Consider the single–input single –output discrete time system given by the input/output difference equation.

$y\left( kT+nT \right)+\underset{i=0}{\overset{n-1}{\mathop \sum }}\,{{a}_{i}}y\left( kT+iT \right)=\underset{i=0}{\overset{m}{\mathop \sum }}\,{{b}_{i}}x\left( kT+iT \right)~~~~~~~~~~~~\left( 1 \right)$

In (1), T is a fixed real number, k is a variable that takes its values from the set of integers, and the coefficients a_{0}, a_{1},a_{2},…,a_{n-1} and b_{0}, b_{1},…,b_{m} are constants. The discrete time signal x(kT) is the input applied to the system and the discrete time signal y(kT) is the output response of the system resulting from input x(kT) with the n initial conditions y(-T), y (-2T)………..y (-nT). Note that we are taking initial time to be zero.

The integer n in (1) is the order of the input-output difference equation. If the coefficient b in (1) is non-zero, the system defined by (1) is causal if and only if the integer m in (1) is less than or equal to the order n; that is, ** **

To see this, if we set k=0 in (1) and solve for y (nT), we have

$y\left( nT \right)=-{{a}_{0}}y\left( 0 \right)-{{a}_{1}}y\left( T \right)-\ldots -{{a}_{0n-1}}y\left( nT-T \right)+{{b}_{0}}x\left( 0 \right)+{{b}_{1}}y\left( T \right)+\ldots +{{b}_{m}}x\left( mT \right)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\left( 2 \right)$

From (2) we see that the output y (nT) at time nT depends on the value x (mT) of the input at time mT. Hence if m>n, the output y (nT) at time nT depends on the value of the input at future time mT, and therefore the system is non-causal.

We can rewrite the input/output difference equation (1) in the form.

$y\left( kT+nT \right)=-\underset{i=0}{\overset{n-1}{\mathop \sum }}\,{{a}_{i}}y\left( kT+iT \right)+\underset{i=0}{\overset{m}{\mathop \sum }}\,{{b}_{i}}x\left( kT+iT \right)~~~~~~~~~~~\left( 3 \right)$

From (3), we see that y(kT+nT) is a function of y(kT+iT) for i=0,1,…,n-1 and a function of x(kT+iT) for i=0,1,…,m.

Hence the system defined by (1) or (3) is finite dimensional with order or dimension equal to n. Furthermore, since (1) is a linear difference equation with constant coefficients, the system defined by (1) is linear and time invariant in nature.

**Different Equation Solution by Recursion**

Linear input/output different equations can be solved by a direct numerical procedure. More precisely, the output y (kT) for some finite range of integer values of k can be computed recursively as follows:

First, we rewrite (1) in the form (3). Setting k=-n in (3), we have

$y\left( 0 \right)=-\underset{i=0}{\overset{n-1}{\mathop \sum }}\,{{a}_{i}}y\left( -nT+iT \right)+\underset{i=0}{\overset{m}{\mathop \sum }}\,{{b}_{i}}x\left( -nT+iT \right)$

Or

$y\left( 0 \right)=-{{a}_{0}}y\left( -nT \right)-{{a}_{1}}y\left( -nT+T \right)-\ldots -{{a}_{n-1}}y\left( -T \right)+{{b}_{0}}x\left( -nT \right)+{{b}_{1}}x\left( -nT+T \right)+\ldots +{{b}_{m}}x\left( -nT+mT \right)$

So the output y (0) at time 0 is a linear combination of y (-nT), y (-nT+T)… y (-T) and X (-nT), X (-nT+T)… x (-nT+mT).

Setting k= -n+1 in (3), We Have

$y\left( T \right)=-\underset{i=0}{\overset{n-1}{\mathop \sum }}\,{{a}_{i}}y\left( -nT+T+iT \right)+\underset{i=0}{\overset{m}{\mathop \sum }}\,{{b}_{i}}x\left( -nT+T+iT \right)$

Or

$y\left( T \right)=-{{a}_{0}}y\left( -nT+T \right)-{{a}_{1}}y\left( -nT+2T \right)-\ldots -{{a}_{n-1}}y\left( 0 \right)+{{b}_{0}}x\left( -nT+T \right)+{{b}_{1}}x\left( -nT+2T \right)+\ldots +{{b}_{m}}x\left( -nT+mT+T \right)$

Hence the output y(T) at time T is a linear combination of y(-nT+T), y(-nT+2T),…,y (0) and x(-nT+T), x(-nT+2T),…,x (-nT+T+mT).

If we continue, it is clear that the next value of the output is a linear combination of the n past values of the output and m+1 values of the input. At each step of the computation, it is necessary to store only the n past values of the output (plus, of course, the input values). This process is called nth-order recursion.

Here the term recursion refers to the property that the next value of the output is computed from the n previous values of the output (plus input values). The discrete time system given by (3) is sometimes called a recursive discrete-time system since its output can be computed recursively.

It is important to note that the recursive structure is a result of the assumption that the system is finite dimensional. If the system is infinite dimensional, the output response cannot be computed recursively. Thus infinite-dimensional discrete-time systems are sometimes referred to as non-recursive systems.

