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.

Applications

Multidimensional systems or m-D systems are the necessary mathematical background for modern digital image processing with many applications in biomedicine, X-ray technology and satellite communications. There are also some studies combining m-D systems with partial differential equations (PDEs).

Linear multidimensional state-space model

A state-space model is a representation of a system in which the effect of all "prior" input values is contained by a state vector. In the case of an m-d system, each dimension has a state vector that contains the effect of prior inputs relative to that dimension. The collection of all such dimensional state vectors at a point constitutes the total state vector at the point.

Consider a uniform discrete space linear two-dimensional (2d) system that is space invariant and causal. It can be represented in matrix-vector form as follows:

Represent the input vector at each point $(i,j)$ by $u(i,j)$ , the output vector by $y(i,j)$ the horizontal state vector by $R(i,j)$ and the vertical state vector by $S(i,j)$ . Then the operation at each point is defined by:

{\begin{aligned}R(i+1,j)&=A_{1}R(i,j)+A_{2}S(i,j)+B_{1}u(i,j)\\S(i,j+1)&=A_{3}R(i,j)+A_{4}S(i,j)+B_{2}u(i,j)\\y(i,j)&=C_{1}R(i,j)+C_{2}S(i,j)+Du(i,j)\end{aligned}} 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.

Multidimensional transfer function

A discrete linear two-dimensional system is often described by a partial difference equation in the form: $\sum _{p,q=0,0}^{m,n}a_{p,q}y(i-p,j-q)=\sum _{p,q=0,0}^{m,n}b_{p,q}x(i-p,j-q)$ 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.

Realization of a 2d transfer function

Often an image processing or other md computational task is described by a transfer function that has certain filtering properties, but it is desired to convert it to state-space form for more direct computation. Such conversion is referred to as realization of the transfer function.

Consider a 2d linear spatially invariant causal system having an input-output relationship described by:

$Y(z_{1},z_{2})={\sum _{p,q=0,0}^{m,n}b_{p,q}z_{1}^{-p}z_{2}^{-q} \over \sum _{p,q=0,0}^{m,n}a_{p,q}z_{1}^{-p}z_{2}^{-q}}X(z_{1},z_{2})$ 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.

Example: all zero or finite impulse response

$Y(z_{1},z_{2})=\sum _{p,q=0,0}^{m,n}b_{p,q}z_{1}^{-p}z_{2}^{-q}X(z_{1},z_{2})$ 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:

$A_{1}={\begin{bmatrix}0&0&0&\cdots &0&0\\1&0&0&\cdots &0&0\\0&1&0&\cdots &0&0\\\vdots &\vdots &\vdots &\ddots &\vdots &\vdots \\0&0&0&\cdots &0&0\\0&0&0&\cdots &1&0\end{bmatrix}}$ $A_{2}={\begin{bmatrix}0&0&0&\cdots &0&0\\0&0&0&\cdots &0&0\\0&0&0&\cdots &0&0\\\vdots &\vdots &\vdots &\ddots &\vdots &\vdots \\0&0&0&\cdots &0&0\\0&0&0&\cdots &0&0\end{bmatrix}}$ $A_{3}={\begin{bmatrix}b_{1,n}&b_{2,n}&b_{3,n}&\cdots &b_{m-1,n}&b_{m,n}\\b_{1,n-1}&b_{2,n-1}&b_{3,n-1}&\cdots &b_{m-1,n-1}&b_{m,n-1}\\b_{1,n-2}&b_{2,n-2}&b_{3,n-2}&\cdots &b_{m-1,n-2}&b_{m,n-2}\\\vdots &\vdots &\vdots &\ddots &\vdots &\vdots \\b_{1,2}&b_{2,2}&b_{3,2}&\cdots &b_{m-1,2}&b_{m,2}\\b_{1,1}&b_{2,1}&b_{3,1}&\cdots &b_{m-1,1}&b_{m,1}\end{bmatrix}}$ $A_{4}={\begin{bmatrix}0&0&0&\cdots &0&0\\1&0&0&\cdots &0&0\\0&1&0&\cdots &0&0\\\vdots &\vdots &\vdots &\ddots &\vdots &\vdots \\0&0&0&\cdots &0&0\\0&0&0&\cdots &1&0\end{bmatrix}}$ $B_{1}={\begin{bmatrix}1\\0\\0\\0\\\vdots \\0\\0\end{bmatrix}}$ $B_{2}={\begin{bmatrix}b_{0,n}\\b_{0,n-1}\\b_{0,n-2}\\\vdots \\b_{0,2}\\b_{0,1}\end{bmatrix}}$ $C_{1}={\begin{bmatrix}b_{1,0}&b_{2,0}&b_{3,0}&\cdots &b_{m-1,0}&b_{m,0}\\\end{bmatrix}}$ $C_{2}={\begin{bmatrix}0&0&0&\cdots &0&1\\\end{bmatrix}}$ $D={\begin{bmatrix}b_{0,0}\end{bmatrix}}$ 