Refinement equation and orthogonal wavelets basis

Question

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.

Answer 1

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

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