# Tag Info

5

Yes indeed! In theory as long as the wavelet is orthogonal, the sum of the squares of all the coefficients should be equal to the energy of the signal. In practice, one should be careful that: the decomposition is not "expansive", i.e. the number of samples and of coefficients is the same. wavelet filter coefficients are not re-scaled, as happens in some ...

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Assume the original sampling rate is $F_s$ and that the filters are perfect brickwall Halfband filters. After the low pass filtering the frequency content is from $0 \rightarrow \frac{F_s}{4}$. Because of the new bandwidth, the sampling rate can be reduced to $F_s/2$. For the high-pass filter, the frequency content lies in the range $\frac{F_s}{4} \... 3 The formalism of the continuous wavelet transform is relatively flexible. To make a practical tool out of it, you ought to discretize it, and here come the pain. It is quite easy to discretize it in a very redundant way (a discrete wavelet frame), but the price to pay when analyzing high-dimensional signals (2D images, 3D volumes) can quickly become too high:... 2 [Good question, that made me rethink of stuff I believed natural. I shall incorporate them in future lectures] Downsampling the highpass (and the lowpass) provide you with a critically sampled filterbank, in other words, not redundant. Critical sampling is a price to pay in some applications like compression, and in the context of finding an orthogonal ... 2 As illustrated by the simplified figure and the Matlab code below, such a graph can be obtained with a standard discrete wavelet decomposition, possibly with a periodic extension to preserve the size, over 3 to 5 levels of decomposition. Wavelet coefficients are simply concatenated in the standard low-frequency /high-frequency fashion, with scaling ... 2 Right now, I read the live ECG data into a 5 second buffer, and then perform the wavelet filter on the buffer (as described above). This seems rather inefficient, because I have to filter the entire buffer again when new data comes in. You cannot avoid the filtering / reconstruction steps over the buffer with this kind of processing. Is it possible to ... 2 They do mean the pseudo-frequency of the wavelet which is not dependent on the signal being analyzed. The misleading terminology that they use seems to come from from one of the references, Han, P. (2013), Investigation of ULF seismo-magnetic phenomena in Kanto, Japan during 2000–2010, PhD thesis, Chiba University, Chiba, Japan. Quoting an article of a ... 2 First, one should be cautious about processing short signals like these. I am unsure about the length of 3 for a3 and d3. Now, the many questions: I would not call them "frequency plot", but stacked subband plots. They are just illustrations of the behavior of the wavelet coefficients. For each subband, different scalings may be used. Another ... 1 With such short (known) stimuli, does it make sense to use generic analysis methods, or is it «better» (less work, results that are more usable) to tailor your processing to the transmitted processing (matched filter?) or apply a parametric model and some numerical optimization? 1 A lifting scheme is a method for splitting a sequence of discrete samples into downsampled subsequences, so that you can predict a subsequence from the other, and update the former later, using possibly nonlinear predict and update operators. This corresponds to a generalization of one level of discrete wavelet transform, usually obtained fully linearly by ... 1 This question has multiple facets (after comments), so I will focus on the principal. First, regarding coefficient localization: a discrete wavelet coefficient depends on several signal samples. The number of coefficients influenced by a single sample in a continuous wavelet representation typically depends on the properties of the mother wavelet and the ... 1 In one level DWT, each output of the low-pass or a high-pass can indeed be considered as signals. Thus each of those signals are subsampled by a factor of 2, and the same two-filter-subsampling is iterated on the low-pass output, several times (wavelet decomposition) at$L$levels. Each final output of the different branches could still individually be ... 1 The discrete wavelet transform should denote "the operations" that, applied to some data, yield a discrete wavelet decomposition. The first one can be seen as a matrix operator, while the second relates to the actual wavelet coefficients, or the structure of thereof, that you would obtain after the application of the first one. In everyday language, they ... 1 The classical discrete wavelet transform is critically decimated. In other words, it should preserve the quantity of "samples". In other words, apart from data border effects, the number of wavelet coefficients should be the same as the numer of data samples. More generally, a critically-sampled, analysis multi-band filter bank with$M$channels ($M=2$for ... 1 The whole idea is that every level splits the image information in two equal halves, which can be represented by half the number of coefficient as the previous level, each. Otherwise, you'd not be doing much of a decomposition, would you? So, since you only need half of the coefficients (the rest is redundant), you just keep half of the coefficients. That'... 1 Your time-frequency grid is Mondrian-shaped: essentially, rectangles supported on dyadic splits of the time or the frequency axes. Hence, you can easily start for any reversible time-frequency tiling, and further slit any of its rectangle onto another invertible tiling. Some call that hierarchical, or nested time-scale/time-frequency decompositions. As long ... 1 Treating boundaries correctly is non trivial for wavelets and filter banks in general. Symmetric/anti-symmetric filters (such as with biorthogonal wavelets as the 5/3 or the 9/7) help a little. Yet, with dyadic sub-sampling, without playing too heavily on polyphase matrices and symmetries, I would suggest the following procedure: choose a level$L$for the ... 1 No, by essence, as you guessed correctly. There is a direct counting argument: given a$2N\times 2N$image, the first level of the DWT generates three$N\times N$wavelet subbands. And through the pigeonhole principle, you cannot have a bijection between a$4N^2$and a$3N^2\$ coefficient set. As for the contourlet, you can resort to undecimated wavelet ...

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Standard 2-band discrete wavelet transforms have some subsampling, thus they tend to be much less precise in localization. Plus, their shapes are limited, and the ones with finite support are slightly dissymmetric (all but Haar), which is problematic for symmetric pulse/edge detection. If you don't care about computations, and are interested in ...

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