Noise Figure of an Attenuator

Consider an attenuator as shown in Figure 1, whose conversion loss is L_{dB}.
Assume that input and output ports are perfectly matched and the attenuator is in thermal equilibrium with its surroundings.

Figure 1. Noise figure of attenuator

Figure 1. Noise figure of attenuator

Noise power at the input of attenuator due to a perfectly matched source at temperature(T_o) is given by

    \[N_i = kT_oB\]

k : Boltzman constant, 1.38 \times 10^{-23} J/K
T : Absolute temperature
B : Bandwidth of the circuit

The input noise power spectral density is

    \[{N_i \over B} = kT_o\]

When a noisy signal pass through the attenuator, noise power gets attenuated in the same way as the signal power. So power spectral density of noise at output is

    \[{N_o \over B} = {N_i \over LB} = {kT_o \over LB}\]

where L= 10^{L_{dB}/10}

Only part of the noise come out, and the rest is dissipated in the attenuator. Under the assumption that the attenuator is in thermal equilibrium, no heat flows out on the attenuator due to temperature gradient.

The heat dissipated in the attenuator is equivalent to excess noise power(N_x) in the attenuator.

(1)   \begin{eqnarray*} {N_x \over B} &=& {N_i \over B} - {N_o \over B} \\ &=& {N_i \over B}\left( 1 - {1\over L}\right) = kT_o \left(1 - {1\over L}\right) \\ \end{eqnarray*}

Noise factor of attenuator,

(2)   \begin{eqnarray*} F &=& 1 + {N_x \over G N_i} = 1 + {N_x \over G kT_o B} \\ &=& 1 + {1 \over G kT_o} kT_o \left(1 - {1\over L}\right) \\ &=& 1 + L \left(1 - {1\over L}\right) = L \end{eqnarray*}

Noise figure of attenuator,

(3)   \begin{equation*} \boxed{NF = 10 \log_{10} F = 10 \log_{10} L} \end{equation*}

Therefore noise figure of attenuator is same as conversion loss.

Leave a Comment

This site uses Akismet to reduce spam. Learn how your comment data is processed.