Question

Determine the decibel values of thermal noise power (N) and thermal noise power density (No) given the following: T=292 degrees Kelvin, B = 40MHz. (note: brackets "[ ]" represent decibel values)
a. [N] = +128 dBW, [No]= +204dBW

b. [N] = -128 dBW, [No] = -204 dBW

c. [N] = -128 dBm, [No]= -204 dBm

d. none of the above are correct

Answer 1

Answer: d. None of the above are correct.

Step-by-step explanation: Noise is a superfluous random alteration in an eletrical signal. There are different types of noises created by different devices and process. Thermal noise is one of them. It is unavoidable because is created by the agitation of the charge carriers, due to temperature, inside an eletrical conductor at equilibrium and is present in all eletrical circuits.

The formula to find the thermal noise power (N) is: N = $k_{b}$.T.B, where:

$k_{b}$ is Boltzmann constant (1.38.$10^{-23}$J/K);

T is temperature in Kelvin;

B is the bandwith;

Calculating the thermal noise power:

N = 1.38.$10^{-23}$·292·40

N = 16118.4.$10^{-23}$ dBm

The thermal noise power [N] = 16118.4.$10^{-23}$ dBm

Noise power density or simply Noise density (N₀) is the noise power per unit of bandwith and its SI is watts per hertz.

For thermal noise, N₀ = kT, where

k is the Boltzmann constant in J/K;

T is the receiver system noise temperature in K;

N₀ = 1.38.$10^{-23}$ . 292

N₀ = 402.96.$10^{-23}$ W/Hz

The thermal noise power density [N₀] = 402.96.$10^{-23}$ W/Hz