Page 290 - Applied Numerical Methods Using MATLAB

VECTOR DIFFERENTIAL EQUATIONS 279
we can take the inverse Laplace transform of Eq. (6.5.5) to write

t
x(t) = φ(t)x(0) + φ(t) ∗ Bu(t) = φ(t)x(0) + φ(t − τ)Bu(τ) dτ (6.5.8)
0

For Eq. (6.5.3), we use Eq. (6.5.6) to find

−1 −1
φ(t) = L {[sI − A] }

−1 −1
s 0 0 1 −1 s −1
−1
= L − = L
0 s 0 −1 0 s + 1

1 s + 1 1
−1
= L
s(s + 1) 0 s

−t
1/s 1/s − 1/(s + 1) 1 1 − e
−1
= L = −t (6.5.9)
0 1/(s + 1) 0 e
and use Eqs. (6.5.8), (6.5.9), and u(t) = u s (t) = 1 ∀ t ≥ 0toobtain
t
−t −(t−τ)
1 1 − e 1 11 − e 0
x(t) = −t + −(t−τ) 1 dτ
0 e −1 0 e 1
0
−t t −t
e τ − e −(t−τ) t − 1 + 2e
= −t + −(t−τ) = −t (6.5.10)
−e e 1 − 2e
0
Alternatively, we can directly take the inverse transform of Eq. (6.5.5) to get
−1 −1
X(s) = [sI − A] {x(0) + [sI − A] BU(s)}
1
1 s + 11
1
0
= +
s(s + 1) 0 s −1 1 s
2
1 s + 1 1 s 1 s + 1
= = (6.5.11)
2
2
s (s + 1) 0 s −s + 1 s (s + 1) s(1 − s)
2
s + 1 1 1 2 −t
X 1 (s) = = − + , x 1 (t) = t − 1 + 2e (6.5.12a)
2
s (s + 1) s 2 s s + 1
1 − s 1 2 −t
X 2 (s) = = − , x 2 (t) = 1 − 2e (6.5.12b)
s(s + 1) s s + 1
which conforms with Eq. (6.5.10).
The MATLAB program “nm651_2.m” uses a symbolic computation routine
“ilaplace()” to get the inverse Laplace transform, uses “eval()”to evaluate