Noise in Analog Systems Questions and Answers

Answer: B

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

Thermal (Johnson) noise has a power spectral density of 4kTR that is flat over all frequencies used in electronics. A flat spectrum, like white light containing all colours, is why it is called white noise, so the assertion is true.

The reason is also true: passing white noise through a network shapes its output spectrum as \(S_{out}(f) = |H(f)|^{2} S_{in}(f)\), giving coloured noise. But that describes what happens after filtering, not why thermal noise is called white, so it does not explain the assertion.

Answer: B

Statements a, b and c are correct, so the answer is (a), (b) and (c).

Shot noise arises because current is carried by discrete charges that cross a potential barrier at random times rather than simultaneously (a); it is a random fluctuation that accompanies any direct current crossing a barrier, as in diodes and transistors (b). Johnson (thermal) noise arises from the random thermal motion of electrons, which have kinetic energy, in any conductor (c).

Statement d is false. The mean of thermal noise voltage is zero, but its root-mean-square value is finite and measurable:

\[ V_{rms} = \sqrt{4kTRB} \]

so the option 'd only' and any option omitting b or c are wrong.

Answer: A

The equivalent noise temperature is \(T_{0}(F - 1)\).

The noise figure \(F\) is defined with the source at the reference temperature \(T_{0}\) (290 K). The noise the network adds, referred to its input, can be expressed as an extra temperature \(T_{e}\), so

\[ \begin{aligned} F &= \frac{T_{0} + T_{e}}{T_{0}} = 1 + \frac{T_{e}}{T_{0}} \\ \Rightarrow T_{e} &= T_{0}(F - 1) \end{aligned} \]

A noiseless network has \(F = 1\) and \(T_{e} = 0\), as it should. \(T_{0}(F + 1)\) and \(T_{0}F\) would give a noiseless network a non-zero noise temperature, and \(T_{0}/F\) would make a noisier network appear cooler.