B and C matrices for real modal representation of a 2x2 linear system with complex eigenvalues

Question

What are the allowable transformations on the $B$ and $C$ matrices in a linear state-space system that preserve input-to-output behavior without changing the $A$ and $D$ matrices?

I'm working with the usual state-space representation of linear system:

$$\begin{eqnarray} \frac{d}{dt}x &=& Ax + Bu \cr y &=& Cx + Du \end{eqnarray}$$

where I want to use a modal representation, namely matrix A has only real elements, and is block-diagonal with 1x1 blocks and 2x2 blocks of the form $\begin{bmatrix}a&b\cr-b&a\end{bmatrix}$ to represent complex eigenvalues. This is so that the matrix $A$ is well-conditioned.

In my case $u$ and $y$ are scalars, so $A$ is $n \times n$, $B$ is $n \times 1$, $C$ is $1 \times n$, and $D$ is $1 \times 1$. Also in my case, all the eigenvalues are unique so I don't have to worry about it being a defective matrix.

The only problem is that the writeups of modal representations always use a pure diagonal matrix, which will have complex entries for matrices with complex eigenvalues. I can figure out how to use a similarity transform with unitary matrix $T$ such that

$$\begin{eqnarray} A'&=&T^{-1}AT \cr B'&=&T^{-1}B \cr C'&=&CT \cr D'&=&D \end{eqnarray}$$

and with a suitable choice of T, I can get an $A$ matrix the way I want.

My problem is with the B and C matrices; $T$ has complex entries and seems to produce complex values for B and C. I know there are some degrees of freedom in choosing B and C (for example, I can multiply all entries of B by some constant K and divide all entries of C by K, and I get the same externally controllable + observable behavior from $u$ to $y$), but I can't figure out what I'm allowed to do to B and C if A is fixed.

I should be able to figure out how to solve the problem if I can understand the 2x2 case.

For example

$$\begin{eqnarray} \frac{d}{dt}\begin{bmatrix}x_1 \cr x_2\end{bmatrix} &=& \begin{bmatrix}-\sigma & \omega \cr -\omega & -\sigma\end{bmatrix}\begin{bmatrix}x_1 \cr x_2\end{bmatrix} + \begin{bmatrix}b_1 \cr b_2 \end{bmatrix}u \cr y &=& \begin{bmatrix}c_1 & c_2 \end{bmatrix}\begin{bmatrix}x_1 \cr x_2\end{bmatrix} + Du \end{eqnarray}$$

where $A$ has eigenvalues $-\sigma \pm j\omega$; for $D=0$ this can be used to represent the dynamic system $\frac{d^2y}{dt^2} + 2\sigma\frac{dy}{dt} + \sigma^2 + \omega^2 = k_1\frac{du}{dt} + k_2u$. There's only two degrees of freedom $(k_1,k_2)$ in the observable/controllable system, but four degrees of freedom in choosing the B and C matrices.

The transfer function is $C(sI-A)^{-1}B+D$, so something that maintains the value of $C(sI-A)^{-1}B$ will be ok. In this case as long as $C\begin{bmatrix}s+\sigma & \omega \cr -\omega & s+\sigma\end{bmatrix}B$ is preserved, the system behavior is the same.

Answer 1

Hmm. Here goes another attempt to muddle my way through this...

For pure-diagonal modal representations with diagonal matrix $A$, the state variables are all decoupled, and therefore the entries of the B and C matrices can be independently scalable:

In $H(s) = C(sI-A)^{-1}B$, the matrix $(sI-A)^{-1}$ is diagonal and this matrix product $H(s)$ (which is scalar for the case where $u$ and $y$ are scalar) looks like $\sum\limits_j b_jc_jf_j(s)$. I have the freedom to set $b'_j = k_jb_j$ for any $k_j$ of my choice, and then this causes $c'_j = \frac{1}{k_j}c_j$ to maintain the same result. For conjugate pair eigenvalues $\lambda_{j_1}, \lambda_{j_2}$, the products $b_{j_1}c_{j_1}$ and $b_{j_2}c_{j_2}$ must also be conjugate pairs if the resulting $H(s)$ is real.

If I use $T=\frac{1}{\sqrt{2}}\begin{bmatrix}1&j\cr j&1\end{bmatrix}$, then the I just need to choose $k_1$ and $k_2$ so that the new matrix $B'=\frac{1}{\sqrt{2}}\begin{bmatrix}1&-j\cr -j&1\end{bmatrix}\begin{bmatrix}k_1b_1\cr k_2b_2\end{bmatrix} = \frac{1}{\sqrt{2}}\begin{bmatrix}k_1b_1 -jk_2b_2\cr k_2b_2 -jk_1b_1\end{bmatrix}$ has real entries.

Setting imaginary parts to zero, this means $k_{i1} b_{r1}+k_{r1}b_{i1}-k_{r2}b_{r2}+k_{i2}b_{i2} = 0$ and $k_{i2} b_{r2}+k_{r2}b_{i2}-k_{r1}b_{r1}+k_{i1}b_{i1} = 0$. Not sure where to go from here. I should be able to arbitrarily choose $|k_1|=|k_2|=1$ so $k_{r1}=\cos\theta_1$, $k_{r2}=\cos\theta_2$, $k_{i1}=\sin\theta_1$, $k_{i2}=\sin\theta_2$, and this will give me two equations with two unknowns but I am not sure how to solve it.

Solution outlined here: constrained complex number equation requiring imaginary part to be zero, but my hypothesis that I can choose $|k_1|=|k_2|=1$ is false.