In Figure 1 the filter is constrained to be of the form (1,a_{1},a_{2}).
missif
Figure 1 Top is known data. Middle includes the interpolated values. Bottom is the filter with the leftmost point constrained to be unity. |
The result is pleasing in that the interpolated traces have the same general character as the given values. The filter came out slightly different from the (1,0,-1) that I suggested based on a subjective analysis for Figure . Curiously, constraining the filter to be of the form (a_{-2},a_{-1},1) in Figure 2 yields the same interpolated missing data as in the previous figure.
backwards
Figure 2 Top is known data. Middle includes the interpolated values. Bottom is the filter with the rightmost point constrained to be unity. |
I understand the sum squared of the coefficients of A(Z)Y(Z) is the same as that of A(1/Z)Y(Z) but don't see why that would imply the same interpolated data.
Changing the filter constraint to be of the form (a_{-1}, 1, a_{1}) yields a much different interpolation shown in Figure 3. The interpolated data is much smoother than the given data. This is because the missing data is produced from an interpolation-error filter instead of a prediction-error filter. A fact not evident from the figures is that the value of the output power was smaller for interpolation than for prediction. I cannot explain why the filter turned out to be imperceptibly different from a second difference operator, but somewhere I think I have a proof that it is a symmetrical operator.
center
Figure 3 Top is known data with missing data represented by zeros. Middle includes the interpolated values. Bottom is the filter. |