Answer to Question #249266 in Electric Circuits for TANTAY

Question

Find the input thermal noise voltage of a receiver with a bandwidth of 3.33 kHz, with an input resistance of 42 Ω, and a temperature of 29oC.

Answer 1

Noise voltage are unwanted and disturbing voltage in an electronic system. It is given by:

$V^2 = 4kT \int^{f2}_{f1} Rdf$

where V = RMS input noise voltage

$f_2-f_1 = Band \; width = 3.33 \;kHz = 3.33 \times 10^3 \;Hz \\ k = Boltzmann \;constant = 1.38 \times 10^{-23} \;J/K \\ T = Temperature = 29+273 = 302 \;K R = Resistance = 42 \; \Omega$

Putting all these values we get

$V = (4kTR(f_2 -f_1))^{0.5} \\ V = (4 \times 1.38 \times 10^{-23} \times 302 \times 42 \times 3.33 \times 10^3)^{0.5} \\ V = 482.86 \times 10^{-10} \;V$