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I am numerically computing the growth of an oscillatory instability in a fluid system. Suppose for simplicity that the function $f(x)$ [defined on a finite interval] has oscillations of different modes and different amplitudes, but is otherwise centered about $f = 0$.

As a measure of the growth of the oscillations, I have been numerically computing maxes and mins, and then averaging the heights. So for instance,

$$ \text{Average oscillations} = \frac{1}{N} \sum_{j=1}^N f(x^\text{max}_i), $$

where $x^\text{max}_i$ are the locations of the $N$ maxima within the domain.

  1. Is there an easier way to measure this notion of average oscillatory height? Are there other norms that might work to give us a 'feel' for the value?

  2. What if $f(x)$ is not centered about zero? Suppose that $f(x) = x + \cos(x)$. I'd like to say that the average heights of the oscillations are one---assuming the oscillations are closely spaced, then I can just compute the distance between max and min values, but again, this requires me to code an automatic extrema-finding routine. Is there an easy norm to apply for this case as well?

I am hoping that there might be some commonly used norms (e.g. in fields like signal processing).

Edit: I've decided to accept the answer which hinted at the use of the RMS. In practice, my question was decidedly vague (in particular, I probably did not mean to say, 'norm'. But the answer was enough for me to go forwards.

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    $\begingroup$ I have an answer, but before I do I must ask: do you know how to find the spectrum of the signal? $\endgroup$
    – rurouniwallace
    Aug 18, 2013 at 18:45
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    $\begingroup$ If you want a norm that doesn't require computing maxes and mins, why not $$\frac1{b-a} \int_a^b f^2(x)\,dx?$$ This is a good expression for the "power" carried by a given oscillation $\endgroup$
    – Ben Grossmann
    Aug 18, 2013 at 18:47
  • $\begingroup$ I don't understand item 2. Are you looking for the frequency components in your data? As for item 1, you can measure the power over a sliding window (i.e., convolution) and then compute the amplitude of the sinusoid that would give you that same power over that same period of time. $\endgroup$
    – AnonSubmitter85
    Aug 19, 2013 at 2:10
  • $\begingroup$ And when you say norm, do mean a true norm where $\Vert x \Vert = 0 \Leftrightarrow x=0$, $\Vert x + y\Vert \leq \Vert x \Vert + \Vert y \Vert$, and $\Vert \lambda x \Vert = \lambda \Vert x \Vert$? Or are you using it in a looser sense to means some sort of measurement or estimate? $\endgroup$
    – AnonSubmitter85
    Aug 19, 2013 at 6:55

1 Answer 1

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Generally the root mean square will be used:

$$f_{rms}=\sqrt{\frac{1}{T}\int_0^Tf^2(t)\,dt}$$

However, you can skip the step of calculating the square root, since without the square root the quantity is still a norm.

This still work if the signal is not centered around zero.

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  • $\begingroup$ @Ataraxis: One question---you mention this is applicable for functions not centered around zero. But this yields vastly different answers between say, f = x + sin(x) and f = sin(x). I would say that any measure of the oscillations should yield similar values for these two functions, no? $\endgroup$
    – TSGM
    Aug 18, 2013 at 20:40
  • $\begingroup$ @TSGM Hmm...I'm not entirely sure how you would handle that. If it's a constant, you can just take the derivative of the signal, but it's not as simple if the offset isn't constant...I'll open up a question about it. $\endgroup$
    – rurouniwallace
    Aug 18, 2013 at 21:36
  • $\begingroup$ $f_{rms}(u) = \left( {{1}\over{T}} \int_0^T f^2(t-u) dt \right)^{1/2}$ $\endgroup$
    – AnonSubmitter85
    Aug 19, 2013 at 2:03

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