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The CWT is not, per se, superior to STFT. Due to its variations in scale, it can be better, for instance, when: you address natural signals where transients are shorter than more stationary parts, you do not know the appropriate scale of observation. which is neither the case from your signal: precise frequency on the same support. On the one hand, you ...


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[Beginning of the story] Remember the discrete wavelet curse: in 1D, with 2-scale or dyadic wavelets, you cannot have finite support, realness, orthogonality and linear phase (symmetry/antisymmetry) at the same time, except for the Haar wavelet, which lacks of regularity and overlap. You have to lift one constraint to have the other fulfilled. For instance : ...


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Depends on what kind of “sense” you want. A 1 cycle sine wavelet might give you good time resolution for locating a mostly positive to negative zero-crossing waveform slice, but almost no frequency resolution, as the nearest orthogonal sinewave is a whole octave higher. Add more cycles and you get more frequency resolution, but less time resolution. A ...


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With all due caution, no in both cases (title and body question). I'll start with the second one. Continuous wavelets use all dilations of the mother wavelet, which are not accessed with the STFT The STFT is complex in general, and the windowed sine is not. For the first one: I never tried it, and do not remember having seen it in use, and one should ...


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No, since the STFT gives you complex values by not only correlating to sines of different periods, but also cosines.


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Not really. Unless you fully understand the math behind the DFT, you are likely to run into problems like circular vs linear convolution, time domain aliasing, excessive time domain ringing, etc. For "light weight" frequency selectivity application, time domain filter is typically a lot easier.


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They are equivalent, as long as $j_1$ (first formula) and $-j_2$ (second formula) span the same integer set. They are, if $j_1\in\mathbb{Z}$ and $j_2\in\mathbb{Z}$, or if $j_1\in\mathbb{N}^+$ (positive integers) and $j_2\in\mathbb{N}^-$ (negative integers). They are, under the change of variable: $t\to 2^{2j}x$. Let us remind that if $\psi(u)$ is of unit ...


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They’re equivalent, but the distinction seems to usually come from which definition of multiresolution analysis one uses - ascending or descending chains of detail/approximation spaces.


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I don't think there is any single best number for normalisation because it depends on the structure of the values in your lattice. In the simplest case where all values are equal, the predict operation zeros the black points and the update does not change the white points. Because each predict-update pair halves the number of nonzero points, multiplying ...


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