The continuous wavelet transform as a maximum entropy solution of the corresponding inverse problem

with T (z) = D 1 (z)D 2 (z 01)0 D 2 (z)D 1 (z 01): The sign of Q(z) is the same as that of Re fG(z)=D1(z)g on the unit circle; an identical consideration can be made for R(z) and D 2 (z): Therefore, G(z) can be computed by finding Q(z) = Q(z 01) and R(z) = R(z 01) positive on the unit circle and such that R(z)D 1 (z) 0 Q(z)D 2 (z) = 0 8z such that T (z) = 0; jzj 1: The roots of T (z) outside the unit circle can be readily included in R(z) and Q(z) (and, by symmetry, the roots inside the unit circle). However, given the positivity of R(e j!) and Q(e j!) for all !, the zeros of T (z) on the unit circle must be canceled out by solving the interpolation problem R(z) Q(z) = D 2 (z) D 1 (z) 8z such that T (z) = 0; jzj = 1: With D1(z) = (1 0 M z 01) N and D2(z) = (1 + M z 01) N , those roots are only two, namely, z = 1 and z = 01: The interpolation can be carried out by the algorithm presented in [8], and the degree of C(z) = G(z M) can be checked to be N M [10]. It can be proved that there does not exist C(z) such that C(z)=A 3 (z) is SPR for all A 3 (z) with their roots in : However, the region 2 = fz: p z 2 g, which characterizes the uncertainty in the roots of D(z) as in (10), for N = 3; M = 2 can be enclosed in a region that admits two extreme polynomials, namely, D 1 (z) = 1 and D2(z) = (100:81z 01) 3 , which satisfy (9). Such a region is the intersection of the circles centered at 0.5 and 00.095 and with radii 0.5 and 0.905, respectively. See [9] for more details on these types of regions. An appropriate compensator C(z) is obtained following the steps shown in Section IV: C(z) = 101:33z 02 +0:36z 04 : The true plant is given by H(z) = (1=A 3 (z)) with The input is zero-mean, unit variance white noise filtered by S(z) = (1=A s (z)) with The spectrum of u(1) is especially significant for those ! such that Re (1=A 3 (e …