0
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$
    – Pete W
    May 22 at 13:32
  • $\begingroup$ 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? $\endgroup$ May 22 at 13:37
  • $\begingroup$ 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. $\endgroup$
    – Pete W
    May 22 at 14:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.