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

The input thermal noise voltage is

$$V=\sqrt{4kTBR}=\\=\sqrt{4(1.38·10^{-23})(29+273)3330·42}=\\\space\\=4.83·10^{-8}\text{ V, or 48.3 nV}.$$