Exercise 4. Solve the homogeneous difference equations.
Exercise 4
(a).
with
.
Hint. Get
.
Solution 4 (a).
See text and/or instructor's solution manual.
Answer.
.
Remark. It is our goal to learn Z-transforms and use residues.
Solution.
Method 1. The characteristic
equation
has roots
and
.
The general solution is
.
Solve the linear system
and get
and
.
Therefore,
.
Method
2. Take the
z-transform of both sides and use the initial conditions
:
,
and then get
.
New solve for
and
obtain:
.
Using
Tables. Using
Table
9.1 and the
formula
we
get:
Remark. The details for the partial fraction expansion are at the bottom of the page.
Therefore,
.
We are done.
Alternative
Solution Using Tables. Using
ordinary partial fractions and Table
9.1
and the formula
we
get:
Since
and
we
can write
and
for
.
Therefore, the solution has the following form:
We are done.
Aside. The commands for the ordinary partial fraction expansion are:
Here
and
, and
we can corroborate this solution.
We are really done.
Using
Residues. Calculate
the residues
at
the poles
.
Therefore,
We are really really done.
Aside. We can let Mathematica double check our work.
Aside. The Maple
commands are similar
![]()
![]()
![]()
![]()
We are really really really done.
Aside. We can use Mathematica's Rsolve subroutine.
Aside. The Maple
command is similar
![]()
We are really really really really done.
Aside. We can graph some of the terms in the sequence.
![[Graphics:../Images/ZTransformDEModHome_gr_850.gif]](../Images/ZTransformDEModHome_gr_850.gif)
The
sequence
.
We are really really really really really done.
The Details for the Partial Fractions.
Aside. How can we
expand
into
the proper partial fractions?
It is natural to use the standard partial fraction expansion and the
command:
However, as we have seen, this will produce a solution involving
the
functions.
This can be overcome if we use a special partial fraction expansion
that is easier to use with Table
9.1.
Method
(i). Use the following algebra
steps
Method
(ii). Find the linear combination
of
,
.
Equate the numerators
,
and solve the linear system
![[Graphics:../Images/ZTransformDEModHome_gr_860.gif]](../Images/ZTransformDEModHome_gr_860.gif)
and get
.
Therefore, the desired partial fraction form is
.
Aside. The Mathematica commands for Method (ii) are
Method
(iii). (For distinct
real roots) First make the
substitution
in
and
get
.
Then use the standard procedure for expanding in partial
fractions
.
Then make the substitution
in
and
get
Therefore, the desired partial fraction form is
.
Aside. The Mathematica commands for Method (iii) are
Now we have the desired partial fraction form:
.
This solution is complements of the authors.
(c) 2008 John H. Mathews, Russell W. Howell