![]()
![]()
for
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
.
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
.
![[Graphics:Images/ZTransformDEMod_gr_32.gif]](ztransform/ZTransformDEMod/Images/ZTransformDEMod_gr_32.gif)
Example
9.13. Solve
with
.
![[Graphics:Images/ZTransformDEMod_gr_53.gif]](ztransform/ZTransformDEMod/Images/ZTransformDEMod_gr_53.gif)
Example
9.14. Solve
with
.
![[Graphics:Images/ZTransformDEMod_gr_76.gif]](ztransform/ZTransformDEMod/Images/ZTransformDEMod_gr_76.gif)
Higher Order
Difference Equations
The general form of a
order linear constant coefficient
difference equation (LCCDE), is
(9-14) ![[Graphics:Images/ZTransformDEMod_gr_105.gif]](ztransform/ZTransformDEMod/Images/ZTransformDEMod_gr_105.gif)
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
.
![[Graphics:Images/ZTransformDEMod_gr_152.gif]](ztransform/ZTransformDEMod/Images/ZTransformDEMod_gr_152.gif)
Example 9.16
(a). Use
z-transform
methods to solve
with
.
9.16
(b). Use
z-transform
methods to solve
with
.
![[Graphics:Images/ZTransformDEMod_gr_218.gif]](ztransform/ZTransformDEMod/Images/ZTransformDEMod_gr_218.gif)
Example 9.17
(a). Use
z-transform
methods to solve
with
.
9.17
(b). Use
z-transform
methods to solve
with
.
![[Graphics:Images/ZTransformDEMod_gr_284.gif]](ztransform/ZTransformDEMod/Images/ZTransformDEMod_gr_284.gif)
Example
9.18. Solve
with
.
![[Graphics:Images/ZTransformDEMod_gr_356.gif]](ztransform/ZTransformDEMod/Images/ZTransformDEMod_gr_356.gif)
Example
9.19. Solve
with
.
![[Graphics:Images/ZTransformDEMod_gr_402.gif]](ztransform/ZTransformDEMod/Images/ZTransformDEMod_gr_402.gif)
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
.
![[Graphics:Images/ZTransformDEMod_gr_454.gif]](ztransform/ZTransformDEMod/Images/ZTransformDEMod_gr_454.gif)
Figure
9.2. A typical solution
to
.
![[Graphics:Images/ZTransformDEMod_gr_456.gif]](ztransform/ZTransformDEMod/Images/ZTransformDEMod_gr_456.gif)
Figure
9.3. A typical solution
to
.
Exercises for Section 9.2. Second Order Difference Equations
The Next Module for Z-Transforms is
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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