 Thermal noise worked example

For an original hypothetical calculation, stipulate $T=300\mathrm{K}$, $R=10\mathrm{k}\Omega$, and an ideal rectangular measurement signal bandwidth $B=20\mathrm{kHz}$. Using $k_{\mathrm{B}}=1.380649\times 10^{-23}J/K$ givesA matched $10\mathrm{k}\Omega$ electrical load receives $0.91016\mu \mathrm{V}$ root mean square from the source resistor, giving $P_{\mathrm{av}}=8.283894\times 10^{-17}\mathrm{W}$. These are stipulated model predictions, not laboratory measurements. Increasing $R$ raises the open-circuit voltage noise but does not increase the matched available power. Restricting the measurement signal bandwidth reduces root mean square noise as its square root; it improves signal-to-noise ratio only if the desired signal is preserved.