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This question is related to a series of questions I have asked about the units of PSD and ESDs. I include it as a separate question as it may have worth in isolation.

As I understand it to compute the power spectrum density from some transient signal one takes the approach of computing the truncated Fourier transform $$\mathcal{F}_T(\omega) = \int_{-T/2}^{+T/2} e^{-i \omega t} f(t) \ \ dt$$ Taking the absolute value square and then taking the limit as $T\rightarrow \infty$ $$\text{PSD} = S_{x}(\omega) = \lim_{T\rightarrow\infty}\frac{1}{T}|\mathcal{F}_T(\omega)|^{2}$$ When ignoring the normal dimensioning of units as described here by a very nice answer by Matt L., and also descibed here and in the Wikipedia article here, we get the familiar units of a PSD of $\rm{[V^{2}/Hz]}$.

My question

So that is the background and I believe I follow it (please say if there is a mistake or something overtly wrong!)

But If I want to analytically derive a function for a PSD or $S_{x}(\omega)$, I always end up with zero as the $\lim_{T\rightarrow\infty}$ kills the expression because of the factor of $1/T$. Is this the correct approach to analytically deriving a function or lineshape for a PSD with correct units? The end goal is to use it to fit a function to spectra acquired with an FFT spectrum analyser.

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  • $\begingroup$ @robertbristow-johnson could you provide a correction? Is this the limits, normalisation, or both? $\endgroup$ – Q.P. Nov 10 '19 at 16:12
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    $\begingroup$ I edited the question. I think I fixed what RBJ correctly identified as inaccurate. $\endgroup$ – Matt L. Nov 10 '19 at 16:14
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Energy signals, i.e., signals with finite energy

$$\int_{-\infty}^{\infty}|f(t)|^2dt<\infty$$

have zero power and, consequently, a power spectrum that is equal to zero. They do have an energy density spectrum, which is the squared magnitude of their Fourier transform.

You will only get a non-zero power spectrum, according to the definition in your question, for power signals, which have a finite non-zero power defined by

$$\overline{f^2(t)}=\lim_{T\to\infty}\frac{1}{T}\int_{-T/2}^{T/2}|f(t)|^2dt$$

I.e., power signals satisfy

$$0<\overline{f^2(t)}<\infty$$

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  • $\begingroup$ Thanks @Matt L. so you are saying that in the case where $T\rightarrow\infty$ one always ends up with a zero-valued spectrum? So what would you say is the best approach to produce a function for a say $60$ $\rm{s}$ averaged Fourier transform simply take $T\rightarrow 60$? I think I am still clearly not understanding something, could you perhaps expand your answer a little more, perhaps in line with what I am trying to achieve? $\endgroup$ – Q.P. Nov 10 '19 at 16:18
  • $\begingroup$ @Q.P.: No, it depends on the signal whether the limit $T\to\infty$ is zero or not. As I said, for energy signals (having finite energy) the limit is zero, but for power signals (e.g., a sinusoid, the step function, or any periodic signal, etc.), the limit is finite but non-zero. $\endgroup$ – Matt L. Nov 10 '19 at 16:21
  • $\begingroup$ okay so this isn't a clear cut case of following a recipe. In my case I am interested in a simple well known system where $f(t) = \exp(-t/\tau) \cos(\omega_0 t)$ when I calculate the Fourier transform and get the familiar Lorentzian in units of $V /Hz$ and then $V^{2} / Hz^{2}$ after taking asb-square, I still end up with 0 for the power spectrum density, even if I set my limits to my averaging time. I realise this may be somewhat painful, but an explicit example would be greatly appreciated. $\endgroup$ – Q.P. Nov 10 '19 at 16:26
  • $\begingroup$ @Q.P.: That signal is an energy signal, so its power spectrum is indeed zero. As soon as you're able to compute the Fourier transform (without any Dirac impulses or other funny stuff) then you have an energy signal with zero power (otherwise the Fourier transform wouldn't exist in the conventional sense). $\endgroup$ – Matt L. Nov 10 '19 at 16:32
  • $\begingroup$ Okay, so then how do we reconcile the analytic approach, if I know my system is essentially a damped oscillator where I measure a power spectrum density on a FFT spectrum analyser? This is my overarching motivation. As my data is a PSD I feel I need to produce a function $\sqrt{|F(\omega)|^{2} + e_{n}^{2}}$ where $e_{n}$ is the Johnson noise of my analysers input impedance and defines my signals noise floor? I need to find a way of defining $F(\omega)$ which is consistent with a spectral power signal -- I hope that goes some way of explaining my confussions. $\endgroup$ – Q.P. Nov 10 '19 at 16:38

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