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In this exercise the state space representation of the imaged system is asked for. $G_1(s) = \frac{s-1}{s+2} = 1 - \frac{3}{s+2} G_2(s)=\frac{1}{s-1}$ I can see that $G_1(s)$ is "able to leap" (hope it is the correct translation of sprungfähig), because nominator and denominator have the same order.

So for the system matrix I get $A = \begin{pmatrix} -2 & 3 \\1 & 0 \end{pmatrix}$
That should be correct.
But I am not sure with B and C.
Can I get both by looking at the image? Because that is what I did and it looks plausible.
$B = \begin{pmatrix} -3 \\ 1 \end{pmatrix}$ $C = \begin{pmatrix} 0 \\1 \end{pmatrix}$

And does the output y look like this, because of the leapable ability? $ y= \begin{pmatrix} 0 & 1 \end{pmatrix} x + d = \begin{pmatrix} 0 & 1 \end{pmatrix} x + 1$

Differential equations: $\frac{dx_1}{dt} = -2x_1+3x_2-3r$ $\frac{dx_2}{dt} = x_1+r$

Matrix B
Matrix B is the control matrix, and determines how the system input affects the state change. If the state change is not dependent on the system input, then B will be the zero matrix.

Matrix C
Matrix C is the output matrix, and determines the relationship between the system state and the system output.

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    I don't know how you computed the feedback loop but you can check the validity of your representation by the resulting transfer function. Besides I think you would end up with a first order system since there is a pole-zero cancellation..2011-10-07

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$ G_1G_2 = \frac{1}{s+2} $ Then, the transfer function from $r$ to $y$ is given by $ G_{yr} = \frac{G_1 G_2}{1+G_1G_2} = \frac{1}{s+3} $ A state space representation for this is $ \left[ \begin{array}{c|c} A &B\\ \hline C& D \end{array}\right]= \left[ \begin{array}{c|c} -3 &1\\ \hline 1& 0 \end{array}\right] $

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    I assume you were referring to this answer [here](http://math.stackexchange.com/questions/70037/how-to-obtain-a-possible-state-space-representation-of-this-2nd-order-transfer-fu). But I don't know how you got the $A$ matrix values out of y(t) and u(t).2011-10-08