Question

Thermal noise is independent of

• A. Bandwidth
• B. Centre frequency
• C. Temperature
• D. Boltzmann's constant

Hint:

Thermal noise power is \(P_n = kTB\) (available noise power from a resistor).
Thermal noise voltage (rms) is \(v_n = \sqrt{4kTRB}\).
Thermal noise has a flat power spectral density over a very wide range of frequencies (white noise), meaning its power per unit bandwidth is constant and does not depend on the specific frequency or centre frequency.

Answer 1

The Correct Option is B

Thermal noise (also known as Johnson-Nyquist noise) is electronic noise generated by the thermal agitation of charge carriers (usually electrons) inside an electrical conductor at equilibrium, which happens regardless of any applied voltage.

The formula for the mean square thermal noise voltage across a resistor \(R\) over a bandwidth \(B\) at temperature \(T\) is:

\(\overline{v_n^2} = 4kTRB\) 

where:

• \(k\) is Boltzmann's constant (\(1.38 \times 10^{-23}\) J/K)
• \(T\) is the absolute temperature in Kelvin (K)
• \(R\) is the resistance in Ohms (\(\Omega\))
• \(B\) is the bandwidth in Hertz (Hz) over which the noise is measured.

The noise power spectral density is:

\(S_n(f) = 2kTR\) (for one-sided PSD) or \(S_n(f) = kTR\) (for two-sided PSD, over positive and negative frequencies). This power spectral density is flat, meaning it is independent of frequency ("white noise") up to very high frequencies (terahertz range at room temperature).

From the formula \(\overline{v_n^2} = 4kTRB\), we can see that thermal noise depends on:

• Bandwidth (B) - Option (a)
• Temperature (T) - Option (c)
• Boltzmann's constant (k) - Option (d) (It's a fundamental constant, but noise power is proportional to it)
• Resistance (R) (not listed as an option for independence)

Since the power spectral density of thermal noise is flat ("white"), it does not depend on the specific centre frequency of the bandwidth \(B\), as long as \(B\) is within the range where the noise is white. The total noise power depends on the width of the bandwidth \(B\), not where that bandwidth is centered (e.g., a 1 kHz bandwidth centered at 1 MHz will have the same thermal noise power as a 1 kHz bandwidth centered at 10 MHz, assuming \(R\) and \(T\) are the same).

Therefore, thermal noise is independent of the centre frequency.

Final Answer:

Centre frequency