Question #249266
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.
1
Expert's answer
2021-10-10T16:05:56-0400

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

V2=4kTf1f2RdfV^2 = 4kT \int^{f2}_{f1} Rdf

where V = RMS input noise voltage

f2f1=Band  width=3.33  kHz=3.33×103  Hzk=Boltzmann  constant=1.38×1023  J/KT=Temperature=29+273=302  KR=Resistance=42  Ωf_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(f2f1))0.5V=(4×1.38×1023×302×42×3.33×103)0.5V=482.86×1010  VV = (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


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