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## Method of Least Squares

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**Least Squares**• Method of LeastSquares: • Deterministicapproach • Theinputs u(1), u(2), ..., u(N) areappliedtothesystem • Theoutputs y(1), y(2), ..., y(N) areobserved • Find a model whichfitstheinput-outputrelationto a (linear?) curve, f(n,u(n)) • ‘best’ fit byminimisingthesum of thesqures of thedifference f - y**Least Squares**• The curve fitting problem can be formulated as • Error: • Sum-of-error-squares: • Minimum (least-squares of error) is achieved when the gradient is zero observations model variable**Problem Statement**• Fortheinputstothesystem, u(i) • Theobserveddesiredresponse is, d(i) • Relation is assumedto be linear • Unobservablemeasurementerror • Zeromean • White**Problem Statement**• Design a transversalfilterwhichfindstheleastsquaressolution • Then, sum of errorsquares is**Data Windowing**• We will express the input in matrix form • Depending on the limits i1 and i2 this matrix changes Covariance Method i1=M, i2=N Prewindowing Method i1=1, i2=N Postwindowing Method i1=M, i2=N+M1 Autocorr. Method i1=1, i2=N+M1**Principle of Orthogonality**• Error signal • Least squares (minimum of sum of squares) is achieved when • i.e., when • The minimum-error time series emin(i) is orthogonal to the time series of the input u(i-k) applied to tap k of a transversal filter of length M for k=0,1,...,M-1 when the filter is operating in its least-squares condition. !Time averaging! (For Wiener filtering) (this was ensemble average)**Corollary of Principle of Orthogonality**• LS estimate of the desired response is • Multiply principle of orthogonality by wk* and take summation over k • Then • When a transversal filter operates in its least-squares condition, the least-squares estimate of the desired response -produced at the output of the filter- and the minimum estimation error time series are orthogonal to each other over time i.**Energy of Minimum Error**• Due to the principle of orthogonality, second and third terms are orthogonal, hence where • , when eo(i)= 0 for all i, impossible • , when the problem is underdetermined fewer data points than parameters infinitely many solutions (no unique soln.)!**Normal Equations**Principle of Orthogonality Minimum error: • Hence, Expanded system of the normal equations for linear least-squares filters. → z(-k), 0 ≤k ≤M-1 time-average cross-correlation bw the desired response and the input (t,k), 0≤(t,k) ≤M-1 time-average autocorrelation function of the input**Normal Equations (Matrix Formulation)**• Matrix form of the normal equations for linear least-squares filters: • Linear least-squares counterpart of the Wiener-Hopf eqn.s. • Here and z are time averages, whereas in Wiener-Hopf eqn.s they were ensemble averages. (if -1 exists!)**Minimum Sum of Error Squares**• Energy contained in the time series is • Or, • Then the minimum sum of error squares is**Properties of the Time-Average Correlation Matrix **• Property I: The correlation matrix is Hermitian symmetric, • Property II: The correlation matrix is nonnegative definite, • Property III: The correlation matrix is nonsingular iff det() is nonzero • Property IV: The eigenvalues of the correlation matrix are real and non-negative.**Properties of the Time-Average Correlation Matrix **• Property V: The correlation matrix is the product of two rectangular Toeplitz matrices that are Hermitian transpose of each other.**Normal Equations (Reformulation)**• But we know that which yields • Substituting into the minimum sum of error squares expression gives then ! Pseudo-inverse !**Projection**• The LS estimate of d is given by • The matrix is a projection operator • onto the linear space spanned by the columns of data matrix A • i.e. the space Ui. • The orthogonal complement projector is**Projection - Example**• M=2 tap filter, N=4 → N-M+1=3 • Let • Then • And orthogonal**Uniqueness of the LS Solution**• LS always has a solution, is that solution unique? • The least-squares estimate is unique if and only if the nullity (the dimension of the null space) of the data matrix A equals zero. • AKxM, (K=N-M+1) • Solution is unique when A is of full column rank, K≥M • All columns of A are linearly independent • Overdetermined system (more eqns. than variables (taps)) • (AHA)-1 nonsingular → exists and unique • Infinitely many solutions when A has linearly dependent columns, K<M • (AHA)-1 is singular**Properties of the LS Estimates**• Property I: Theleast-squaresestimate is unbiased, providedthatthemeasurementerrorprocesseo(i) has zeromean. • Property II: Whenthemeasurementerrorprocesseo(i) is whitewithzeromeanandvariance2, thecovariancematrix of theleast-squaresestimateequals2-1. • Property III: Whenthemeasurementerrorprocesseo(i) is whitewithzeromean, theleastsquaresestimate is thebestlinearunbiasedestimate. • Property IV: Whenthemeasurementerrorprocesseo(i) is whiteandGaussianwithzeromean, theleast-squaresestimateachievestheCramer-Raolowerboundforunbiasedestimates.**Computation of the LS Estimates**• The rank (W) of an KxN (K≥N or K<N) matrix A gives • The number of linearly independent columns/rows • The number of non-zero eigenvalues/singular values • The matrix is said to be full rank (full column or row rank) if • Otherwise, it is said to be rank-deficient • Rank is an important parameter for matrix inversion • If K=N (square matrix) and the matrix is full rank (W=K=N) (non-singular) inverse of the matrix can be calculated, A-1=adj(A)/det(A) • If the matrix is not square (K≠N), and/or it is rank-deficient (singular), A-1 does not exist, instead we can use the pseudo-inverse (a projection of the inverse), A+**SVD**• We can calculate the pseudo-inverse using SVD. • Any KxN matrix (K≥N or K<N) can be decomposed using the Singular Value Decomposition (SVD) as follows:**SVD**• The system of eqn.s, • is overdetermined if K>N, more eqn.s than unknowns, • Unique solution (if A is full-rank) • Non-unique, infinitely many solutions (if A is rank-deficient) • is underdetermined if K<N, more unknowns than eqn.s, • Non-unique, infinitely many solutions • In either case the solution(s) is(are) where**Computation of the LS Estimates**• Find the solution of (A: KxM) • If K>M and rank(A)=M, ( ) the unique solution is • Otherwise , infinitely many solutions, but pseudo-inverse gives the minimum-norm solution to the least squares problem. • Shortest length possible in the Euclidean norm sense.