# Tag Info

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On the one hand with the DWT, only a restricted choice of wavelets is available: those that implement 2-band perfect reconstruction (Daubechies, Symmlets, Coiflets, Spline). They are non-redundant, and often orthogonal or close to orthogonal, which simplifies some computations, inversion or statistical analysis, for instance. Yet, they are not quite shift-...

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The CWT & DWT implementations differ in how they discretize the scale parameter used to stretch or shrink copies of the basic wavelet. The finer grain scale parameter in the CWT can be useful for applications that require a very high-fidelity signal analysis, for example, where localization of transients or precise characterizations of signal ...

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A standard continuous wavelet transformation (the one that produce a 2D scale/shift map) is a linear operator. It produces real or complex coefficients that are related to the amplitude on "how a given wavelet at specific shift and scale matches the signal". These coefficients are (most generally) homogeneous with the signal's amplitude. This being ...

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It's rather translation equivariant: $$\text{CWT}_{s, t}x(t - t_0) = \text{CWT}_{s, t - t_0}x(t) \tag{1}$$ and $$\langle x(t−t_0),\psi(t) \rangle= \langle x(t), \psi(t+t_0)\rangle \tag{2}$$ That is, when a signal is shifted, its representation is also shifted but not modified (like LTI). This makes its derived features, such as energy and norm, or most ...

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Fundamentally: DWT is orthogonal, CWT is redundant. Former packs the most information per sample, latter spreads out its decomposition. As will be explained: CWT yields vastly superior analysis information. CWT decomposes a signal using a redundant "dictionary" of time-frequency atoms (borrowing analogy from S. Mallat). In language, a rich ...

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I can't comment because of reputation but if i'm not mistaken, the need for DWT to be supplied an input that is length power of 2 is independent of it's subsequent levels being downsampled by a power of two. The input does not have to be a power of two but you'll have to deal with problems at the edges of the signal then. Here is a good link for reference ...

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If the discrete wavelet transform can be implemented with a FIR filter bank, with appropriate extensions, yes, up to numerical precision, coefficients will be the same. If the discrete wavelet transform possesses a non finite support, then a FIR filter bank implementation would require filter truncation, and the results may differ. On those case, like for ...

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