Exercise 9. Use
residues to find
.
9 (c).
. Hint.
.
Solution 9 (c).
See text and/or instructor's solution manual.
Answer.
.
Aside. We
could use Table
9.1 of z-transforms and get
the solution
Remark. The details for the partial fraction expansion are at the bottom of the page.
Solution. Using
residues we get
and
Thus,
Therefore,
.
We are done.
Aside. We can let Mathematica double check our work.
Aside. We can use Mathematica's Limit and Residue subroutines.
The
Maple code using limits is similar
![]()
![]()
![]()
The
Maple code using residues is similar
![]()
![]()
![]()
We can
use Maple's subroutine to find the inverse.
![]()
We are really done.
Aside. We can graph some of the terms in the sequence.
![[Graphics:../Images/ZTransformIntroModHome_gr_566.gif]](../Images/ZTransformIntroModHome_gr_566.gif)
The
sequence
.
We are really really done.
The Details for the Partial Fractions.
Aside. How
can we expand
into
the proper partial fractions?
It is natural to try the command:
But this is not the desired form for using Table 9.1 of z-transforms.
Method
(i). Use the following algebra
steps
Method
(ii). Find the linear combination
of
.
Equate the numerators
,
and solve the linear system
![[Graphics:../Images/ZTransformIntroModHome_gr_575.gif]](../Images/ZTransformIntroModHome_gr_575.gif)
and get
.
Therefore, the desired 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 form is
.
Aside. The Mathematica commands for Method (iii) are
Now use the substitutions
,
and
.
Therefore, the desired form is
.
This solution is complements of the authors.
(c) 2008 John H. Mathews, Russell W. Howell