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\title{{\LARGE \textbf{Wavelets vs. Exponentials}}}
\author{\large TOM SCHOUTEN}
\maketitle


\begin{abstract}

If the discrete Fourier transform and Z--transform are related to shift--invariant
operators, what can we say about the discrete wavelet transform?

\end{abstract}

\section{Introduction}

Relating polynomials to filters can be done using the Z--transform for infinite signals,
and the ring of polynomials modulo $z^N-1$ which is isomorphic to the ring of diagonal
matrices. Since the Z--transform is a limit case of the ring of polynomials for $N \to \infty$, 
we work with the latter.

The general matrix representation of this ring is $QCQ^{-1}$. Take $Q=F$ the Fourier transform.
The columns of $F$ are the normalized $W_k / \sqrt{N}$ with $W_k = (1, w_k, \ldots, w_k^{N-1})$
and $w_k = e^{jk/N}$, the $N$th roots of unity.

This gives the ring of circulant matrices, which can be related to the ring of polynomials modulo
$z^N-1$ by associating $a(z)$ with $A = a(Z)$, with $Z$ the circulant shift down operator

\begin{equation}
Z = \left[
\begin{array}{ccccc}
0 & 0 & \hdots & 0 & 1 \\
1 & 0 & \hdots & 0 & 0 \\
0 & 1 & \hdots & 0 & 0 \\
\vdots & \vdots &    & \vdots & \vdots \\
0 & 0 & \hdots & 1 & 0 \\
\end{array}
\right].
\end{equation}

The coefficients of the polynomial $a(z)$ are then the elements of the first column of $A$.
The diagonal elements of $C$ are in this case given by the DFT of this column, or
$c_k = a(w_{-k})$ corresponding to the eigenvector $W_k$. 
For $N \to \infty$ this gives infinite dimensional hankel matrices and the Z--transform.

\def\W{\widetilde{W}}

What happens if we replace the matrix $Q$ in the ring representation to a biorthogonal
wavelet basis? This is represented by the matrices $\W$ and $W$ satisfying

\begin{equation}
\W^TW = \W W^T = I.
\end{equation}

The isomorphism then becomes $A = W C \W^T$. 
To relate these matrices to the polynomials modulo $z^N-1$ we need to find a $Z = W C_z \W ^T$ such
that $Z^n = I$.

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