Homogeneous Difference Equations
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)
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
There are three types of solutions to (9-8) which are determined by the nature of the roots in (9-9).
(i) If then
we have real distinct roots and , and
(ii) If then
we have a double real root , and
(iii) If then
we have complex roots and , and
The solution for case (iii) can also be written as the following linear combination
where and .
Caution. Be sure to use the value of arctan that lies in the range .
Remark About Stability
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 .
Explore Solution 9.12.
Example 9.13. Solve with .
Explore Solution 9.13.
Example 9.14. Solve with .
Explore Solution 9.14.
The general form of a order linear constant coefficient difference equation (LCCDE), is
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
This formula can be expressed in a recursive form:
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
for solving difference equations. Applying the linearity
and time delay shift property of the
to equation (9-15), we obtain
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:
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.
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
first order linear constant coefficient difference equation
with the initial conditions (and implicitly we have ).
(i) Using the time forward
and take the z-transform of each term and get the equation
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
(iv) Alternate calculation using
residues. Perform steps (i)
and (ii) listed above. Then
where are the poles of .
real coefficients. Hence, if , and if
are poles, then we can use the computational fact:
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).
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).
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).
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 .
Explore Solution 9.18.
Example 9.19. Solve with .
Explore Solution 9.19.
Convolution for Solving a Non-homogeneous
(i) Solve the homogeneous equation and get .
(ii) Use the transfer function .
(iii) Construct the particular solution using convolution
(iv) The total solution to the nonhomogeneous difference equation is
(a). Find the general solution
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).
Exercises for Section 9.2. Second Order Difference Equations
The Next Module for Z-Transforms is
Introduction to Filtering
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This material is coordinated with our book Complex Analysis for Mathematics and Engineering.
(c) 2012 John H. Mathews, Russell W. Howell