1. Progress on fitting the model to the December xBSM filter data
Dan Peterson,

2. A = ∫ exp(i 2π P(y) /λ T(y) ) dy
Have in the past fit the unknown performance of the detector to (Exray)2.31
But, I have found that a factor of E1 is a necessary normalization due to the way we integrate the amplitude:
A = ∫ exp(i 2π P(y) /λ T(y) ) dy
I will ignore the complicated transparency for now,
and integrate over the boundaries of a simple slit.
A = ∫ exp(i 2π P(y) /λ ) dy
The path length, P(y), is the sum of 2 hypotenuse:
S (source)
O (observer) y Hs Ho

3. The path length difference (compared to S+O ) is:
P(y) = ( S2 + y2 )1/2 + ( O2 + y2 )1/2 - ( S + O )
≈ ½ ( 1/S + 1O ) y2
≡ 1/s y2
s= 2 (S O) /(S + O), ( for S=4m and O=10m, s = ~ 5.7m )
Then the amplitude is:
A = ∫ exp(i 2π/(sλ) y2 ) dy
= (sλ)½ ∫ exp(i 2π x2 ) dx , where x=y/(sλ) ½
now, as written, the integral in unit-less .
The integration method introduces a factor of λ ½ into the amplitude;
it must be removed by multiplying the intensity by E1 .
The performance of the detector is approximately (Exray)1.31 , not (Exray)2.31 .
That makes the power function approximation easier to accept.

4. December 2012 data
Model, power function detector response, E1.15
({model}-[data})/{data}
({model}-[data})/{data}
New fitted detector response

5. A new applied weight is derived from the data.
The previously used power function, E1.15, provides a match to the data,
in which, comparing the filter transmissions,
the maximum value of {model-data}/{data} ) = 0.56
and the RMS is
The data-driven determination of the applied weight provides a match with
the maximum value of {model-data}/{data} ) = 0.14
and RMS =
The power function is
scaled to have value=1 at 4 keV.
The custom function is scaled to have the same average of the
10 values.