AE-777A (Optimal Space Flight Control)
Quiz No. 1 (Solution)
For a system governed by the following differential equations:
ξ˙1 = ξ2
1
ξ˙2 = η − ξ23
3
where ξ1 (t) ∈ R and ξ2 (t) ∈ R are state variables, and η(t) ∈ R is the control
input:
(a) Linearize the system about the reference solution, ξ1 = c = const., ξ2 =
0, and determine the state-space coefficient matrices, A and B of the
linearized system.
(b) Find the state-transition matrix of the linearized system.
(c) Investigate the stability of the linearized system.
(d) Investigate the controllability of the linearized system.
(e) Is the linearized system observable with y = ξ2 being the only output?
(f) If possible, design a state-feedback regulator for the linearized system such
that the closed-loop characteristic polynomial is the following:
s2 + 2s + 2 = 0
1
Solution:
(a) The state equations can be expressed in the following vector form:
ξ˙ = f (ξ, η)
(1)
where ξ = (ξ1 , ξ2 )T , and
f (ξ, η) =


ξ2



− 13 ξ23 + η

(2)
To linearize the state equations about the reference solution, ξr = (c, 0)T ,
ηr = 0, we carry out their following truncated Taylor series expansion
about the reference solution:
ẋ = Ax + Bu ,
where
x=


 x1 

x2
= ξ − ξr =

(3)


 ξ1 − c 
ξ2

(4)

and u = η − ηr = η, with A, B being the following Jacobian matrices:


0
1
∂f

(ξr , ηr ) = 
A =
∂ξ
2
0
−ξ2 ξ =0
2


0
1

= 
(5)
0
0
 
0
∂f
(6)
(ξr , ηr ) =  
B=
∂η
1
(b) We calculate the state-transition matrix as follows:

eAt
s
−1
0
s
= L−1 (sI − A)−1 = L−1 

= L−1 

1
s
1
s2
0
1
s
1
t
0
1
= 
−1




,
2
(t ≥ 0) .
(7)
(c) Stability analysis of the linearized system reveals the following characteristic equation:


s
−1
 = s2 = 0 ,
det(sI − A) = det 
(8)
0
s
whose roots are
s1,2 = 0 .
(9)
Since both the eigenvalues of A are zeros, the system is unstable.
(d) Controllability is analyzed using the following test matrix for controllability:
P
=
(B, AB)

0
= 
1
1


(10)
0
whose rank equals two, the order of the system. Hence, the system is
controllable.
(e) Observability with only the state variable, ξ2 = x2 = y, being measured
as the output yields C = (0, 1), and is analyzed using the following test
matrix for observability:
N
(C T , AT C T )


0
0

= 
1
0
=
(11)
whose rank equals one, which is less than the order of the system. Hence,
the system is unobservable.
(f) The linear state-feedback control law is given by:
u = −Kx
(12)
where x = (x1 , x2 )T is the state vector and K = (k1 , k2 ) is the regulator
gain matrix. The linearized state equations are obtained in Part (a) to be
the following:
ẋ = Ax + Bu ,
(13)
where the state coefficient matrices are the following:
0
1
0
A=
B=
0
0
1
(14)
Substituting Eq.(12) into Eq.(13) we have the following closed-loop state
equation:
ẋ = (A − BK)x ,
(15)
whose characteristic polynomial is given by
3
det(sI − A + BK)
s
k1
=
det
=
s2 + k2 s + k1
−1
s + k2
(16)
Comparing Eq.(18) with the given characteristic polynomial, s2 + 2s + 2,
we get
k1 = k2 = 2 ,
(17)
or K = (2, 2).
Note: The same result as Eq.(17) is obtained by Ackermann’s formula (see
Lecture 4):
K = (â − a)(P W )−1
(18)
4