Passive Approximation with High-Order B-Splines

Abstract

Convex optimization has emerged as a well-suited tool for passive approximation. Here, it is desired to approximate some pre-defined non-trivial system response over a given finite frequency band by using a passive system. This paper summarizes some explicit results concerning the Hilbert transform of general B-splines of arbitrary order and arbitrary partitions that can be useful with the convex optimization formulation. A numerical example in power engineering is included concerning the identification of some model parameters based on measurements on high-voltage insulation materials.

Appendix

The discontinuity behavior of linear B-splines N _0,2(x) with knot values N _0,2(x ₁)=1 and N _0,2(x ₀) = N _0,2(x ₂) = 0 is given by

$$\displaystyle \begin{gathered} N_{0,2}^{(1)}(x_0+)=\frac{1}{x_1-x_0}, \end{gathered} $$

(21)

$$\displaystyle \begin{gathered} N_{0,2}^{(1)}(x_1+)=-\frac{1}{x_2-x_1}. \end{gathered} $$

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The discontinuity behavior of quadratic B-splines N _0,3(x) with knot values N _0,3(x ₁) = (x ₁ − x ₀)∕(x ₂ − x ₀), N _0,3(x ₂) = (x ₃ − x ₂)∕(x ₃ − x ₁), and N _0,3(x ₀) = N _0,3(x ₃) = 0 is given by

$$\displaystyle \begin{gathered} N_{0,3}^{(2)}(x_0+)=\frac{2}{(x_2-x_0)(x_1-x_0)}, \end{gathered} $$

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$$\displaystyle \begin{gathered} N_{0,3}^{(2)}(x_1+)= -\frac{2}{(x_2-x_0)(x_2-x_1)}-\frac{2}{(x_3-x_1)(x_2-x_1)}, \end{gathered} $$

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$$\displaystyle \begin{gathered} N_{0,3}^{(2)}(x_2+)=\frac{2}{(x_3-x_1)(x_3-x_2)}. \end{gathered} $$

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The discontinuity behavior of cubic B-splines N _0,4(x) with knot values

$$\displaystyle \begin{gathered} N_{0,4}(x_1)=\frac{(x_1-x_0)^2}{(x_3-x_0)(x_2-x_0)}, \end{gathered} $$

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$$\displaystyle \begin{gathered} N_{0,4}(x_2)=\frac{(x_2-x_0)(x_3-x_2)}{(x_3-x_0)(x_3-x_1)}+\frac{(x_4-x_2)(x_2-x_1)}{(x_4-x_1)(x_3-x_1)}, \end{gathered} $$

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$$\displaystyle \begin{gathered} N_{0,4}(x_3)=\frac{(x_4-x_3)^2}{(x_4-x_1)(x_4-x_2)}, \end{gathered} $$

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and N _0,4(x ₀) = N _0,4(x ₄) = 0 is given by

$$\displaystyle \begin{aligned} N_{0,4}^{(3)}(x_0+)=\frac{6}{(x_3-x_0)(x_2-x_0)(x_1-x_0)}, \end{aligned} $$

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$$\displaystyle \begin{aligned} \begin{array}{rcl} N_{0,4}^{(3)}(x_1+)=-\frac{6}{(x_3-x_0)(x_2-x_0)(x_2-x_1)} \\ \qquad -\frac{6}{(x_3-x_0)(x_3-x_1)(x_2-x_1)} -\frac{6}{(x_4-x_1)(x_3-x_1)(x_2-x_1)}, \vspace{-6pt}\end{array} \end{aligned} $$

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$$\displaystyle \begin{aligned} \begin{array}{rcl} N_{0,4}^{(3)}(x_2+)=\frac{6}{(x_3-x_0)(x_3-x_1)(x_3-x_2)} \\ \qquad +\frac{6}{(x_4-x_1)(x_3-x_1)(x_3-x_2)} +\frac{6}{(x_4-x_1)(x_4-x_2)(x_3-x_2)}, \end{array} \end{aligned} $$

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$$\displaystyle \begin{aligned} N_{0,4}^{(3)}(x_3+)=-\frac{6}{(x_4-x_1)(x_4-x_2)(x_4-x_3)}. \end{aligned} $$

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