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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 = .
. 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
We consider the the block-diagonal matrices ..
A0 , B = , A=
. C= .. .
.. . , D= .. .
.. . , 4 Lecture 3 and the shift operator ..
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)
, A := ZAZ = A0 A1 ..
. which allows us to identify the appearance of shif...
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- Winter '12