Multidimensional system

In mathematical systems theory, a multidimensional system or m-D system is a system in which not only one independent variable exists (like time), but there are several independent variables. Important problems such as factorization and stability of m-D systems (m > 1) have recently attracted the interest of many researchers and practitioners. The reason is that the factorization and stability is not a straightforward extension of the factorization and stability of 1-D systems because, for example, the fundamental theorem of algebra does not exist in the ring of m-D (m > 1) polynomials.

In mathematical systems theory, a multidimensional system or m-D system is a system in which not only one independent variable exists (like time), but there are several independent variables. Important problems such as factorization and stability of m-D systems (m > 1) have recently attracted the interest of many researchers and practitioners. The reason is that the factorization and stability is not a straightforward extension of the factorization and stability of 1-D systems because, for example, the fundamental theorem of algebra does not exist in the ring of m-D (m > 1) polynomials. R(i+1,j) & = A_1R(i,j) + A_2S(i,j) + B_1u(i,j) \\ S(i,j+1) & = A_3R(i,j) + A_4S(i,j) + B_2u(i,j) \\ y(i,j) & = C_1R(i,j) +C_2S(i,j) + Du(i,j) \end{align} where A_1, A_2, A_3, A_4, B_1, B_2, C_1, C_2 and D are matrices of appropriate dimensions. These equations can be written more compactly by combining the matrices: \begin{bmatrix} R(i+1,j) \\ S(i,j+1) \\ y(i,j) \end{bmatrix} = \begin{bmatrix} A_1 & A_2 & B_1 \\ A_3 & A_4 & B_2 \\ C_1 & C_2 & D \end{bmatrix} \begin{bmatrix} R(i,j) \\ S(i,j) \\ u(i,j) \end{bmatrix} Given input vectors u(i,j) at each point and initial state values, the value of each output vector can be computed by recursively performing the operation above. \sum_{p,q=0,0}^{m,n}a_{p,q}y(i-p,j-q) = \sum_{p,q=0,0}^{m,n}b_{p,q}x(i-p,j-q) where x(i,j) is the input and y(i,j) is the output at point (i,j) and a_{p,q} and b_{p,q} are constant coefficients. To derive a transfer function for the system the 2d Z-transform is applied to both sides of the equation above. \sum_{p,q=0,0}^{m,n} a_{p,q}z_1^{-p}z_2^{-q}Y(z_1,z_2) = \sum_{p,q=0,0}^{m,n}b_{p,q}z_1^{-p}z_2^{-q}X(z_1,z_2) Transposing yields the transfer function T(z_1,z_2) : T(z_1,z_2) = {Y(z_1,z_2) \over X(z_1,z_2)} = {\sum_{p,q=0,0}^{m,n}b_{p,q}z_1^{-p}z_2^{-q} \over \sum_{p,q=0,0}^{m,n}a_{p,q}z_1^{-p}z_2^{-q}} So given any pattern of input values, the 2d Z-transform of the pattern is computed and then multiplied by the transfer function T(z_1,z_2) to produce the Z-transform of the system output. Y(z_1,z_2) = {\sum_{p,q=0,0}^{m,n}b_{p,q}z_1^{-p}z_2^{-q} \over \sum_{p,q=0,0}^{m,n}a_{p,q}z_1^{-p}z_2^{-q}}X(z_1,z_2) Two cases are individually considered 1) the bottom summation is simply the constant 1 2) the top summation is simply a constant k . Case 1 is often called the "all-zero" or "finite impulse response" case, whereas case 2 is called the "all-pole" or "infinite impulse response" case. The general situation can be implemented as a cascade of the two individual cases. The solution for case 1 is considerably simpler than case 2 and is shown below. Y(z_1,z_2) = \sum_{p,q=0,0}^{m,n}b_{p,q}z_1^{-p}z_2^{-q}X(z_1,z_2) The state-space vectors will have the following dimensions: R (1 \times m),\quad S (1 \times n),\quad x (1 \times 1) and y (1 \times 1) Each term in the summation involves a negative (or zero) power of z_1 and of z_2 which correspond to a delay (or shift) along the respective dimension of the input x(i,j) . This delay can be effected by placing 1 ’s along the super diagonal in the A_1 . and A_4 matrices and the multiplying coefficients b_{i,j} in the proper positions in the A_2 . The value b_{0,0} is placed in the upper position of the B_1 matrix, which will multiply the input x(i,j) and add it to the first component of the R_{i,j} vector. Also, a value of b_{0,0} is placed in the D matrix which will multiply the input x(i,j) and add it to the output y . The matrices then appear as follows: A_1 = \begin{bmatrix}0 & 0 & 0 & \cdots & 0 & 0 \\ 1 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 1 & 0 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 0 & \cdots & 1 & 0 \end{bmatrix} A_2 = \begin{bmatrix}0 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 0 \end{bmatrix} A_3 = \begin{bmatrix} b_{1,n} & b_{2,n} & b_{3,n} & \cdots & b_{m-1,n} & b_{m,n} \\ b_{1,n-1} & b_{2,n-1} & b_{3,n-1} & \cdots & b_{m-1, n-1} & b_{m,n-1} \\ b_{1,n-2} & b_{2,n-2} & b_{3,n-2} & \cdots & b_{m-1, n-2} & b_{m,n-2} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ b_{1,2} & b_{2,2} & b_{3,2} & \cdots & b_{m-1,2} & b_{m,2} \\ b_{1,1} & b_{2,1} & b_{3,1} & \cdots & b_{m-1,1} & b_{m,1} \end{bmatrix} A_4 = \begin{bmatrix}0 & 0 & 0 & \cdots & 0 & 0 \\ 1 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 1 & 0 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 0 & \cdots & 1 & 0 \end{bmatrix} B_1 = \begin{bmatrix}1 \\ 0 \\ 0\\ 0\\ \vdots \\ 0 \\ 0 \end{bmatrix} B_2 = \begin{bmatrix} b_{0,n} \\ b_{0,n-1} \\ b_{0,n-2} \\ \vdots \\ b_{0,2} \\ b_{0,1} \end{bmatrix} C_1 = \begin{bmatrix} b_{1,0} & b_{2,0} & b_{3,0} & \cdots & b_{m-1,0} & b_{m,0} \\ \end{bmatrix} C_2 = \begin{bmatrix}0 & 0 & 0 & \cdots & 0 & 1 \\ \end{bmatrix} D = \begin{bmatrix}b_{0,0} \end{bmatrix}

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