4
$\begingroup$

[Thank you, @leftaroundabout for your helpful comments to my related question. I'm trying to ask a less confusing question here.]

$0\leq\phi\leq 2\pi$ is a parameter that theoretically describes a system. $\alpha=\sin(\phi)$ and $\alpha' = \cos(\phi)$ are observables.

In a measurement, $\phi(t)$ is a continuous periodic function of the time $t$. $\alpha$ can be directly measured and $A_i$ is a noisy over-sampled digital representation of such a measurement. $\beta$ is a different observable, the relation between $\beta$ and $\phi$ shall be studied. $B_i$ is a sampled synchronously with $A_i$ and noisy.

Three questions are interesting and the answer shall be found from $A_i$ and $B_i$:

  1. What is $\delta = \beta(\phi=-\pi/2) - \beta(\phi=+\pi/2)$? This the difference of the $\beta$ values corresponding to the extrema of $\alpha$.

  2. Is $\beta = a * \alpha + c$ a valid model and what is the coefficient $a$?

  3. Can we give a confidence interval for $a$ or $\delta$?

Inspection of the data shows that a better model for $B_i$ is $\beta' = a * \alpha + a' * \alpha' + c$. Does this affect the recipe to find $\delta$ or $a$?

A third alternative model is $\beta'' = f(\phi) + c$. How to reliably find $\delta$?

The nature of the noise is unknown. The noise spectrum has both frequencies less then the modulation frequency of $\phi$ and higher frequencies. A Fourier filter (e.g. lowpass) renders $A_i$ and $B_i$ reasonably smooth.

$\endgroup$
  • $\begingroup$ @yoda I have updated the original questions such that both questions are different. This questions works with a specific theoretical model, which the original question does not. I'm interested in both answers. $\endgroup$ – Jan Sep 18 '11 at 20:08
  • $\begingroup$ Is the period of $\phi(t)$ known accurately? $\endgroup$ – Jason R Sep 19 '11 at 13:42
  • $\begingroup$ Also, are the characteristics of the noise known? For example, is it white or colored? Is it zero-mean? Is it stationary? $\endgroup$ – Jason R Sep 20 '11 at 0:52
  • $\begingroup$ Do you need to do this in a real-time (i.e. streaming data) fashion, or are you doing offline analysis of an already-acquired dataset? $\endgroup$ – Jason R Sep 20 '11 at 1:02
2
$\begingroup$

All right, we can cope with this one.

So first, we should like to have some knowledge about the relation of $i$ and $\phi$, $$ \phi\colon\ \mathbb{Z} \to S^1,\qquad i\ \mapsto\ \phi_0+\frac1{\nu_{\!s}\cdot\omega_0}i =: \varphi + \lambda i\quad {}_{^{(\operatorname{mod}2\pi)}} $$ Where we can essentially forget about the $\operatorname{mod}2\pi$, just view it as $S^1=\mathbb{R}/\!{\sim}$.
There are multiple ways to determine the most likely values of $\varphi$ and $\lambda$; what I'd do is a least-square fit in those variables and $A_M$, perhaps also an offset $A_O$ such that $$ \zeta := \sum_i \bigl(A_{\!M}\cdot\sin\,\phi(i) + A_O - A_i\bigr)^2 $$ is minimized. A fitting algorithm like this can be somewhat slowish if you want it to work reliable, I don't know if that's a problem.

Once we have this relation $\phi(i)$, we can just read out the $B_i$ for which $i$ is close to $\tfrac\pi2$ resp. close to $-\tfrac\pi2$. We can simply average over all next-neighbours to these real values, this way we're quite immune from nonlinear response, but rather more subject to noise than with using some kind of interpolation. You want to do it this way if you have a very large amount of data, spanning many periods of $\phi$. If you don't, better use some interpolation between the $B_i$ with $\phi(i)\approx\pm\tfrac\pi2$. I'd try a Friedrichs function as convolution kernel.

$\endgroup$
  • $\begingroup$ This is a very good answer and a method that will certainly work. (The relation between $\phi$ and the index $i$ is not known to be of the form you assume, but I can adopt.) Furthermore, it feels inefficient to discard most of $B$, but I realize that's beyond the scope of this question. Answer accepted. Thank you. $\endgroup$ – Jan Sep 20 '11 at 20:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.