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From the definition, we will use wavelet for expanding signals the same way Fourier series use sinusoids:

$$f(t) = \sum a_{l}ψ_l(t)$$ The definition then defines that if $ψ_l(t)$ form an orthonormal basis for the function space then $$a_l = \, < f(t), ψ_l(t) >$$ How is this second equation derived?
Here is the link for the definition: Introduction to Wavelets

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  • $\begingroup$ Hi William! In general, it's a good idea to include a link to the reference for definitions, as they can vary from textbook to textbook, paper to paper. $\endgroup$
    – Jdip
    Nov 2, 2022 at 21:55
  • $\begingroup$ @Jdip Thank you for your reminder. I added the link to the article with the definitions listed in the question. $\endgroup$
    – WilliamW
    Nov 2, 2022 at 23:23
  • $\begingroup$ The question's unclear, definitions like this aren't "derived", they "are" because we say so, like "Let x=1". What exactly are you looking for? $\endgroup$ Nov 3, 2022 at 3:33

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I'm not sure what you're missing. First, it appears that the author is using $\left \langle x(t), y(t) \right \rangle$ to mean the inner product of two functions over the interval: $\left \langle x(t), y(t) \right \rangle = \int x(t) y(t) dt$.

Second, they mention the basis functions being orthonormal, but they only define the "ortho" half. "Ortho-" is a prefix meaning "square" or "at right angles to". In vector spaces and (I think I'm using the term right) function spaces, a pair of things are orthogonal (literally at right angles to each other) if their inner product is equal to zero. They are orthonormal to one another if each one's inner product with itself is equal to 1 (i.e., it's length = 1).

So, if you have some set of functions $\psi_k(t)$ that are defined on an interval on the real number line, and each one is orthonormal to all the rest, and there's an infinite number of them* then you can represent any analytic function $f(t)$ as a weighted sum of the basis functions. This is the source of their equation $$f(t) = \sum_k a_k \psi_k(t). \tag 1$$

Note that if you choose $\psi_k(t)$ correctly you've just rediscovered the Fourier series -- any sort of transform that uses basis functions like this is a generalization of the Fourier series.

Now, all of the babbling so far has been to get you ready for this: if you start with $$\left \langle f(t), \psi_k(t) \right \rangle \tag 2$$ and substitute in (1), you get $$\left \langle \sum_k a_k \psi_k(t), \psi_k(t) \right \rangle. \tag 3$$

Keeping in mind that $\langle \rangle$ is shorthand for an integration, you can switch the order of the summation and the integration:

$$\left \langle f(t), \psi_k(t) \right \rangle = a_k \sum_j \left \langle \psi_k(t), \psi_j(t) \right \rangle. \tag 4$$

Long story short, every instance inside the sum in (4) where $j \ne k$ is equal to zero, and for $j = k$ it is equal to one. So you end up with $$\left \langle f(t), \psi_k(t) \right \rangle = a_k, \tag 4$$ and the page's assertion is proved**.

* Hopefully I have all these conditions right

** Engineer-proved. I'm sure I'm cutting mathematical corners here.

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