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I haven't been doing much signal processing over the years and I'm wondering if my logic is sound. I'd appreciate any input.

So, I have a data set with independent variable ($A$) in Frequency ($1\textrm{ kHz}$-$101\textrm{ kHz}$) and the dependent variable ($B$) with units $\textrm{ Pa}$ (Pascals). These are experimental data.

I currently have a graph of $B$ ($\textrm{ Pa}$) versus $A$($\textrm{ Hz}$). However, I'm required to represent my data in graph of $\frac{\textrm{ Pa}}{\sqrt{\textrm{Hz}}}$ versus $\textrm{ Hz}$. ($C$ being the new variable with units of $\frac{\textrm{ Pa}}{\sqrt{\textrm{Hz}}}$, Amplitude Spectral Density, or the square root of the Power Spectral Density)

So my question is:

  • Would it be grossly inaccurate to just average my dependent variable $A$ (with Units of $\textrm{ Pa}$) over the square-root bandwidth ($100\textrm{ kHz}$) of data $B$ to derive $C$?
  • Or should I simply (not so simple,but) ifft(A). Then use PSD = [fft(x).*conj(fft(x))], and take the square root of PSD?
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  • $\begingroup$ Frequency is not a unit. Maybe $\frac{\textrm{ Pa}}{\sqrt{\textrm{Hz}}}$ versus Frequency is just: $\frac{\textrm{ Pa}}{\sqrt{\textrm{Hz}}}$ versus $\textrm{Hz}$. $\endgroup$ – Gilles Mar 15 '16 at 0:41
  • $\begingroup$ Thank you for replying and edit. Much appreciated :). Yes, the independent variable A is a frequency term with units of Hz (or kHz). My question is how to derive the Amplitude Spectral Density from the given data? is it even possible? $\endgroup$ – user2531389 Mar 15 '16 at 7:54

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