
C.Flicker noise
D.Thermal noise

Answer: C

Flicker noise, with a spectral density falling roughly as one over frequency, dominates below a corner that may be anywhere from hertz to kilohertz depending on the device. Thermal and shot noise are essentially flat and take over above that corner. And partition noise occurs where current divides between electrodes. Knowing where each mechanism dominates tells a designer which device parameter to optimise for a given application. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted.

Discuss in Forum

Which of these is an external noise source?

A.Thermal noise in the input resistor
B.Shot noise in the first transistor stage
C.Flicker noise in the mixer's bias network
D.Atmospheric static

Answer: D

Atmospheric static, along with galactic and man-made noise, arrives at the antenna from outside the receiver, whereas the other three are generated within it. The distinction matters because external noise can sometimes be reduced by siting, shielding or directional antennas, while internal noise can only be reduced by better devices. At medium wave the external contribution usually dominates, which is why a very low noise front end buys little there.

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10.

Why does a very low noise front end buy little at medium wave?

A.External noise already dominates
B.The bandwidth is too narrow for noise to matter
C.Thermal noise falls with frequency in that band
D.Antennas are inefficient enough to mask it

Answer: A

At those frequencies atmospheric and man-made noise arriving at the antenna exceed anything the receiver adds, so improving the receiver's own noise figure changes the signal-to-noise ratio hardly at all. Bandwidth affects both contributions equally. Thermal noise density is flat with frequency. And antenna inefficiency attenuates signal and external noise together. Knowing which noise dominates decides whether an improvement is worth paying for.

Discuss in Forum

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Noise in Analog Systems Questions and Answers

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