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Signal Processing

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  • $\begingroup$ Thanks for your reply. I think I understand it. I added an update to the question to clarify that the FFT I'm referring to is abs(X)/N. So when I talk about normalizing that to spectral density, I'm dividing by sqrt(Fs/N), which yields the square root of your result above. So it seems that we've arrived at the same result (yours is for power, mine for amplitude), which still leaves me puzzled about what they've done in the Ultrasensitive Inverse Weak-Value Tilt Meter paper. $\endgroup$ Commented Sep 16, 2020 at 18:15
  • $\begingroup$ @user3308243 ok that makes more sense and in which case the units would be rad/root-Hz following my explanation. I will look at the paper closer, to your question. $\endgroup$ Commented Sep 16, 2020 at 19:17
  • $\begingroup$ @user3308243 So I am confused where your question is. They say the sampling rate is 1KHz, and they say the peak of the PSD is is 1.6 nrad / sqrt(1kHz) which is what we would expect as proper units for the square root of the PSD as I explained. Not sure why you mention it should be 1.6 nrad? A density would be per Hz (power is per Hz, square root of power is per root-Hz.). Maybe I didn't read enough of the paper, can you clarify? $\endgroup$ Commented Sep 16, 2020 at 19:23
  • $\begingroup$ Ultimately the plot in Figure 4 looks right to me in that the horizontal axis is in Hz and therefore the root-PSD would be rad/root-Hz. $\endgroup$ Commented Sep 16, 2020 at 19:25
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    $\begingroup$ Ah, yes. Thank you very much. I think I'm starting to see where my confusion is coming from. You are being very clear, and helpful, and the amount of information you've provided above is invaluable. $\endgroup$ Commented Sep 21, 2020 at 3:16