# Tag Info

8

Yes and no. At the theoretical level, both cases can use similar techniques and frameworks (an excellent example being Gaussian process regression). The critical difference is the assumptions used to prevent overfitting (regularization): In the functional case, there is usually some assumption of smoothness, in other words, values occurring close to each ...

7

Wavelets are useful to detect singularities in a signal (see for example the paper here (see figure 3 for an illustration) and the references mentioned in this paper. I guess singularities can sometimes be an anomaly? The idea here is that the Continuous wavelet transform (CWT) has maxima lines that propagates along frequencies, i.e. the longer the line ...

7

Functional Data often involves different question. I've been reading Functional Data Analysis, Ramsey and Silverman, and they spend a lot of times discussing curve registration, warping functions, and estimating derivatives of curves. These tend to be very different questions than those asked by people interested in studying high-dimensional data.

6

The list in the presentation that you reference seems fairly arbitrary to me, and the technique that would be used will really depend on the specific problem. You will note however that it also includes Kalman filters, so I suspect that the intended usage is as a filtering technique. Wavelet transforms generally fall under the subject of signal processing, ...

4

Most commonly used and implemented discrete wavelet basis functions (as distinct from the CWT described in Robin's answer) have two nice properties that make them useful for anomaly detection: They're compactly supported. They act as band-pass filters with the pass-band determined by their support. What this means in practical terms is that your discrete ...

4

I think this is a good question and I don't kown much about implementations. Since wavelet is 'mutli-resolution' you have two types of solutions (which are somehow connected): Modify your signal for example extend you signal over the actual boundary to have meaningfull coefficients. Exemples of that are : periodic wavelet on the interval Zero padding ...

2

Two options: One way to get the amplitude at an arbitrary frequency (say 5.3 Hz) would be to resample the signal at a sampling rate such that the base frequency calculated by the wavelet transform would be 5.3 Hz (instead of 1.0 Hz). A more appropriate way for a frequency range (say the 8-13 Hz alpha rhythm) is to discard the wavelet transform, filter the ...

2

As you probably already know, " Detecting the period of a signal with a length of only a few periods " is difficult. Off the top of my head, there are several approaches you could use: Use some sort of short-time Fourier transform, such as the Gabor transform. I hear rumors that library for doing Gabor transforms in the R language exists.b If I were doing ...

1

I don't know enough about the work to give a complete answer, but the number in detail[2,] is one of the numbers denoted $a_{p,q}$ or $b_{q+1,r}$ on page 17 of the paper. I am not sure which, but you should be able to find out by working through an example. To get this, you can look at the reconstr.bu function, which reconstructs the original signal from ...

1

I don't think that you can compare using wavelets to the 1st example. In the original thesis, it (domainogram) is described as a type of heat map with color levels proportional to probabilities of local enrichment levels of a chromatin component related to the width of neighbours sampled. A wavelet spectrogram does not measure probabilities so much as break ...

1

The first step would be a wavelet transform. See (e.g.) the wavelet package. Once you have your wavelet coefficients, you will need to decide which and how to thresh to achieve compression. Unlike, fourier transforms, using a low-pass filter, will not guarantee any optimality. You could use some arbitrary threshing on the coefficients by trial and error. ...

1

In each level, frequency range will be divided by two, I mean, in node $()$ fs is 0-1Khz, in node $(0)$ fs is 0Hz-500Hz(1KHz/2) and in node $(1)$ fs is 500Hz-1000Hz. in node $(0,0)$ fs is 0-250 and in node $(0,1)$ fs is 250-500; in node $(1,0)$ fs is 500-750 and in node $(1,1)$ fs is 750-1000, and so on. You can follow this rule to reach to proper level ...

1

The correlation coefficient of two sets of values is one number. The auto-correlation of one set of values is a function (see e.g. http://en.wikipedia.org/wiki/Autocorrelation ). Let's call the argument of the function t (looks like it's the $\tau$ in your question), then the value of the auto-correlation function at t is the correlation coefficient of the ...

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