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If the dimension of the independent variable $x$ of $f(x)$ is length, i.e. $[x]=L$, then what is the dimension of Daubechies 4 (hereafter D4) wavelet and scaling function?

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Generally yes, discrete wavelets are often implemented as dimensionless filters, like discrete averagers or differences.

Additionally, wavelets are non dependent on the variable. If this is pertinent to your application, it is possible to equip the wavelet with dimensions: morally, the scaling coefficients ought to be of the same dimension as the variable, yet the detail coefficients can be thought as homogeneous to derivatives (at level 1), second derivatives (at level 2) etc. Consequently, with the proper normalization, they could be expressed in $\textrm{m}/\textrm{s}$, $\textrm{m}/\textrm{s}^2$, etc. according to each level.

If I detail the computation for the analysis/synthesis, decomposed onto the primal wavelet space ($\phi$ and $\psi$) and reconstructed in the dual wavelet space ($\tilde{\phi}$ and $\tilde{\psi}$):

$$f(x)=\sum <f,\phi_{j,k}>\tilde{\phi}_{j,k}(x) + \sum <f,\psi_{n,k}>\tilde{\psi}_{n,k}(x)\,.$$

The dimensional analysis entails that products or convolutions of $\phi$ and $\tilde{\phi}$, and of $\psi$ and $\tilde{\psi}$ should be unit-less. In other words, I would choose them with "inverse" dimensions, and this could be done per scale.

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    $\begingroup$ Thanks for your comment. We know that the signal $f(x)$ could be reconstructed from the inner product $\langle f,\phi_{M,n}\rangle$ and $\langle f,\psi_{k,n}\rangle$, i.e. $f(x) = \sum_{n\in \mathbf{Z}}\langle f,\phi_{M,n}\rangle \phi_{M,n} + \sum_{k=-\infty}^M\sum_{n\in\mathbf{Z}}\psi_{k,n}\langle f,\psi_{k,n}\rangle$. According to dimensional analysis and $\langle f,\phi_{M,n}\rangle = \int f(x)\phi^*_{M,n}(x)dx$, the dimension of $phi_{M,n}$ should be $1/[x]^{1/2}$. So is my thought corrected? $\endgroup$
    – Wang Yun
    Mar 9 at 1:01
  • $\begingroup$ I think you are mixing the "free variable" $x$ and the function or signal $f$ for the dimensional analysis. I have updated my answer $\endgroup$ Mar 10 at 9:01

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