TVSC_Lecture_3

# However we aim to derive a purely algebraic

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Toeplitz Operator. To this end we combine the set of all time-varying state-space realization matrices into block diagonal matrices .. .. .. .. . . . . , B = , C = , D = . Ak Bk Ck Dk A= .. .. .. .. . . . . 3 Lecture 3 Note that the dimension of the individual block matrices may change from time index to time index. Correspondingly, we combine all input-, output- and state-signals to form the vectors . . . . . . . . . x = xk , u = uk , y = yk , . . . . . . . . . whose length may also change from time step to time step. Using these block-diagonal matrices, the vectors and the causal shift-down operator Z we can rewrite the state space equations (1) as Z −1 x y = A·x+B·u = C ·x+D·u (4) Starting out from equation (4) we can eliminate the state vector x to arrive at a description of the output signal y only depending on the input signal u and the state space realization x = y = −1 (1 − ZA) ZBu −1 D + C (1 − ZA) ZB u. (5) Looking at this result we can identify a representation of the input-output, or Toeplitz operator in terms of the linear fractional transformation (LFT) −1 T = D + C (1 − ZA) ZB , (6) which is given in terms of the state space realization matrices and the shift operator. This representation of the Toeplitz operator has an obvious structural resemblance with the standard transfer function representation for time-invariant state-space systems. However, note that Z is the matrix representation of the shift operator and not a complex variable; the state-space matrices A, B , C and D are block-diagonal matrices. Construction of Transfer Operator A look at Equation (6) does not immediately reveal that this linear fractional transformation based on block-diagonal matrices does actually represent the Toeplitz operator as shown in Equation (3). In this section we actually plug in in the block-diagonal matrices in (6) to verify the correctness of this representation. We consider the the block-diagonal matrices .. .. . . B0 A0 , B = , A= A1 B1 A2 B2 .. .. . . C= .. . C0 C1 C2 .. . , D= .. . D0 D1 D2 .. . , 4 Lecture 3 and the shift operator .. . .. .0 10 Z= 1 0 1 . .. . .. . We can savely assume that the dimensions of the block-entries of all matrices are compatible to make the matrix multiplications meaningful. We ﬁrst have a closer look at the middle part of Equation (6), which can be represented by the Neuman expansion, which comparable to geometric series for scalar quantities −1 (1 − ZA) 2 3 = 1 + ZA + (ZA) + (ZA) + . . . . The series expansion is based on powers of the term ZA. Let’s investigate how these powers can be calculated by checking the example 3 (ZA) = ZAZAZA = (ZAZ T )(ZZAZ T Z T )(ZZZAZ T Z T Z T )ZZZ , where I have inserted various copies of the identity in terms of 1 = Z T Z to arrive at the right hand side of the equation. We will simplify notation for further manipulations by observing the shift of a block-diagonal matrix by one block in the south-east direction is given by .. . A−1 (1) T , A := ZAZ = A0 A1 .. . which allows us to identify the appearance of shif...
View Full Document

## This note was uploaded on 07/07/2013 for the course EI 2012 taught by Professor Tum during the Winter '12 term at TU München.

Ask a homework question - tutors are online