Digital Signal Processing Reference
In-Depth Information
dP
dt
+ 2P = 0 + 2P = 2,
or
y
p
(t) = P = 1.
From (3.51), the general solution is
y(t) = y
c
(t) + y
p
(t) = Ce
-2t
+ 1.
We now evaluate the coefficient
C
. The initial condition is given as
y(0) = 4.
The
general solution
y(t)
evaluated at
t = 0
yields
y(0) = y
c
(0) + y
p
(0) = [Ce
-2t
+ 1]
t = 0
= C + 1 = 4 Q C = 3.
The total solution is then
y(t) = 1 + 3e
-2t
.
This solution is verified with the MATLAB program
dsolve('Dy+2*y=2, y(0)=4')
ezplot(y)
If this example is not clear, the reader should study Appendix E.
■
Verification of the response of Example 3.11
EXAMPLE 3.12
We now verify the solution in Example 3.11 by substitution into the differential equation.
Thus,
dy(t)
dt
y= 1 + 3e
-2t
=-6e
-2t
+ 2(1 + 3e
-2t
) = 2,
+ 2y(t)
and the solution checks. In addition,
y(0) = (1 + 3e
-2t
)
t = 0
= 1 + 3 = 4,
and the initial condition checks. Hence, the solution is verified.
■
Consider the natural response for the
n
th-order system
n
d
k
y(t)
dt
k
m
d
k
x(t)
dt
k
[eq(3.50)]
a
k
=
a
b
k
.
a
k= 0
k= 0
