Noise in Analog Systems Questions and Answers

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.

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.

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.

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.

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.

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.

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.

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.

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.

Answer: B

Noise factor is the ratio by which a stage worsens the signal-to-noise ratio, referenced to a standard temperature so devices can be compared, and noise figure is the same quantity in decibels. Absolute noise power depends on bandwidth as well and is a different measure. Gain and saturation are separate specifications. And measurement bandwidth is a condition rather than the figure. Every noise specification names either a degradation or a power, and the two must not be confused.