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I want to use waverec to evaluate a linear combination of the scaled and shifted wavelet of the form

$$ \sum_{i=1}^n\sum_{j=0}^{2^n-1}d_{nj} 2^{-n}\psi_{ij} (\frac{t-j}{2^n}) +c_{00}\phi(t) $$ at $t=k2^{-n} $ for $k=0\dots2^{n}-1$. I know how do it for the Haar wavelet, namely

[c,l]=wavedec(zeros(1,8),3,'haar')
waverec(2^(3/2)*[c00 d10 d20 d21 d30 d31 d32 d33],l,'haar')

But how can you do the same thing for Daubechies wavelets? The problem being that the array c does not have 8 entries but 28:

[c,l]=wavedec(zeros(1,8),3,'db4')

c =

  Columns 1 through 14

     0     0     0     0     0     0     0     0     0     0     0     0     0     0


  Columns 15 through 28

     0     0     0     0     0     0     0     0     0     0     0     0     0     0


l =

     7     7     7     7     8

The Question is related to Number of Daubechies coefficients, but I do not understand the answer.

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  • $\begingroup$ Where is the $i$ in the equation after the sigmas? $\endgroup$ – Spacey Nov 27 '12 at 0:52
  • $\begingroup$ Sorry, just edited the post. In the index of the basis function. Thank you $\endgroup$ – warsaga Nov 28 '12 at 11:35
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The wavdec function does not produce a DWT which is non-redundant. Instead, wavdec takes you N levels down the decomposition process. In other words, you shouldn't expect to get 8 coefficients back.

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