Digital Signal Processing Reference
In-Depth Information
For
t 7 0,
the inverse Laplace transform of this vector yields
1
2
- e
-t
1
2
e
-2t
+
l
-1
[(s
I
-
A
)
-1
B
U(s)] =
B
R
.
e
-t
- e
-2t
The state transition matrix was derived in Example 8.4. Hence, the complete solution of the
state equations is, from (8.28) and Example 8.4,
x
(t) =≥(t)
x
(0) +
l
-1
[(s
I
-
A
)
-1
B
U(s)]
1
2
- e
-t
+
1
2
e
-2t
2e
-t
- e
-2t
e
-t
- e
-2t
x
1
(0)
x
2
(0)
=
B
RB
R
+
B
R
,
-2e
-t
+ 2e
-2t
-e
-t
+ 2e
-2t
e
-t
- e
-2t
and the state variables are given by
1
2
x
1
(t) = (2e
-t
- e
-2t
)x
1
(0) + (e
-t
- e
-2t
)x
2
(0) +
1
2
e
-2t
1
2
+ [2x
1
(0) + x
2
(0) - 1]e
-t
- e
-t
+
=
1
2
]e
-2t
.
+ [-x
1
(0) - x
2
(0) +
In a like manner,
x
2
(t) = (-2e
-t
+ 2e
-2t
)x
1
(0) + (-e
-t
+ 2e
-2t
)x
2
(0)
+ e
-t
- e
-2t
= [-2x
1
(0) - x
2
(0) + 1]e
-t
+ [2x
1
(0) + 2x
2
(0) - 1]e
-2t
.
The output is given by
y(t) = 4x
1
(t) + 5x
2
(t) = 2 + [-2x
1
(0) - x
2
(0) + 1]e
-t
+ [6x
1
(0) + 6x
2
(0) - 3]e
-2t
.
The first-row element of can be calculated directly with the MATLAB program
S=dsolve('Dx1 = x2,Dx2 = -2*x1-3*x2,x1(0) = x10,x2(0) = x20')
S.x1
S.x2
≥(t)
x
(0)
The second-row element can be calculated in the same manner.
■
The solution of state equations is long and involved, even for a second-order
system. The necessity for reliable machine solutions, such as digital-computer simula-
tions, is evident. Almost all system analysis and design software have simulation capa-
bilities; the simulations are usually based on state models. As an example, a S
IMULINK
simulation is now discussed.
EXAMPLE 8.6
S
IMULINK
simulation
The system of Example 8.5 was simulated with S
IMULINK
. The block diagram from the simula-
tion is given in Figure 8.6(a), and the response is given in Figure 8.6(b). From Example 8.5,
we see that the two system time constants are 0.5 s and 1 s. Hence, the transient part of the
response becomes negligible after approximately four times the larger time constant, or
y(t)
