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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

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  • \$\begingroup\$ The spectral shape determines everything, for which you have not defined \$\endgroup\$ – Sunnyskyguy EE75 May 21 at 0:16
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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.

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    \$\begingroup\$ Shouldn’t the lower end of the integral be $$f_0-B/2$$? \$\endgroup\$ – Jonas Schäfer May 20 at 16:28
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    \$\begingroup\$ @JonasSchäfer Yes, thank you! I'll immediately correct the typo. \$\endgroup\$ – Daniele Tampieri May 20 at 16:29
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If you had a noise value of N volts per \$\sqrt{Hz}\$ then that would be a noise voltage of \$N\cdot \sqrt{124}\$ volts in your 124 Hz bandwidth. Reversing the operation means dividing by \$\sqrt{124}\$.

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It's not a completely trivial question. Measurement of noise power is more difficult than measurement of tones.
Apparent noise power is affected by:

  • Detector (if any) type and its response
  • Non-linearities in the system (log amplifier in a spectrum analyser)
  • Bandwidth of the filter used (noise equivalent bandwidth is not the same as -3 dB bandwidth)

For example - here's a diagram of one channel, and an adjacent one (dashed). The scalloping loss may not be equivalent to the adjacent channel leakage.

enter image description here

So when converting between "noise seen in a certain 124 Hz channel", and "noise power density in \$\text{dBm}/{\text{Hz}}\$ or \$V/{\sqrt{Hz}}\$ you might need to correct for more than just the bandwidth. This is the case whether this is an FFT channel or a spectrum analyser setting.

If you're using a spectrum analyser, it will have a mode called "marker noise" which compensates as best as possible for these effects, to give you the effective noise power per Hz in a signal-free channel. If you're doing it yourself, I recommend reading Agilent application note 1303 Spectrum Analyzer Measurements and Noise: Measuring Noise and Noise-like Digital Communications Signals with a Spectrum Analyzer

This application note has some great background to noise measurement in different bandwidths. It's aimed at spectrum analysers but many of the principles are useful to other measurement systems.

Found all over the web, but not on the official site as far as I can see. Here is one copy.

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