Electronic – resolution bandwidth

noiseresolutionspectrum analyzer

I have the following trivial question:

I have acquired a noise spectrum with a resolution bandwidth of \$124\mathrm{Hz}\$.
I would like to convert it to \$\mathrm{V/\sqrt{Hz}}\$. Shall I simply multiply or divide the signal by \$\sqrt{124}\$??


Best Answer

Yes: however this requires a few words of explanation. Noise voltage has not the usual meaning since its exact value is obviously not exactly known. What can be measured and has a well defined and predictable behavior is the noise power adsorbed by a well defined load \$Z_L\$, in a frequency bandwidth \$B\$ centered around a frequency \$f_o\$, which has the following form $$ P(B,f_o)=\frac{\langle V_n^2(f_o)\rangle}{Z_L}=\frac{1}{Z_L}\int\limits_{f_o-B/2}^{f_o+B/2}S(f)\mathrm{d}f $$ i.e. it is the power spectral density \$S(f)\$ which gives the expected value. Since you have acquired a noise spectrum with a resolution bandwidth of \$B=124\mathrm{Hz}\$ the measured value you have acquired is precisely $$ \langle V_n^2(f_o)\rangle=\int\limits_{f_o-B/2}^{f_o+B/2}S(f)\mathrm{d}f\implies S(f)\simeq \frac{\langle V_n^2(f)\rangle}{B} $$ However, instead of \$S(f)\$ it is customary preferred to give the noise voltage density \$v_n(f)\$ $$ v_n(f)=\sqrt{S(f)}=\frac{\langle V_n^2(f)\rangle^\frac{1}{2}}{\sqrt{B}} $$ In sum, if your spectrum analyzer gives you already a power spectrum in which each spectral line is expressed as a voltage, dividing it by the square root of its resolution bandwidth gives you the sought for noise voltage density.