# Symmetric discrete wavelets with rational coefficients (which aren't the Legall wavelet)?

The Legall wavelet is famous for example being in the lossless mode for the JPEG2000 image coding standard due to for example having rational filter taps for both highpass and lowpass filter, making perfect numeric precision possible throughout the whole filter bank. (And even without the use of floating point arithmetics if that is important).

One forward ("primal") functions: The backward ("dual") functions:

Do there exist other symmetric wavelets with the same property of rational coefficients (preferrably possible to write on common denominator of $2^k$ for some $k$)?

If we let the Haar lowpass mask be $R(\omega)=\frac{1}{2}(1+e^{-i\omega}),$ then $B_{k}(\omega)=\frac{1}{2^{k}}\sum_{j=0}^{k}\binom{k}{j}e^{-ij\omega}=R^{k}(\omega)$ is the $k$th order B-spline, and clearly has dyadic rational coefficients. It is easy to find the highpass mask by considering $B_{k}(\omega+\pi)$, but it will take a little work to find the dual lowpass filter.
To do this, we need to solve $$B_{k}(\omega)\overline{D(\omega)}+B_{k}(\omega+\pi)\overline{D(\omega+\pi)}\equiv 1,$$ where $D(\omega)$ is our dual lowpass filter, and $D(\omega+\pi)$ will be the dual highpass filter. If we write $D(\omega)=\sum_{j=-N}^{N}a_{j}e^{-ij\omega},$ the equation above will give several relations between the $a_{j},$ to which you should add the conditions for being lowpass ($\sum_{j}a_{j}=1$), and whatever accuracy number you want $D$ to have ($D^{(m)}(\pi)=0$ for $0\leq m\leq r,$ where $r$ is the desired accuracy number).
Typically, you would want as much accuracy as $B_{k}$ has, and the least possible support subject to these other constraints, and with these choices, the dual filter is unique.
• Right, if you want symmetry, you’ll have to choose an even value of $k$, and you should enforce this while finding the dual filters also. – RideTheWavelet Nov 20 '17 at 16:55