Example 9.11. Given
the repeated dosage drug level model
with the initial condition
.
9.11 (d). Use convolution to find the
solution.
Explore Solution 9.11 (d).
The solution to the homogeneous equation
is
.
![[Graphics:../Images/ZTransformIntroMod_gr_615.gif]](../Images/ZTransformIntroMod_gr_615.gif)
The transfer function is
and
the unit-sample response is
.
![]()
The input sequence is
and
it's z-transform is
.
![[Graphics:../Images/ZTransformIntroMod_gr_623.gif]](../Images/ZTransformIntroMod_gr_623.gif)
The particular solution is calculated with the
formula
as
follows:
![[Graphics:../Images/ZTransformIntroMod_gr_626.gif]](../Images/ZTransformIntroMod_gr_626.gif)
Details
![[Graphics:../Images/ZTransformIntroMod_gr_628.gif]](../Images/ZTransformIntroMod_gr_628.gif)
In convolution form
and
we have
![]()
![[Graphics:../Images/ZTransformIntroMod_gr_632.gif]](../Images/ZTransformIntroMod_gr_632.gif)
Remark. The
particular solution
obtained
by using convolution has the initial condition
.
The total solution to the nonhomogeneous difference equation
is
![[Graphics:../Images/ZTransformIntroMod_gr_638.gif]](../Images/ZTransformIntroMod_gr_638.gif)
Now we compute
, and
solve for the constant
which
will produce the proper initial condition. Therefore,
,
which can be arranged to yield
.
![[Graphics:../Images/ZTransformIntroMod_gr_644.gif]](../Images/ZTransformIntroMod_gr_644.gif)
We can explore the situation when
,
and
.
![[Graphics:../Images/ZTransformIntroMod_gr_649.gif]](../Images/ZTransformIntroMod_gr_649.gif)
![]()
Figure 9.1. The
solution to
with
.