Numerical recipes

14.8 Savitzky-Golay Smoothing Filters 651
M nL nR Sample Savitzky-Golay Coefficients
2 2 2 −0.086 0.343 0.486 0.343 −0.086
2 3 1 −0.143 0.171 0.343 0.371 0.257
2 4 0 0.086 −0.143 −0.086 0.257 0.886
2 5 5 −0.084 0.021 0.103 0.161 0.196 0.207 0.196 0.161 0.103 0.021 −0.084
4 4 4 0.035 −0.128 0.070 0.315 0.417 0.315 0.070 −0.128 0.035
4 5 5 0.042 −0.105 −0.023 0.140 0.280 0.333 0.280 0.140 −0.023 −0.105 0.042
nL + nR +1points in the moving window, and then set gi to be the value of that polynomial
at position i. (If you are not familiar with least-squares fitting, you might want to look ahead
to Chapter 15.) We make no use of the value of the polynomial at any other point. When we
move on to the next point fi+1, we do a whole new least-squares fit using a shifted window.
All these least-squares fits would be laborious if done as described. Luckily, since the
process of least-squares fitting involves only a linear matrix inversion, the coefficients of a
fitted polynomial are themselves linear in the values of the data. That means that we can do
all the fitting in advance, for fictitious data consisting of all zeros except for a single 1, and
then do the fits on the real data just by taking linear combinations. This is the key point, then:
There are particular sets of filter coefficients cn for which equation (14.8.1) “automatically”
accomplishes the process of polynomial least-squares fitting inside a moving window.
To derive such coefficients, consider how g0 might be obtained: We want to fit a
polynomial of degree M in i, namely a0 + a1i + ···+ aMi M to the values f−nL ,...,fnR .
Then g0 will be the value of that polynomial at i =0, namely a0. The design matrix for
this problem (§15.4) is
Aij = i j
i = −nL,...,nR, j =0,...,M (14.8.2)
and the normal equations for the vector of aj’s in terms of the vector of fi’s is in matrix notation
(A T · A) · a = A T · f or a =(A T · A) −1 · (A T We also have the specific forms
· f) (14.8.3)
�
A T �
· A =
ij
nR� AkiAkj =
nR� k i+j
(14.8.4)
and
�
A T �
· f =
j
k=−n L
n R�
k=−n L
Akjfk =
k=−n L
n R�
k=−n L
k j fk
(14.8.5)
Since the coefficient cn is the component a0 when f is replaced by the unit vector en,
−nL ≤ n1,
the array c must be multiplied by k! to give derivative coefficients. For derivatives, one
usually wants m =4or larger.
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