**Difference Equation Example**

Suppose that the discrete-time system is given by the second-order difference equation

$y\left( kT+2T \right)-1.5y\left( kT+T \right)+y\left( kT \right)=2x\left( xT \right)~~~~~~~~~~~~~\left( 4 \right)$

Let us take the input x (kT) to be the discrete-time unit step function u (kT). We shall take the initial conditions to be

y (-1)=1, y(-2T)=2

Now rewriting (4) in the form

$y\left( kT+2T \right)=1.5y\left( kT+T \right)-y\left( kT \right)+2x\left( xT \right)~~~~~~~~~~~~~~~~~~~~~~\left( 5 \right)$

We shall compute the output values y (0), y (T), y (2T), y (3T) by solving (5) recursively. First, setting k=-2 in (5), we have

$y\left( 0 \right)=1.5y\left( -T \right)-y\left( -2T \right)+2x\left( -2T \right)=\left( 1.5 \right)\left( 1 \right)-2+2\left( 0 \right)=-0.5$

Setting k=-1 in (5), we get

$y\left( T \right)=1.5y\left( 0 \right)-y\left( -T \right)+2x\left( -T \right)=\left( 1.5 \right)\left( -0.5 \right)-1+2\left( 0 \right)=-1.75$

Continuing, we have

$y\left( 2T \right)=1.5y\left( T \right)-y\left( 0 \right)+2x\left( 0 \right)=\left( 1.5 \right)\left( -1.75 \right)+0.5+2\left( 1 \right)=-0.125$

$y\left( 3T \right)=1.5y\left( 2T \right)-y\left( T \right)+2x\left( T \right)=\left( 1.5 \right)\left( -0.125 \right)+1.75+2\left( 1 \right)=3.5625$

**Difference Equation Solution By Analytical Expression**

The recursion process will yield the values of y(kT) for any finite range of integer values of k. if we want an analytical expression for the solution y(k) to (1) or (3) that is valid for all integer k ⦥0, we must use techniques for solving linear difference equations. Here we shall derive the solution to the general first–order linear different equation

$y\left( kT+T \right)=-ay\left( kT \right)+{{b}_{0}}x\left( kT \right)+{{b}_{1}}x\left( kT+T \right)~~~~~~~~~~~~~\left( 6 \right)$

Setting k=-1 in (6), we have

$y\left( 0 \right)=-ay\left( -T \right)+{{b}_{0}}x\left( -T \right)+{{b}_{1}}x\left( 0 \right)~~~~~~~~~~~~\left( 7 \right)$

Setting k=0 in (6) gives

$y\left( T \right)=-ay\left( 0 \right)+{{b}_{0}}x\left( 0 \right)+{{b}_{1}}x\left( T \right)~~~~~~~~~~~\left( 8 \right)$

Inserting the expression (7) for y(0) into (8), we have

$y\left( T \right)={{a}^{2}}y\left( -T \right)-a{{b}_{0}}x\left( -T \right)-a{{b}_{1}}x\left( 0 \right)+{{b}_{0}}x\left( 0 \right)+{{b}_{1}}x\left( T \right)={{a}^{2}}y\left( -T \right)-a{{b}_{0}}x\left( -T \right)+\left( {{b}_{0}}-a{{b}_{1}} \right)x\left( 0 \right)+{{b}_{1}}x\left( T \right)~~~~~~~~~~~~~~\left( 9 \right)$

Setting k=1 in (6) gives

$y\left( 2T \right)=-ay\left( T \right)+{{b}_{0}}x\left( T \right)+{{b}_{1}}x\left( 2T \right)~~~~~~~~~~~~~~~~~~~~\left( 10 \right)$

Inserting the expression (9) for y(T) into (10), we have

$y\left( 2T \right)=-{{a}^{3}}y\left( -T \right)+{{a}^{2}}{{b}_{0}}x\left( -T \right)-a\left( {{b}_{0}}-a{{b}_{1}} \right)x\left( 0 \right)-a{{b}_{1}}x\left( T \right)+{{b}_{0}}x\left( T \right)+{{b}_{1}}x\left( 2T \right)$

$=-{{a}^{3}}y\left( -T \right)+{{a}^{2}}{{b}_{0}}x\left( -T \right)-a\left( {{b}_{0}}-a{{b}_{1}} \right)x\left( 0 \right)+\left( {{b}_{0}}-a{{b}_{1}} \right)x\left( T \right)+{{b}_{1}}x\left( 2T \right)~~~~~~~~~~\left( 11 \right)$

From the pattern in (9) and (11), we have that

$y\left( kT \right)={{\left( -a \right)}^{k+1}}y\left( -T \right)+{{\left( -a \right)}^{k}}{{b}_{0}}x\left( -T \right)+\underset{i=0}{\mathop{\overset{k-1}{\mathop{\sum }}\,}}\,\left[ {{\left( -a \right)}^{k-1-i}}\left( {{b}_{0}}-a{{b}_{1}} \right)x\left( iT \right) \right]+{{b}_{1}}x\left( kT \right)~~~~~~~~~~~~~~~~~~~~~~~~\left( 12 \right)$

Equation (7) and (12) give the complete output response y (kT) for k≥0 resulting from initial conditions y(-T), initial input x(-T) and input x (kT) applied for k=0,1,2,…….

Analytical expressions for the solution of second-order and higher-order linear constant-coefficient difference equations can be derived in terms of matrix equations.