Periodic Array
The fundamental equations are in ..\..\Fittery\Complex\WeightedFourierFit\Weighted
Fourier Fit.doc equation (28). The derivatives of the full 2 are
ME
 W  n  m  P n  X m  R n
m, n
m M B
MB  n  ME
(1)
(2)
MC  M E  M B  1
Note that MB is frequently less than 0, e.g MB = -Mc/2 and ME = Mc/2-1.
MB  n  ME
 W n

W p  n   W  n  M 
n  ME
(3)
 W n  M 
n  MB

Suppose that the penalty term has made it possible to extend the periodic set of W’s so
that (1) becomes
ME
 W  n  m  P n  X m  R n
p
m M B
m,n
MB  n  ME
(4)
Multiply by Wp-1(k-n) and sum on n
ME
ME
ME
  W  k  n W  n  m   W  k  n  P n X m  W  k  n  R n
1
p
m M B n M B
p
1
p
m,n
1
p
MB
(5)
Define
ME
X p  k   Wp1  k  n  R  n 
(6)
MB
So that (5) becomes
ME
 
m M B
k ,m
 Wp1  k  m  P  m X  m  X p  k 
k ,m
 Wp1  k  m  P  m X  m  X p  k   0 M B  k  M E (8)
(7)
Or
ME
 
m M B
Write this as a set of functions fk(X).
ME
 
m M B
k ,m
 
 Wp1  k  m  P  m X  m  X p  k   f k X
MB  k  ME
(9)
This is ready for ..\..\optimization\solving\Newton.doc#Newton_Raphson. The term
X[m] = m+jm for which the derivatives are needed. This is easy since nothing
depends on them except the X. Thus the complex partial is with respect to X[n].
f k  X 
  k ,n  Wp1  k  n  P  n M B  k  M E (10)
X  n
This leads to
ME
 
mM B
k ,m
 

 Wp1  k  m  P  m  X it  m  X it 1   f k X  0
Assuming that the diagonal term dominates
f k X it 1
X it  k   X it 1  k  
(12)
1  W p1  0 P  k 


(11)