Module

for

Homogeneous Difference Equations

9.2  Second Order Homogeneous Difference Equations

Before proceeding with the
z-transform method, we mention a heuristic method based on substitution of a trial solution.  Consider the second order homogeneous linear constant-coefficient difference equation (HLCCDE)

(9-8)            ,

where
are constants.  Using the trial solution  ,  and direct substitution into (9-8) produces the equation  ,  and dividing through by produces the characteristic polynomial    and characteristic equation

(9-9)            .

There are three types of solutions to (9-8) which are determined by the nature of the roots in (9-9).

Case (i)  If     then we have real distinct roots     and  ,  and

(9-10)            .

Case (ii)  If      then we have a double real root   ,  and

(9-11)            .

Case (iii)  If      then we have complex roots     and  ,  and

(9-12)            .

The solution for case (iii) can also be written as the following linear combination

(9-13)            ,

where    and  .

Caution.  Be sure to use the value of arctan that lies in the range  .

Stability depends on the location of the roots of the characteristic polynomial.  Without loss if generality, if    then both roots lie inside the unit circle and the solutions are asymptotically stable and tends to zero as  .   If   and   then a root lies on the unit circle and the solutions are stable.  If    then there is an unstable solution.  If    then at least one root lies outside the unit circle and there is an unstable solution.

The following subroutine uses the characteristic equation to construct solutions to a second order homogeneous difference equation.

Mathematica Subroutine (Solution of a Difference Equation).

Example 9.12.  Solve     with  .

Solution 9.12.

Explore Solution 9.12.

Example 9.13.  Solve     with  .

Solution 9.13.

Explore Solution 9.13.

Example 9.14.  Solve     with  .

Solution 9.14.

Explore Solution 9.14.

Higher Order Difference Equations

T
he general form of a order linear constant coefficient difference equation (LCCDE), is

(9-14)

where
and . The sequence is given and the sequence is output.  The integer    is the order of the difference equation.  The compact form of writing this difference equation is

(9-15)            .

This formula can be expressed in a recursive form:

(9-16)            .

This form of the LCCDE explicitly shows that the present output    is a function of the past output values  ,  for  ; and the present input  ,  and the previous inputs    for  .

Now we would like to emphasize the method of z-transforms for solving difference equations.  Applying the linearity and time delay shift property of the z-transform to equation (9-15), we obtain

(9-17)            .

This can be rearranged as    and then solved for the quotient .  The sequence   can be used to construct a particular solution to (9-14), i.e.  .  This solution can be expressed using the convolution sum, as follows:

(9-18)            .

Remark.  This particular solution    does not involve initial conditions for (9-14).  We will illustrate how to use convolution at the end of this section.

Difference Equations with Initial Conditions

Often times a difference equation involves only one input on the right hand side of (9-14) and we write

then we could shift the index and use the form

.

Consider the first order linear constant coefficient difference equation (LCCDE)

(9-19)

with the initial conditions    (and implicitly we have  ).

Step (i)  Using the time forward properties
,
,

and take the z-transform of each term and get the equation

(9-20)            .

Step (ii)  Solve equation (9-20) for  .

Step (iii)  Use partial fractions to expand    in a sum of terms, and look up the inverse z-transform(s) using Table 1, to get the solution

.

Step (iv)  Alternate calculation using residues.  Perform steps (i) and (ii) listed above.  Then find    using residues

.

where    are the poles of  .

Remark  The function    has real coefficients.  Hence, if , and if and are poles, then we can use the computational fact:

(9-21)

We now show how to obtain answers to Examples 9.12-9.14 using z-transform methods.

The following subroutine uses z-transforms to construct solutions to a second order homogeneous difference equation.

Mathematica Subroutines (Solution of a Difference Equation).

Example 9.15 (a).  Use z-transform methods to solve     with  .

9.15 (b).  Use z-transform methods to solve     with  .

Solution 9.15 (a).

Solution 9.15 (b).

Explore Solution 9.15 (a).

Explore Solution 9.15 (b).

Example 9.16 (a).  Use z-transform methods to solve     with  .

9.16 (b).  Use z-transform methods to solve     with  .

Solution 9.16 (a).

Solution 9.16 (b).

Explore Solution 9.16(a).

Explore Solution 9.16 (b).

Example 9.17 (a).  Use z-transform methods to solve     with  .

9.17 (b).  Use z-transform methods to solve     with  .

Solution 9.17 (a).

Solution 9.17 (b).

Explore Solution 9.17 (a).

Explore Solution  9.17 (b).

Example 9.18.  Solve     with  .

Solution 9.18.

Explore Solution 9.18.

Example 9.19.  Solve     with  .

Solution 9.19.

Explore Solution 9.19.

Convolution for Solving a Non-homogeneous Equation

(i)    Solve the homogeneous equation      and get  .

(ii)    Use the transfer function   .

(iii)    Construct the particular solution using convolution

,   or

.

(iv)    The total solution to the nonhomogeneous difference equation is

.

Example 9.20 (a).  Find the general solution to  .

9.20 (b).  Find the general solution to  .

Figure 9.2.  A typical solution to  .

Figure 9.3.  A typical solution to  .

Solution 9.20 (a).

Solution 9.20 (b).

Explore Solution 9.20 (a).

Explore Solution 9.20 (b).

The z-Transform

(c) 2006 John H. Mathews, Russell W. Howell