1
$\begingroup$

In some papers, the dilations and translations of a wavelet function is written as follows:

$$\psi_{j,k}=\frac {1}{\sqrt {2^j}} \psi\left({2^{-j}t-k}\right),\quad\text{where $j$ and $k$ are integers}$$

However, in some literature, the above formula is written in a different way:

$$\psi_{j,k}=\ {\sqrt {2^j}} \psi\left({2^{j}x-k}\right), \quad\text{where $j$ and $k$ are integers}$$

Are these two formula equivalent?

$\endgroup$

2 Answers 2

1
$\begingroup$

Yes, they are equivalent, since $j\in\mathbb{Z}$. Note that your first definition becomes equal to the second when $j$ is substituded by $-j$. It is just a convention, if you consider a positive $j$ to be a up- or downscaling.

So, when writing a paper or reading a book, it is always important to understand the definition of the transform used.

$\endgroup$
0
0
$\begingroup$

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 norm(in $L_2$ norm), $\frac{1}{\sqrt{a}}\psi(u/a)$ is of unit norm as well, and the translation $u\to u-k$ does not change it.

The choice of notation depends on the community: maths, signal processing, functional analysis.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.