# Refinement equation and orthogonal wavelets basis

I have a wavelet function space defined by,

$$\phi(x) = \sqrt(2) \sum_n h_{\phi}(n) \phi(2x-n)$$ .

given the Fourier transform of the function is:

$$\hat{\phi}(\omega) = \frac{1}{\sqrt(2)} \hat{h}_{\phi}(\omega) \cdot \hat{\phi}(\frac{\omega}{2})$$

why does the sum of the dc component in jumps of $$2\pi$$ equals to 1, i.e.:

$$\sum_{l=-\infty}^{\infty} |\hat{\phi}(\omega +2\pi l)|^2 = 1$$

or even its necessarily constant. Either I am missing something trivial, or whatever, but I can't seem to agree with it.

thanks.

This follows from the orthonormality of shifted versions of $$\phi(x)$$:

$$\langle \phi(x),\phi(x+n)\rangle =\delta[n]\tag{1}$$

where $$\delta[n]$$ is the unit impulse. Taking the discrete-time Fourier transform (DTFT) of $$(1)$$ gives the equation in your question.

Note that the left-hand side of $$(1)$$ is the auto-correlation of $$\phi(x)$$ evaluated at integers $$n$$. The Fourier transform of the auto-correlation is $$\left|\hat{\phi}(\omega)\right|^2$$, where $$\hat{\phi}(\omega)$$ is the Fourier transform of $$\phi(x)$$. Sampling the auto-correlation results in the infinite sum of shifted spectra.

EDIT: The equation can be derived as follows: define the auto-correlation of $$\phi(t)$$ as

$$p(\tau)=\int_{-\infty}^{\infty}\phi(t)\phi(t+\tau)dt\tag{2}$$

Realizing that $$p(\tau)=(\phi_{-} \star \phi)(\tau)$$ (where $$\phi_{-}(t)=\phi(-t)$$), the Fourier transform of $$p(\tau)$$ is

$$P(\omega)=\left|\hat{\phi}(\omega)\right|^2\tag{3}$$

Sampling $$p(\tau)$$ at integer $$n$$ results in a periodic continuation of the spectrum:

$$\textrm{DTFT}\{p(n)\}=\sum_{k=-\infty}^{\infty}P(\omega+2\pi k)\tag{4}$$

With $$p(n)=\langle \phi(x),\phi(x+n)\rangle$$ and with $$(3)$$ we finally get

$$\textrm{DTFT}\left\{\langle \phi(x),\phi(x+n)\rangle\right\}=\sum_{k=-\infty}^{\infty}\left|\hat{\phi}(\omega+2\pi k)\right|^2\tag{5}$$

• according to my math, – tnt1674 Jan 3 at 18:50
• according to my math, 1 should be equal to $\frac{1}{2\pi} \int_{0}^{2\pi} \sum_{l=-\infty}^{\infty} |\hat{\phi}(\omega +2\pi l)|^2 e^{j n \omega } d\omega$ in many of the papers, people pull the term $\sum_{l=-\infty}^{\infty} |\hat{\phi}(\omega +2\pi l)|^2$ out of the integral since its constant. I fail to precept why its known to be constant. The rest is clear to me. – tnt1674 Jan 3 at 19:04
• @tnt1674: I don't see where your math comes from, but I've added the derivation to my answer. – Matt L. Jan 3 at 19:36