ECE 530 – Analysis Techniques for Large-Scale Electrical Systems Prof. Hao Zhu Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign 11/16/ Lecture 21: State Estimation; QR Decomposition

State Estimation (SE) The goal of SE is to find the ML estimator for states x with given measurements and their noise levels Noise level  i weighs each measurement according to the ML principle (maximizing the likelihood function) Measurement noise is assumed Gaussian We can still use WLS if noise is not Gaussian – but sacrificing the statistical optimality For example, outliers (bad measurements) often motivate the least absolute-value (LAV) estimation 2

SE Iterative Solution Algorithm Sequential linearization based on the gradient of J(x) Use Newton's method to solve for x for which the gradient is zero 3 This is exactly the least-squares form developed earlier with H T R -1 H an n by n matrix. This could be solved with Gaussian elimination, but this isn't preferred because of the numerical issues

Example: Two Bus Case Assume a two bus case with a generator supplying a load through a single line with x=0.1 pu. Assume measurements of the p/q flow on both ends of the line (into line positive), and the voltage magnitude at both the generator and the load end. So B 12 = B 21 = We need to assume a reference angle unless directly measuring 

Example: Two Bus Case Let 5 We assume an angle reference of   

Example: Two Bus Case With a flat start guess we get 6

Example: Two Bus Case 7

QR Factorization We will introduce another factorization/ decomposition technique for non-square (sparse) matrices First, let us normalize the measurement matrix by splitting the R -1 matrix as QR factorization for the m by n matrix H' is with Q an m by m orthonormal matrix and U an upper triangular matrix (most books use Q R but we use U to avoid confusion with the previous R) 8

QR Factorization We then have But since Q is an orthonormal matrix, Hence we have And 9

QR Factorization Next issue we discuss the QR factorization algorithm to factor any matrix A into an m by m orthonormal matrix Q and an m by n upper triangular matrix U (usually R) Several methods are available including the Householder method and the Givens method Givens is preferred when dealing with sparse matrices An excellent reference is Gene H. Golub and Charles F. Van Loan, Matrix Computations, second edition, Johns Hopkins University Press,

Givens Algorithm The Givens algorithm works by pre-multiplying the desired matrix (A here) by a series of matrices and their transposes, starting with G 1 G 1 T – G k is the identity matrix modified to have two non-ones on the diagonals c = cos(  ), and two non-zero off diagonal elements, one set to -s = -sin(  ), and one to s = sin(  – Each G k is an orthonormal matrix, so pre-multiplying by G k G k T = I ensures the final product is equal to A – G k values are chosen to zero out particular elements in the lower triangle of A 11

Givens Algorithm Algorithm proceeds column by column, sequentially zeroing out elements in the lower triangle of A, starting at the bottom of each column 12

Givens Algorithm 13

Givens G Matrix 14 i j

Small Givens Example Let First we zero out A[2,1], a=1, b=2 giving s= 0.894, c=

Small Givens Example Next zero out A[3,2] with a=1.7889, b=1, giving c= , s= Final solution is 16

Givens Method for SE Example Starting with the H matrix we get To zero out H'[5,1]=1 we have b=100, a=-1000, giving c=0.995, s=

Start of Givens for SE Example Which gives The next rotation would be to zero out element H'[4,1], continuing until all the elements in the lower triangle have been reduced 18

Givens Comments For a full matrix, Givens is O(mn 2 ) since each element in the lower triangle needs to be zeroed O(nm), and each operation is O(n) Computation can be drastically reduced for a sparse matrix since we only need to zero out the elements that are initially non-zero, and any that become non-zero (i.e., the fills) – Also, for each multiply we only need to deal with the nonzeros in the impacted row Givens rotation is commonly used to solve the SE 19