Vrftsjtryk

I have two linear control systems that are represented by their state space models

$$left(
begin{array}{c|c}
A_1 & B_1 \
hline
C_1 & D_1 \
end{array}
right),
left(
begin{array}{c|c}
A_2 & B_2 \
hline
C_2 & D_2 \
end{array}
right)$$

where $A_i$ is the state matrix, $B_i$ is the input matrix, $C_i$ is the output matrix and $D_i$ is the feed-forward matrix. With $sigma(A_1)capsigma(A_2)=emptyset$.

The output of the first system is a vector signal of dimension $n_1$, which is the same dimension of the input signal of the second system.

I have found that the state space representation is given by:
$$Sigma_G: begin{cases}
x'(t) = Ax(t)+Bu(t)\
y(t)=Cx(t)+Du(t)
end{cases}$$
where $A=left(
begin{array}{cc}
A_1 & 0 \
B_2C_1 & A_2 \
end{array}
right)$, $B=begin{bmatrix} B_1\B_2D_1 end{bmatrix}$, $C=[D_2C_1,,,, C_2]$ and $D=[D_2D_1]$.

From all of this, I would like to show that $(A,B)$ is controllable if and only if $(A_1,B_1)$ is controllable and $rank(A_2-lambda I ,,,,,,B_2T_1(lambda))=n_2$ for all $lambda in sigma(A_2)$, where $n_2$ is the dimension of the second original state space and $T_1(s)$ is the transfer matrix of the first original state space.

My attempts/progress:

It is well documented that this transformation matrix, $T_1(lambda)$, is given by $T_1(lambda)=C_1 (lambda I-A_1)^{-1}B_1+D_1$.

For $(A,B)$ to be controllable we could first find $C=rank[B,,,AB,,,A^2B ... A^{n-1}B]$ though I'm not sure what value we should equate this to in order to show controllability. In finding $C$ I had the following workings:

$$AB=begin{bmatrix} A_1B_1\B_2C_1B_1+A_2B_2D_1 end{bmatrix}$$ and $$A^2B=begin{bmatrix} A_1^2B_1\ B_2C_1A_1B_1+A_2(B_2C_1B_1+A_2B_2D_1) end{bmatrix}.$$ Pretty soon I was able to see that the top part of the block Matrix C had the pattern of $$B_1, A_1B_1, A_1^2B_1,...$$ which resembles the condition of $(A_1,B_1)$ being controllable if and only if $$rank[B_1,A_1B_1,A_1^2B_1,...,A_1^{n-1}B_1]=n_1.$$

As for the bottom part of C, I see absolutely no pattern and so I have no idea how to approach the second part of the equivalence.

Is there a connection between solving this question and my observations?

After many wasted hours of failed attempts to solve this, I appreciate any help offered to me.

This is not a complete answer, but I think it can give some insight to the problem.

You cannot separate the top and bottom of the $C$ matrix in general. You have to use the assumption that the spectrum of $A_1$ and $A_2$ are distinct to do so.

We can prove the statement more easily with the Hautus criteria, which is equivalent to
$$begin{bmatrix}w_1^T && w_2^Tend{bmatrix}begin{bmatrix}A_1-lambda I && 0 && B_1 \ B_2 C_1 && A_2-lambda I && B_2 D_1end{bmatrix} = 0 iff begin{bmatrix}w_1^T && w_2^Tend{bmatrix} = 0, ~~ forall lambda in sigma(A) tag{1}$$
if and only if $(A,B)$ is controllable.

First, assume that $(A,B)$ is controllable. Then $(1)$ applies. Since the spectrum of $A_1$ and $A_2$ are distinct and $sigma(A)=sigma(A_1)cupsigma(A_2)$, we can first check for the $lambda in sigma(A_1)$. Since $lambda notin sigma(A_2)$, for any nonzero $w_2$, $w_2^T (A_2-lambda I) neq 0$, which means $(1)$ is satisfied. Now select $w_2=0$, which means
$$w_1^T begin{bmatrix}A_1-lambda I && B_1end{bmatrix} = 0 iff w_1^T = 0, ~~ forall lambda in sigma(A_1)$$
So, $(A_1,B_1)$ is controllable.

Similarly, for $lambdainsigma(A_2)$ and any nonzero $w_1$, $w_1^T(A_1-lambda I)neq0$, hence $(1)$ is satisfied. For $w_1=0$ we need
$$w_2^T begin{bmatrix}A_2-lambda I && B_2 C_1 & B_2 D_1end{bmatrix} = 0 iff w_2^T = 0, ~~ forall lambda in sigma(A_2)$$

Now, we need to show somehow this is equivalent to
$$operatorname{rank}begin{bmatrix}A_2-lambda I & B_2 T_1(lambda)end{bmatrix}=n_2, ~~ forall lambda in sigma(A_2)$$