I am looking at the Kalman decomposition of a linear system into 'controllable' and 'uncontrollble' subspaces. The references I am using are these lecture notes and section 3.3 of 'Robust and Optimal Control' by Zhou and Doyle.
If we have a linear system governed by:
$\dot{x} = Ax + Bu$
Where $x$ is the system variable vector, $u$ is an input vector and $A,B$ are matrices of appropriate dimension, then we can perform a Kalman decomposition, changing the variables to $x' = Tx$, which gives an equation of the form:
$ \frac{dx'}{dt} =
\begin{bmatrix}
A_{11} & A_{12} \\
0 & A_{22} \\
\end{bmatrix} x' + \begin{bmatrix}
B_1 \\
0 \\
\end{bmatrix} u$
We can now write $x' = \begin{bmatrix}
x'_1 \\
x'_2 \\
\end{bmatrix}
$ so:
$ \frac{d}{dt} \begin{bmatrix}
x'_1 \\
x'_2 \\
\end{bmatrix} =
\begin{bmatrix}
A_{11} & A_{12} \\
0 & A_{22} \\
\end{bmatrix} \begin{bmatrix}
x'_1 \\
x'_2 \\
\end{bmatrix} + \begin{bmatrix}
B_1 \\
0 \\
\end{bmatrix} \begin{bmatrix}
u_1 \\
u_2 \\
\end{bmatrix}$
The texts I am using now go on to say that the vector $x_1'$ is controllable and $x_2'$ is not.
It is not clear to me that $x_1'$ is controllable, since its evolution involves the uncontrollable variable $x_2'$ (multiplying out the first matrix gives $A_{11} x_1' + A_{12}x_2' + …$) . The texts I am using go on to show that the pair ($A_{11}, B_1$) is controllable, whilst ignoring the $A_{12}$ term.
My question is this: why can we say that the variable $x_1'$ is controllable, given that the expression for its evolution involves $x_2'$, which is uncontrollable?
Best Answer
Because the $A_{12}$ term is irrelevant for controllability of the $x_1'$ states. To see this write the equation for $x_1'$ $$ \begin{align} \dot{x}_1'(t) &= A_{11} x_1'(t) + A_{12} x_2'(t) + B_1 u_1(t) \tag{1} \\ &= A_{11} x_1'(t) + A_{12} e^{A_{22} t} x_2'(0) + B_1 u_1(t) \end{align}$$ If $x_2'(0) = 0$ then it is obvious. But we can also solve this equation as follows: $$ x_1'(t) = e^{A_{11} t} x_1'(0) + \int_0^t e^{A_{11} (t - \tau)} A_{12} e^{A_{22} \tau} x_2'(0) d\tau + \int_0^t e^{A_{11} (t - \tau)} B_1 u_1(\tau) d\tau $$ Now, to reach an arbitrary final state $x_f$ at time $t_f$, we can select $u_1$ as $$ u_1(t) = B_1^T e^{A_{11}^T (t_f - t)} W_c^{-1} (t_f) \left( x_f - e^{A_{11} t_f} x_1'(0) - \int_0^{t_f} e^{A_{11} (t_f - \tau)} A_{12} e^{A_{22} \tau} x_2'(0) d\tau \right)$$ where $$ W_c (t) = \int_0^t e^{A_{11} (t - \tau)} B_1 B_1^T e^{A_{11}^T (t - \tau)} d\tau $$ is the controllability gramian.
So the system $(1)$ is controllable if and only if $(A_{11}, B_1)$ is controllable. Basically we can cancel out the parts that comes from the initial conditions. This means we can select $x(0) = 0$ without losing generality to obtain the results.