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or I4 = ailAil + aizAi2 + *** + ainAin L IAl = 2aijAij
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where IAl is the determinant of A and adj A is the adjoint of A. These two terms will now be described . The determinant of a matrix IAl is a scalar which is computed from the elements of the matrix as follows:
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where Aij , the cofactor of the element a ij, is computed as Aij = (-l) +jMij The determinant Mij is the minor of the element aij and is defined as follows. If the row and column containing the element a ij are deleted from a square matrix A, the determinant of the resulting matrix, which is an (n - 1) X (n - 1) matrix, is the minor Mij . An alternate expression for the calculation of a determinant which uses the elements of a specific column and its cofactors is as follows: IAl = TaijAij
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A determinant of a matrix with two equal rows or columns is zero.
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STATE-SPACE REPRESENTATION OF PHYSICAL SYSTEMS
We now define the adjoint of a matrix. Let the matrix B be an IZ X n matrix whose elements bij are the cofactors Aji of A, i.e., the transpose of the cofactor matrix. B is the adjoint of A; thus B = adj A = transpose of cofactor matrix or 41 A21 . . . An1
A12 A22 . . . An2
adj A = *
. . .
. . .
. . .
Al, AZ,, . . . Some useful properties of the inverse am (m)- = B- A- (A-l)T = (AT)- (A- )- = A
The derivations of relationships presented here, as well as other properties of matrices, can be found in textbooks on linear algebra (see Anton, 1984).
EXAMPLES
1. Evaluate the determinant of A for the following matrix 2 3 5 A = 1 2 o 1 1 0
For this problem, we use Eq. (28A.2) with i = 1 (i.e., use row 1)
JAI = 2[; 3 -3[i A] +q; ;]
IAl =
2UNO)
- U)(l)1 - 3U)(O) - UWI + X(l)(l) - (OWI
IAl = 2(-1) - 3(-2) + 5(1) = 9
2. Find the inverse of the matrix A=
2 3 1 4
adj A A- = IAl The determinant of A is (Al = (2)(4) - (3)(1) = 5 The matrix of minors is
STATE-SPACE
METHODS
The cofactor matrix is
The adjoint of the matrix, which is the transpose of the cofactor matrix, is adjA = [-: therefore A-l = i [ -: -z] = [-t 3. Obtain the inverse of the matrix 2 3 A = One can show that IA/ = 18 The cofactor matrix is * -5 7 The adjoint matrix is r 1 7 .,-: A- = ; 7 I -5 -5 1 -5 7 1 7 7-5 1 7 -5 1 1 7 1 -5 -57 1 1 2 3 1 13 2 -i]
PROBLEMS
28.1. In the liquid level process shown in Fig. p28.1, the three tanks are interacting. The process may be described by: k=Ax+Bu
where x =
x2 x3
andu=
STATE-SPACE REPRESENTATION OF PHYSICAL SYSTEMS
--+A,=1 -z -. 3 --- Xl R, =l
L$,2 -z -. 3 --- x2
A, =l -z -. - = --- x3
R, 1 R2 =l -E R5
FIGUREP28.1
---3
A=2-3 0 1
determine values of R3, Rq, and Rs. If one of these values of R is negative, what is your interpretation (b) Determine B. 28.2. For the system shown in Fig. P28.2, find A and b in
i = Ax+
The tanks are interacting. The following data apply: Al = 1, A2 = f, R1 = A, R2 = 2, R3 = 1
FIGURE P28-2
CHAPTER
TRANSFER FUNCTION MATRIX
In the previous chapter, we have seen that a linear dynamic system can be expressed in terms of the following equations i = Ax+Bu y = cx where x = column vector of n state variables (X 1~2, . . . ,x,) u = column vector of m inputs or forcing terms (U t ,U 2, . . . ,u m) y = column vector of p outputs (yi,y2, , . . ,y,) A = n X n matrix of coefficients B = it X m matrix of coefficients C = p X n matrix of coefficients One of the objectives of this chapter is to show how one solves Eqs. (29.1) and (29.2) in a systematic manner. Before discussing the solution of the matrix differential equation of Eq. (29. l), consider the scalar differential equation (29.3) dxldt = Ax + Bu In this equation all of the terms are scalars. The solution to Eq. (29.3) can be written as the sum of the complementary function and the particular integral as follows:
x(t)
(29.1) (29.2)
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