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I was supposed to find state space representation and its matrices of this system:

system in s-domain

and I have no idea, how to do this. We were told not to transfer the system to time domain, but I can only do state space representation from time domain schemas.

When I tried to solve this, I got matrices $$A = \left( \begin{array}{ccc@{\ }r} -a & k \\ -b & -p \\ \end{array} \right)$$

$$B = \left( \begin{array}{ccc@{\ }r} 0 \\ b \\ \end{array} \right)$$

$$C = \left( \begin{array}{ccc@{\ }r} 1 & 0 \\ \end{array} \right)$$

$$D = \left( \begin{array}{ccc@{\ }r} 0 \end{array} \right)$$

Is that right? If not, how sould I solve it?

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  • $\begingroup$ You should define what the matrices $A, B, C, D$ refer to in your state-space model. Then it should be pretty easy to tell if you're right. $\endgroup$
    – Jason R
    Commented Apr 13, 2013 at 13:34
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    $\begingroup$ I thought it is globaly used. Exactly what matrices at en.wikipedia.org/wiki/State_space_representation means. $\endgroup$
    – user50222
    Commented Apr 13, 2013 at 14:43
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    $\begingroup$ Would it help if you knew that the transfer function of a system described by $[A,B,C,D]$ is $C(s{\bf I} - A)^{-1}B + D$ ? $\endgroup$
    – Peter K. ♦
    Commented Apr 13, 2013 at 16:55
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    $\begingroup$ You say I have no idea, how to do this and yet you present an answer. Can you detail how you got your answer, and what you don't understand about it? $\endgroup$
    – Peter K. ♦
    Commented Apr 14, 2013 at 13:59
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    $\begingroup$ I went like: $$X_2(s) = (U(s)-X_1(s)) \cdot \frac {b}{s+p} $$ $$X_1(s) = X_2(s) \cdot \frac {k}{s+a}$$ That could mean: $$sX_2(s) + pX_2(s) = bU(s) - bX_1(s) \to \dot x_2(t) = bu(t) - bx_1(t) - px_2(t)$$ $$sX_1(s) + aX_1(s) = kX_2(s) \to \dot x_1(t) = -ax_1(t) + kx_2(t)$$ and output sould be: $$y(t) = x_1(t)$$ that would lead to matrices I wrote. But I don't know, if I can do that this way, or if that is what was the task, cause we were told not to transfer to time domain, but I can't imagine how to do it without transfer I did. $\endgroup$
    – user50222
    Commented Apr 14, 2013 at 14:17

2 Answers 2

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From visual inspection of the given block diagram, we obtain

$$\begin{bmatrix} s+a & -k\\ b & s+p\end{bmatrix} \begin{bmatrix} X_1 (s)\\ X_2 (s) \end{bmatrix} = \begin{bmatrix} 0\\ b\end{bmatrix} U (s)$$

which can be rewritten as follows

$$\left( s \begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix} - \begin{bmatrix} -a & k\\ -b & -p\end{bmatrix} \right) \begin{bmatrix} X_1 (s)\\ X_2 (s) \end{bmatrix} = \begin{bmatrix} 0\\ b\end{bmatrix} U (s)$$

Hence,

$$\mathrm A := \begin{bmatrix} -a & k\\ -b & -p\end{bmatrix} \qquad\qquad \mathrm b := \begin{bmatrix} 0\\ b\end{bmatrix}$$

Since $Y (s) = X_1 (s)$,

$$\mathrm c^\top := \begin{bmatrix} 1 & 0\end{bmatrix} \qquad\qquad d := 0$$

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$L(s)$ being the loop gain, the closed loop is given by $P(s) = L(s)/(1+L(s)$ hence this gives

$$ \begin{align*} P(s) &= \frac{bk}{(s+a)(s+p)+bk} = \frac{bk}{(s+a)(s+p)+bk}\frac{X(s)}{X(s)}\\ &=\frac{bk}{s^2 + (a+p)s + (ap + bk)}\frac{X(s)}{X(s)}\\ \end{align*} $$ Then if you write out the state equations

$$ P(s) = \left[\begin{array}{cc|c}0&1&0\\-(ap+bk)&-(a+p)&b\\\hline k&0&0\end{array}\right] $$

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