University of Trento Automatic Control 1: Solutions Exercise 1 (15 ...

University of Trento
Prof. Alberto Bemporad April 22, 2010
Automatic Control 1: Solutions
Exercise 1 (15 points)
The system equations are
{
A1 ḣ 1 (t) = −(a 12 + a 13 ) √ 2gh 1 (t) + ϕ(t)
A 2 ḣ 2 (t) = −a 23
√
2gh2 (t) + a 12
√
2gh1 (t)
Introducing the state and output variables as x 1 = h 1 , x 2 = h 2 , u = ϕ = u, y = h 2 , we obtain a state-space
representation with ⎧ √ ⎪⎨ ẋ 1 (t) = − a12+a13
A 2gx1
1
(t) + u(t)
√ √A 1
ẋ
⎪ 2 (t) = − a23
⎩
A 2gx2
2
(t) + a12
A 2gx1
2
(t)
y(t) = x 2 (t)
Substituting a 12 = a 23 = a 13 = 1 m 2 , A 1 = 200 m 2 , A 2 = 100 m 2 , g = 10, it yields
{
ẋ 1 (t) = − √ √
20
100 x1 (t) + u(t)
200
ẋ 2 (t) = − √ √
20
100 x2 (t) + √ √
20
100 x1 (t)
Solving ẋ 1 = ẋ 2 = 0 with ū = 10, it yields
Linearizing around these values, we obtain
[
]
A =
− ga13a23(a12+a13)
a 12ūA 2
− g(a12+a13)2
ūA 1
0
ga 12(a 12+a 13)
ūA 2
=
¯x 1 = ¯x 2 = 5/4
[
−
1
50
0
1
50
− 1 50
] [ 1
, B = 200
0
]
, C = [ 0 1 ] , D = 0
The eigenvalues of matrix A are the elements on the diagonal, which have a strictly negative real part.
Therefore, the linearized system is asymptotically stable. Finally, the observability matrix is
[ 1
]
R = 200
− 1
10000
1
0
10000
which is full rank.
stabilizable.
Exercise 2 (13 points)
Therefore the system is completely reachable and, of consequence, controllable and
[ ] 0 1
As for the first request, the reachability matrix is R = which is full rank, making it possible to
1 3
to place the poles of the closed-loop system (A + BK) at any desired position. Given the controller gain
K = [k 1 k 2 ], the characteristic polynomial is
([ ])
λ − 2 −1
p c (λ) = det(I − (A + BK)) = det
= λ 2 + (−k
1 − k 1 λ − 3 − k 2 − 5)λ − k 1 + 2k 2 + 7
2

The desired characteristic polynomial is p d (λ) = (λ + 2) 2 = λ 2 + 4λ + 4. Forcing the equalities between
the coefficients of the two polynomials, it yields k 1 = −15 and k 2 = −9. The controller is then u(t) =
−9x 1 (t) − 15x 2 (t).
[ ] 1 1
As for the second request, the observability matrix is Θ = which is full rank, and then it is possible
1 4
to place the poles of the observer at will. Given an observer gain L = [l 1 l 2 ] ′ , the characteristic polynomial
can be expressed as
([ ])
λ − 2 + l1 l
p c (λ) = det(I − (A − LC)) = det
1 − 1
= λ 2 + (l
1 + l 2 λ − 3 + l 1 + l 2 − 5)λ − 4l 1 − l 2 + 7
2
The desired characteristic polynomial is p d (λ) = (λ + 1)(λ + 2) = λ 2 + 3λ + 2. Forcing the equalities between
the coefficients of the two polynomials, it yields l 1 = −1 and l 2 = 9. The resulting Luenberger observer is
[
] [ ] [ ]
3 2
0 −1
ˆx(t) =
ˆx(t) + u(t) + y(t)
−10 −6
1
9
Placing the poles at (0.5, 0.5) is possible, because the system is completely observable. This would imply an
unstable dynamics of the observation error, which would diverge to infinity, making the observer useless.
Exercise 3 (5 points)
For the Laplace transform F (s) of a continuous-time function f(t):
• Initial value theorem: lim t→0 + f(t) = lim s→∞ sF (s)
• Final value theorem: lim t→+∞ f(t) = lim s→0 sF (s)
For the Zeta transform F (z) of a function f(k):
• Initial value theorem: f(0) = lim z→∞ F (z)
• Final value theorem: lim k→+∞ f(k) = lim z→1 (z − 1)F (z)