How do I find specific transfer functions from a plant of a MIMO system?

Question

Suppose I have a physical system, for example a mass-spring-damper system, which has been written in state space. Now, suppose that the matrices describing the system are \$A,B,C,D\$.

Moreover, let' s say it is a MIMO system, and I have \$6\$ inputs, which are \$u_{1},u_{2},d_{1},d_{2},n_{1},n_{2}\$, where the first two are the outputs of the controller, the second pair are the disturbances and the last ones are the noises.

Ans there are \$4\$ outputs, which are the displacement of a mass, defined as \$z_{1},z_{2}\$ each of which has two components.

So, I define the plant as:

G = ss(A,B,C,D)

now, I have my plant defined. Suppose now I want to find here the transfer functions of the systems, such as the sensitivity function, the complementary sensitivity function and the control sensitivity function, how could I do?

I know that I will have smething like:

\$\begin{bmatrix} z_{11}\\ z_{12}\\ z_{21}\\ z_{2,2} \end{bmatrix} = \begin{bmatrix} & & & \\ & & & \\ & & & \\ & & & \\ & & & \\ & & & \end{bmatrix} \cdot \begin{bmatrix} d_1\\ d_2\\ u_1\\ u_2\\ n_1\\ n_2 \end{bmatrix}\$

and to know the transfer function I need to look at the appropriate entry in the transfer matrix, but how do I know that the order of the inputs and of the outputs is this?

So, what I mean is that I could also have

\$\begin{bmatrix} z_{21}\\ z_{22}\\ z_{11}\\ z_{12} \end{bmatrix} = \begin{bmatrix} & & & \\ & & & \\ & & & \\ & & & \\ & & & \\ & & & \end{bmatrix} \cdot \begin{bmatrix} u_1\\ u_2\\ d_1\\ d_2\\ n_1\\ n_2 \end{bmatrix}\$

or other combinations, so if I do \$G(1,1)\$, in the first case I obtain a transfer function, and in the second case I obtain a different transfer function.

So, how can I do?

Answer 1

For any linear time-invariant system the transfer function will be

$$ G(s) = YU^{-1} = C (sI-A)^{-1}B + D,$$

and the blank matrix in you question will be \$G(s) \in \mathcal{R}^{4 \times 6} \$, with the top element of the output matrix \$ z\$

$$z_{1} = ([C_{1i}] (sI-A)^{-1}B + D)U,$$

where \$ [C_{1i}] \$ is the row vector with the elements in the first row of \$C\$. Do notice that \$z_1(u_1,u_2,\dots,n_2) \$.

If you permute the inputs in \$U\$ it is the same as applying an invertible matrix \$ P\$ to have a \$ \hat{U} = PU\$. So, as long as \$ P\$ is just a permutation (a single 1 per line and no columns has more than a single 1), if \$G(1,1)\$ gives you a TF you will be able to find that same TF somewhere in \$ \hat{G} \$.

Moreover, if you have

$$ \begin{bmatrix} d_1\\ d_2\\ u_1\\ u_2\\ n_1\\ n_2 \end{bmatrix} \rightarrow G,$$

$$ \begin{bmatrix} u_1\\ u_2\\ d_1\\ d_2\\ n_1\\ n_2 \end{bmatrix} \rightarrow \hat{G},$$

The permutation is

$$ P = \begin{bmatrix} 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}$$

and leads to

$$ G(1,1) = \hat{G}(1,3). $$