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EE 616 Computer Aided Analysis of Electronic Networks Lecture 7 Instructor: Dr. J. A. Starzyk, Professor School of EECS Ohio University Athens, OH, 45701 1 Outline 2 Computer Generation of Network Functions Unit circle polynomial interpolation Condition numbers for interpolation Computer Generation of Network Functions We want to obtain a network function in the semi-symbolic form Ns Fs Ds This will be useful for direct evaluation of F(s) at different frequencies (instead of using n3 / 3 operations for LU factorization each time) or inverse Laplace transform to obtain time domain solution. Assume we know LU factorization at frequency si and assume that T = LU has all entries with s in the numerator. Determinant D(si) can be found as n Ds i det Ts i det Ls i L ij j1 3 Computer Generation of Network Functions (cont’d) Solving LUX = we can get X and then F(si). The numerator N(si) can be found as N si Dsi F si Changing si we get sets of pairs (si, N(si)) and (si, D(si)), and we get the coefficients of N and D. This is a wellknown problem of polynomial interpolation. 4 Unit circle polynomial interpolation Assume we have (n + 1) distinct points (xi, f(xi) = yi). We want to find coefficients of the polynomial n f (x) a j x i j0 We have a 0 a 1 x i a 2 x i2 ... a n x in y i or in the matrix form or Xa = y 1 x 0 1 x 1 1 x n x 02 x 12 x 2n x 0n x 1n x nn y0 a 0 y a 1 = 1 a y n n a = X-1y The most numerically stable selection of 5 i 0,1,2,..., n where xi X x ij is on the unit circle. Unit circle polynomial interpolation (cont’d) Let us define and xk = k exp so 2 1 n 1 X x ij ij 1 X X* n 1 Therefore the solution is a X 1 y 1 X * y 1 and 1 n 1 or simply 6 y n 1 ij 1 n aj y k jk n 1 k 0 Using this equation all coefficients of the approximating polynomial can be explicitly calculated. Finding each coefficient requires an addition of n + 1 complex numbers - each one easily obtained from the approximated function values y k . In addition the polynomial interpolation on the unit circle is the most numerically stable algorithm. Condition numbers for interpolation Condition number K X ~ for the matrix X is ~ K x max / min ~ where max, min are the largest and smallest eigenvalues of XX*. K(X) is a measure of perturbations in y. In our case XX * XX 1 n 1 n 1I So all the eigenvalues of XX* are equal to (n + 1) and K(x) = 1. Note that K(x) 1 always. Therefore, the selection of all points on the unit circle yields the best possible accuracy of the polynomial approximation problem. 7 Condition numbers for interpolation - Example Example: Numerical value of the numerator and denominator are N s0 t1 3 Ds0 1 11 N s p 1 3 Ds p 1 7 Find transfer function (first order polynomials N(s) and D(s)). n=1 since so 8 n+1=2 a 0 a 1s T b 0 b1s exp 2 1 1 2 Condition numbers for interpolation - Example 1 1 1 0k a 0 1 Ns k 3 3 0 2 k 0 2 1 1 1 1k a 1 1 Ns k 3 3 3 2 k 0 2 1 1 1 0k b 0 1 Ds k 11 7 2 2 k 0 2 1 1 1 1k b1 1 Ds k 11 7 9 2 k 0 2 And, we obtain 9 3s T 2 9s Condition numbers for various interpolations 10 Growth of error versus degree of the polynomial n 11