Say we have the following system:
x′=Axt+But
x
A
x
t
B
u
t
(1)
x=(
x1
x2
)
x
x1
x2
In this case, an example controllability question could seek
to know if there exists an input u such that:
x1ms=(
10V
1A
)
x
1
ms
10
V
1
A
Instead of deriving the general solution for what is called a system's controllable space,
XcontrXcontr,
we will simply state it and then give a justification for it.
Formally,
XcontrXcontr
is the set of all controllable states. We will define it in terms of a quantity known as the controllability matrix,
CABCAB:
CAB=(
BABA2B…An−1B
)
C
A
B
B
A
B
A
2
B
…
A
n
1
B
(2)
The controllable space can then be found by taking the image of this matrix.
Xcontr=imCAB
Xcontr
im
CAB
(3)To justify this expression, we begin with the
formal matrix equation for a system's state and substitute in
the infinite series definition of the matrix exponential. We
can then extract the AA and BB
matrices into a larger matrix multiplication.
x=∫eA(t−τ)Buτdτ=∫(I+A(t−τ)+A22t−τ2+…)Buτdτ=B∫uτdτ+AB∫t−τ1!uτdτ+A2B∫t−τ22!uτdτ+…=(
BABA2B…An−1B
)(
∫uτdτ
∫(t−τ)uτdτ
⋮
∫t−τnn!uτdτ
)
x
τ
A
t
τ
B
u
τ
τ
I
A
t
τ
A
2
2
t
τ
2
…
B
u
τ
B
τ
u
τ
A
B
τ
t
τ
1
u
τ
A
2
B
τ
t
τ
2
2
u
τ
…
B
A
B
A
2
B
…
A
n
1
B
τ
u
τ
τ
t
τ
u
τ
⋮
τ
t
τ
n
n
u
τ
(4)
As the second term in the multiplication is dependent on
uu,
it can be thought of as a free variable. Therefore, the set
of possible values for
xx
is dependent on the image of first term, which can be seen to be the controllability matrix as defined above.
Continuing the example circuit started above, we can get a better feel for what controllability means. Here are the state equations:
x1′=-1R1Cx1+1R1Cu
x1
-1
R1
C
x1
1
R1
C
u
x2′=−(R2Lx2)+1Lu
x2
R2
L
x2
1
L
u
Pulling the AA and
BB
matrices out of these equations, we can compute the controllability matrix
CAB=(
AAB
)
C
A
B
AAB
.
Note that as it is only a second order system, the controllability matrix is only two-dimensional.
CAB=(
1R1C-1R1C2
1L−R2L2
)
C
A
B
1
R1
C
-1
R1
C
2
1
L
R2
L
2
Immediately, we can understand some things about
the system by looking at the rank of the CC
matrix. Let's look at the determinant:
detC=1LR1C(−R2L+1R1C)
C
1
L
R1
C
R2
L
1
R1
C
If the determinant of the controllability matrix is non-zero, then
Xcontr=imC=ℝ2
Xcontr
im
C
ℝ2
;
the system is completely controllable. For this to happen we'd need to ensure that
R2L≠1R1C
R2
L
1
R1
C
.
However, if this inequality is not satisfied and the
determinant of the controllability matrix is 00,
then we know that it is not full rank. If it is not full rank, then
XcontrXcontr
will not span the entire space and the system is not completely controllable. The physical effect here is resonance in the circuit. This reduces our controllability matrix to only one dimension (the two columns are linearly dependent).
Xcontr=span(
1R1C
1L
)
Xcontr
span
1
R1
C
1
L
