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Using a signal of three joined sinus of equal length at different frequencies (40 Hz, 50 Hz, 80 Hz) I calculate a scalogram of it (scale vertical, time horizontal, The steps of variing the scale parameter are equidistant):

Plot of the Scalogram:

enter image description here

In contrast to my expection that lower frequencies (i.e. higher values for the scale) should be better resolved according to wavelet resolution theory, the first third (corresponding to 40 Hz) is smeared vertically much more than the second third (50 Hz) which in turn is again more smeared than the last third part (80 Hz) of the time duration.

Why?

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migrated from stackoverflow.com May 6 '13 at 17:40

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    $\begingroup$ What did you use to calculate the scalogram? $\endgroup$ – endolith May 7 '13 at 0:06
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The scalogram is probably constructed using frequencies that vary logarithmically. e.g. 1, 2, 4, 8 etc. If you plot the y-axis of your graph on a log-scale then I think they will have the same width. The log-scale makes sense when we consider that our ears perceive sound pitch in this manner. For example, each octave in a musical scale is a doubling of frequency.

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I think the point actually is that the resolition at higher scales become better. Hence the 40 Hz sinus is "visible" to the wavelets over a wider range of scales. The 80 Hz sinus in contrast is only detected at less scales since each scale covers a larger bandwidth.

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