# 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}\$$??

Regards

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.