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Typology: Study notes
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Consider the continuous time LTI system
˙x = Ax + Bu, x 2 Rn, u 2 Rk^ (1)
For the system in (1), the controllability Gramians is given, respectively, by
WC(t 0 , t 1 ) :=
∫ (^) t 1 t 0 e A(t 0 τ ) BB
e A′(t 0 τ ) dτ =
∫ (^) t 1 t 0 0 e At BB
e A′t dt
The controllability matrix of the system in (1) is defined as
C := [B AB A
B A n 1 B]n(kn) (2)
Consider the parallel RC circuit shown in figure 1. The controllability matrix for the
electrical circuit is given by
C = [B AB] =
[ 1 R 1 C 1 ^
R 2 C 2 ^
]
ControllabLe Systems
Consider the following continuous time LTV system
x^ ˙ = A(t)x + B(t)u, x 2 R
n , u 2 R
k (3)
Definition 7.2.
Given two times t 1 > t 0 0, the system in (3), or simply the pair (A(.), B(.)), is (completely
state- ) controllable on [t 0 , t 1 ] if C [t 0 , t 1 ] = Rn^ i.e. if every state can be transferred to the
origin.
Consider the dynamics of the cart-pendulum system discussed in Lecture 1. The
dynamics of the cart-pendulum system is
˙z =
d
dt
2
6 6 6 4
x
θ
x^ ˙
θ^ ˙
3
7 7 7 5
=
2 6 6 6 6 4 x^ ˙
θ^ ˙ Jtmlsθ θ˙^2 kJt ˙x+cmlcθ θ˙+m^2 l^2 gcθ sθ MtJt m^2 l^2 c^2 θ kmlcθ ˙x m^2 l^2 cθ sθ θ˙^2 cMt θ˙ Mtmglsθ MtJt m^2 l^2 c^2 θ
3 7 7 7 7 5
2 6 6 6 6 4 0
0 Jt MtJt m^2 l^2 c^2 θ mlcθ MtJt m^2 l^2 c^2 θ
3 7 7 7 7 5
u
The equilibrium points of the system are [x 0 0 0]
i.e. the pendulum with upright
position and [x π 0 0]
, the pendulum with down position, 8 x 2 R.
Assume that the system stays within a small neighbourhood with the pendulum in the
upright position. Then
sin θ θ, cos θ 1 , θ˙
0
) The linearized dynamics leads to a set of linear state space equation
z^ ˙ =
2
6 6 6 4
0 0 1 0
0 0 0 1
0 m^2 l^2 g MtJt m^2 l^2 kJt MtJt m^2 l^2
cml MtJt m^2 l^2 0 Mtmgl MtJt m^2 l^2
kml MtJt m^2 l^2 cMt MtJt m^2 l^2
3
7 7 7 5
| {z } A
z +
2
6 6 6 4
0
0 Jt MtJt m^2 l^2 ml MtJt m^2 l^2
3
7 7 7 5
| {z } B
u
Summary: Module 12 Lecture 1
▶ Controllability Matrix
▶ Controllable Systems
Contents: Module 12 Lecture 2
▶ Controllability Tests
C = [B AB A
B A
B], where
B =
2
6 6 6 4
0
0 Jt MtJt m^2 l^2 ml MtJt m^2 l^2
3
7 7 7 5
, AB =
2 6 6 6 6 4 Jt JtMt l^2 m^2 lm l^2 m^2 JtMt