0.7 Generalizations of the basic multiresolution wavelet system  (Page 17/28)

Decomposition with this rather trivial operator gives a time-domain description in that the first expansion coefficient ${\alpha }_0$ is simply the first value of the signal, $x\left(0\right)$ , and the second coefficient is the second value of the signal. Using a different set of basis vectors mightgive the operator

which has the normalized basis vectors still orthogonal but now at a 45 ^o angle from the basis vectors in [link] . This decomposition is a sort of frequency domain expansion. The first column vector will simply be theconstant signal, and its expansion coefficient $\alpha \left(0\right)$ will be the average of the signal. The coefficient of the second vector willcalculate the difference in $y\left(0\right)$ and $y\left(1\right)$ and, therefore, be a measure of the change.

Notice that $\mathbf{y}=\left[\begin{array}{c}\hfill 1\\ \hfill 0\end{array}\right]$ can be represented exactly with only one nonzero coefficient using [link] but will require two with [link] , while for $\mathbf{y}=\left[\begin{array}{c}\hfill 1\\ \hfill 1\end{array}\right]$ the opposite is true. This means the signals $\mathbf{y}=\left[\begin{array}{c}\hfill 1\\ \hfill 0\end{array}\right]$ and $\mathbf{y}=\left[\begin{array}{c}\hfill 0\\ \hfill 1\end{array}\right]$ can be represented sparsely by [link] while $\mathbf{y}=\left[\begin{array}{c}\hfill 1\\ \hfill 1\end{array}\right]$ and $\mathbf{y}=\left[\begin{array}{c}\hfill 1\\ \hfill -1\end{array}\right]$ can be represented sparsely by [link] .

If we create an overcomplete expansion by a linear combination of the previous orthogonal basis systems, then it should be possible to have a sparserepresentation for all four of the previous signals. This is done by simply adding the columns of [link] to those of [link] to give

This is clearly overcomplete, having four expansion vectors in a two-dimensional system. Finding ${\alpha }_k$ requires solving a set of underdetermined equations, and the solution is not unique.

For example, if the signal is given by

there are an infinity of solutions, several of which are listed in the following table.

Case 1 is the minimum norm solution of $\mathbf{y}=\mathbf{X}\alpha$ for ${\alpha }_k$ . It is calculated by a pseudo inverse with the Matlab command a = pinv(X)*y . It is also the redundant DWT discussed in the next section and calculated by a = X'*y/2 . Case 2 is the minimum norm solution, but for no more than two nonzero values of ${\alpha }_k$ . Case 2 can also be calculated by inverting the matrix [link] with columns 3 and 4 deleted. Case 3 is calculated the same way with columns 2 and 4deleted, case 4 has columns 2 and 3 deleted, case 5 has 1 and 4 deleted, case 6 has 1 and 3 deleted, and case 7 has 1 and 2 deleted. Cases 3through 7 are unique since the reduced matrix is square and nonsingular. The second term of $\alpha$ for case 1 is zero because the signal is orthogonal to that expansion vector. Notice that the norm of $\alpha$ is minimum for case 1 and is equal to the norm of $\mathbf{y}$ divided by the redundancy, here two. Also notice that the coefficients in cases 2, 3, and4 are the same even though calculated by different methods.

Because $\mathbf{X}$ is not only a frame, but a tight frame with a redundancy of two, the energy (norm squared) of $\alpha$ is one-half the norm squared of $\mathbf{y}$ . The other decompositions (not tight frame or basis) do not preserve the energy.

Next consider a two-dimensional signal that cannot be exactly represented by only one expansion vector. If the unity norm signal is given by