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^[1][2]:

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

${\displaystyle {\begin{array}{rcl}R(i+1,j)=A_{1}R(i,j)+A_{2}S(i,j)+B_{1}u(i,j)\\S(i,j+1)=A_{3}R(i,j)+A_{4}S(i,j)+B_{2}u(i,j)\\y(i,j)=C_{1}R(i,j)+C_{2}S(i,j)+Du(i,j)\end{array))}$

where ${\displaystyle A_{1},A_{2},A_{3},A_{4},B_{1},B_{2},C_{1},C_{2))$ and ${\displaystyle D}$ are matrices of appropriate dimensions.

These equations can be written more compactly by combining the matrices:

${\displaystyle {\begin{bmatrix}R(i+1,j)\\S(i,j+1)\\y(i,j)\\\end{bmatrix)){\begin{bmatrix}A_{1}&A_{2}&B_{1}\\A_{3}&A_{4}&B_{2}\\C_{1}&C_{2}&D\\\end{bmatrix)){\begin{bmatrix}R(i,j)\\S(i,j)\\u(i,j)\\\end{bmatrix))}$

Given input vectors ${\displaystyle u(i,j)}$ at each point and initial state values, the value of each output vector can be computed by recursively performing the operation above.

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

where ${\displaystyle x(i,j)}$ is the input and ${\displaystyle y(i,j)}$ is the output at point ${\displaystyle (i,j)}$ and ${\displaystyle a_{p,q))$ and ${\displaystyle b_{p,q))$ are constant coefficients.

To derive a transfer function for the system the 2d Z-transform is applied to both sides of the equation above.

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

Transposing yields the transfer function ${\displaystyle T(z_{1},z_{2})}$:

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

So given any pattern of input values, the 2d Z-transform of the pattern is computed and then multiplied by the transfer function ${\displaystyle T(z_{1},z_{2})}$ to produce the Z-transform of the system output.

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:

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

Two cases are individually considered 1) the bottom summation is simply ${\displaystyle 1}$ 2)the top summation is simply a constant ${\displaystyle k}$. 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.

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

The state-space vectors will have the following dimensions:

${\displaystyle R(1\times m),S(1\times n),x(1\times 1)}$ and ${\displaystyle y(1\times 1)}$

Each term in the summation involves a negative (or zero) power of ${\displaystyle z_{1))$ and of ${\displaystyle z_{2))$ which correspond to a delay (or shift) along the respective dimension of the input ${\displaystyle x(i,j)}$. This delay can be effected by placing ${\displaystyle 1}$’s along the super diagonal in the ${\displaystyle A_{1))$. and ${\displaystyle A_{4))$ matrices and the multiplying coefficients ${\displaystyle b_{i,j))$ in the proper positions in the ${\displaystyle A_{2))$. The value ${\displaystyle b_{0,0))$ is placed in the upper position of the ${\displaystyle B_{1))$ matrix, which will multiply the input ${\displaystyle x(i,j)}$ and add it to the first component of the ${\displaystyle R_{i,j))$ vector. Also, a value of ${\displaystyle b_{0,0))$ is placed in the ${\displaystyle D}$ matrix which will multiply the input ${\displaystyle x(i,j)}$ and add it to the output ${\displaystyle y}$. The matrices then appear as follows:

${\displaystyle A_{1}={\begin{bmatrix}0&0&0&\cdots &0&0\\1&0&0&\cdots &0&0\\0&1&0&\cdots &0&0\\\vdots &\vdots &\vdots &\vdots &\vdots &\vdots \\0&0&0&\cdots &0&0\\0&0&0&\cdots &1&0\\\end{bmatrix))}$

${\displaystyle A_{2}={\begin{bmatrix}0&0&0&\cdots &0&0\\0&0&0&\cdots &0&0\\0&0&0&\cdots &0&0\\\vdots &\vdots &\vdots &\vdots &\vdots &\vdots \\0&0&0&\cdots &0&0\\0&0&0&\cdots &0&0\\\end{bmatrix))}$

${\displaystyle A_{3}={\begin{bmatrix}b_{1,n}&b_{2,n}&b_{3,n}&\cdots &b_{m-1,n}&b_{m,n}\\b_{1,n-1}&b_{2,n-1}&b_{3,n-1}&\cdots &b_{m-1,n-1}&b_{m,n-1}\\b_{1,n-2}&b_{2,n-2}&b_{3,n-2}&\cdots &b_{m-1,n-2}&b_{m,n-2}\\\vdots &\vdots &\vdots &\vdots &\vdots &\vdots \\b_{1,2}&b_{2,2}&b_{3,2}&\cdots &b_{m-1,2}&b_{m,2}\\b_{1,1}&b_{2,1}&b_{3,1}&\cdots &b_{m-1,1}&b_{m,1}\\\end{bmatrix))}$

${\displaystyle A_{4}={\begin{bmatrix}0&0&0&\cdots &0&0\\1&0&0&\cdots &0&0\\0&1&0&\cdots &0&0\\\vdots &\vdots &\vdots &\vdots &\vdots &\vdots \\0&0&0&\cdots &0&0\\0&0&0&\cdots &1&0\\\end{bmatrix))}$

${\displaystyle B_{1}={\begin{bmatrix}1\\0\\0\\0\\\vdots \\0\\0\\\end{bmatrix))}$

${\displaystyle B_{2}={\begin{bmatrix}b_{0,n}\\b_{0,n-1}\\b_{0,n-2}\\\vdots \\b_{0,2}\\b_{0,1}\\\end{bmatrix))}$

${\displaystyle C_{1}={\begin{bmatrix}b_{1,0}&b_{2,0}&b_{3,0}&\cdots &b_{m-1,0}&b_{m,0}\\\end{bmatrix))}$

${\displaystyle C_{2}={\begin{bmatrix}0&0&0&\cdots &0&1\\\end{bmatrix))}$

${\displaystyle D={\begin{bmatrix}b_{0,0}\end{bmatrix))}$