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
Wp1  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
Wp1  k  n W  n  m  M 
Wp1  k  n W  n  m  M  
M / 2 1

n  M / 2
W p1  k  n  R  n 
(6)
So that defining
X p k  
M / 2 1

n  M / 2
W p1  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 p1  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 p1  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
Wp1  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 / 21
 
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 / 21
 
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.