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Periodic Array The system of equations is1 M / 2 1 m M / 2 W n m X m R n (1) This has an easy inverse if W is periodic so that (2) W n M W n The problem in (1) is that n-m can be > M/2. For m=-N/2 and n=N/2-1 the argument of W is M-1. For this system to be periodic, W(M-1) must equal W(-1). Construct Wp n W n W n M W n M (3) Then W p n M W n M W n 2M W n W n (4) - not a good idea W p n M W n M W n W n 2M W n In this case (1) becomes M / 2 1 m M / 2 Wp n m X m M / 2 1 m M / 2 W n M m X m M / 2 1 m M / 2 W n M m X m R n (5) Multiply by the inverse of Wp(k-n) and sum over n M / 2 1 m M / 2 X m M / 2 1 n M / 2 Wp1 k n Wp n m M / 2 1 m M / 2 M / 2 1 m M / 2 X m X m M / 2 1 n M / 2 M / 2 1 n M / 2 Wp1 k n W n m M Wp1 k n W n m M M / 2 1 n M / 2 W p1 k n R n (6) So that defining X p k M / 2 1 n M / 2 W p1 k n R n (7) Equation (6) becomes X k X p k M / 2 1 m M / 2 X m M / 2 1 n M / 2 W p1 k n W n m M W n m M (8) Mixing The natural way to solve (8) is to start with the first term, and then iterate. But the equation set in (1) can be solved directly if all but the first L terms are taken to be zero. Thus it makes sense to include Xb[k] in the solution. Write (8) as 0 X p k 1 X b k X p k M / 2 1 m M / 2 0 X p m 1 X b m M / 2 1 n M / 2 W p1 k n W n m M W n m M (9) Find 0 and 1 by minimizing 2 0 , 1 0 1 X p k 1 X b k M / 2 1 m M / 2 0 X p m 1 X b m 2 M / 2 1 n M / 2 Wp1 k n W n m M W n m M Owing to the fact that X is a linear function of the ’s, this is a one step procedure. The partial with respect to 0 is 2 0 , 1 0 0 , 1 2 1 1 0 1 X p k 1 X b k M / 2 1 M / 21 1 0 X p m 1 X b m W p k n W n m M W n m M n M / 2 m M / 2 M / 2 1 M / 2 1 1 X p k X p m W p k n W n m M W n m M c.c. m M / 2 n M / 2 0 1 X p k 1 X b k M / 2 1 M / 21 1 0 X p m 1 X b m W p k n W n m M W n m M n M / 2 m M / 2 M / 2 1 M / 2 1 1 X b k X b m W p k n W n m M W n m M c.c. m M / 2 n M / 2 This system equations usually occurs as part of Newton Raphson iteration. In this case the R is the set of residuals which is constantly getting smaller. These interact a bit to the approximation of X below causeing one set to want a band like solution and the next set to want a periodic solution. Also since R is in principly going to zero, highly accurate solutions at each step are not needed.