Timeline for Which time-frequency coefficients does the Wavelet transform compute?
Current License: CC BY-SA 2.5
6 events
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| Nov 14, 2011 at 22:44 | history | migrated | from stackoverflow.com (revisions) | ||
| Feb 15, 2011 at 19:01 | comment | added | endolith | "STFT are both redundant analyses of a signal". I don't think that's true. If you have a 100-point signal, divide it up into chunks of 10 points, then do a 10-point FFT on each, you still have the same information stored in the same amount of samples. | |
| Dec 12, 2009 at 19:58 | comment | added | endolith | From what I understand, CWT has the same limitation, but uses a better trade-off. | |
| Dec 10, 2009 at 16:24 | comment | added | Patrick | The FWT is a critical sampling of the CWT. I'm still trying to understand it better, but I've learned that the STFT and CWT are both Frames. Frame theory is getting beyond me, but one interesting notion is the uncertainty formula, that for the STFT, dw * dt > C (dw is the frequency resolution, and dt is the time resolution). In other words, as you try to better resolve frequency, you lose time resolution. The CWT does not have this limitation. I will keep reading and try and clarify my answer above once I clarify it in my head. | |
| Nov 24, 2009 at 17:11 | comment | added | endolith | "it is actually the minimum sampling in shift/scale, and is not a redundant representation." Ah! I think you're right, and this would explain why it's always compared to the FFT, which is also a minimal representation. | |
| Nov 24, 2009 at 16:33 | history | answered | Patrick | CC BY-SA 2.5 |