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Calculating time-dependent frequency of oscillations

+ 3 like - 0 dislike
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I have a curve (example shown below) which oscillates with a slowly-decreasing amplitude and offset and a slowly-changing frequency. What's a good way to estimate these values as functions of time? Any two of these values can be calculated reliably given the other, but I'm not sure where to start.

Here are some methods I've considered, but I'm not sure how effective they will be:

- estimating the maxima and calculating periods

- doing Fourier decomposition over a sliding window

- fitting a function like \(C(t) + \sum A(t) \sin(\omega t) + B(t)\cos(\omega t) \)

Thanks!

asked Mar 31, 2015 in Computational Physics by lightlike (15 points) [ no revision ]
reopened Mar 31, 2015 by dimension10

I have heared (without fully grasping the method) that wavelet theory can be of use to analyse situations where the amplitudes and frequencies (or wavenumbers) of a signal depend on time (or position). 

@dilaton: Wavelets give a multiresolution analysis appropriate, e.g., for the analysis of sound, music say. My first thought was wavelets, too. But it is not quite the right tool for what this particular question asks - an estimate of how a single (or two) amplitude and frequency change with time. 

1 Answer

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I would use nonlinear least squares to fit an expression $C(t)+A(t) \cos \omega(t)$ with low degree polynomials $C,A,\omega$, trying different degrees for the different pieces and using BIC to decide on the best degree combination. 

You may need to choose good starting values: Begin with $\omega(t)= a+bt$ with $a,b$ estimated from a Fourier transform. With $\omega(t)$ fixed you have a linear least squares problem for the coefficients of $C$ and $A$. Use the result with $\omega(t)= a+bt+0t^2+\ldots$ as a starting point for the nonlinear fit.

There seems to be a second high-frequency component in your data - inspect the residual to see whether this is noise or something significant. If the latter, add an additional term $+A_2(t) \cos \omega_2(t)$ and repeat. Here the starting point is chosen by working on the residual similarly as before.

answered Mar 31, 2015 by Arnold Neumaier (12,355 points) [ no revision ]

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