# State space representation of a product of transfer functions?

In short, I suppose the question boils down to: If we are asked to find the state space representation of $$C(s) \cdot G(s)$$, where C(s) and G(s) are both transfer functions, then do we treat them as two state space models in series or not? If series/not in series, then why?

Example:

If we are given that $$C(s) = \frac{s - 1}{s + 1}$$ and $$G(s) = \frac{1}{s - 1}$$, then let us try two methods I have seen:

1. Just multiplying the two transfer functions

Thus, $$C(s)G(s) = \frac{s - 1}{s^2 - 1}$$ and this can be written as:

$$\dot x(t) = Ax(t) + Bu(t) = \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}x(t) + \begin{pmatrix} 1 \\ 0 \end{pmatrix} u(t)$$

and $$y(t) = Cx(t) = \begin{pmatrix} 1 & -1 \end{pmatrix}x(t)$$

This representation will eventually lead to the fact that there is an unobservable state.

Method 2: Treating the state spaces as in series

High level: find state space representation for both and then treat them in series. Therefore, output $$y_1$$ is the input $$u_2$$

If we take $$C(s)$$ as appearing before $$G(s)$$, then we can find the following state space representation for $$C(s) = \frac{s - 1}{s + 1} = 1 - \frac{2}{s + 1}$$ (functions of $$t$$ have been dropped for simplicity) $$\dot x_1 = - x_1 + u_1$$ and $$y_1 = -2x_1 + u_1$$

Then for $$G(s)$$, we get: $$\dot x_2 = x_2 + u_2$$ and $$y_2 = x_2$$

Then, because they are in series, we let $$u_2 = y_1 = -2x_1 + u_1$$. By stacking them together we can get: $$\begin{pmatrix} \dot x_1 \\ \dot x_2 \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ -2 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} + \begin{pmatrix} 1 \\ 1 \end{pmatrix} u_1$$

and $$y_2 = \begin{pmatrix} 0 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$$

However, this leads to the fact that there is an uncontrollable state.

Which method is correct?

Any help would be greatly appreciated.

• C(s) has a RHP zero and G(s) has a RHP pole, so any cancellation here is considered hazardous, and would indeed create an unobservable state, which shouldn't be "erased" from the system model. May 22 at 13:32
• Thanks @PeteW for the reply! Yep, that is correct (method 1 doesn't have any cancellations). So does that indicate that one of the methods is correct over the other? May 22 at 13:37
• I suppose the model that does include this troublesome internal state is "correct", in the sense that any practical work requires the criterion of internal stability, which is a more strict requirement than having stable poles/eigenvalues in the closed loop response. May 22 at 14:33