Exercise 12
(a). Solve
with
.
Hint. Get
.
Solution 12 (a).
See text and/or instructor's solution manual.
Answer.
.
Alternative
Answer.
.
Remark. The preferred method to use involves Z-transforms, limits and residues.
Solution. 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
formula
from
Table
9.1
and the substitution
and
and
get
Aside. We can verify the substitution.
We are done.
Using
Residues. Calculate
residues of
at
the poles
.
At the conjugate pole we can use the computation
.
Thus,
Therefore,
.
We are done.
Aside. We can let Mathematica double check our work.
Remark. The
following two computations were done using Mathematica 7.
![]()
![]()
![]()
![]()
Remark. The
following computation was done using Mathematica 7.
![]()
![]()
Now use limits.
First, find the limit ![[Graphics:../Images/ZTransformDEModHome_gr_2522.gif]](../Images/ZTransformDEModHome_gr_2522.gif)
Second, find the limit ![[Graphics:../Images/ZTransformDEModHome_gr_2531.gif]](../Images/ZTransformDEModHome_gr_2531.gif)
Then
Aside. The Maple
commands are similar
![]()
![]()
![]()
![]()
We are really done.
Aside. We can use Mathematica's Rsolve subroutine.
Aside. The Maple
command is similar
![]()
For curiosities sake, we can compute some of the terms in the sequence with the various formulae.
We leave it for the reader to prove that the four forms of the answer are equivalent.
We are really really done.
Aside. We can graph some of the terms in the sequence.
![[Graphics:../Images/ZTransformDEModHome_gr_2572.gif]](../Images/ZTransformDEModHome_gr_2572.gif)
The
sequence
.
We are 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
function.
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
Now we have the desired partial fraction form:
.
Method
(ii). Find the linear combination
of
,
.
Equate the numerators
,
and solve the linear system
and get
.
Therefore, the desired partial fraction form is
.
Aside. The Mathematica commands for Method (ii) are
Method
(iii). The
substitution
does
not apply when there are complex roots.
This solution is complements of the authors.
(c) 2008 John H. Mathews, Russell W. Howell