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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.. –  user13838 Oct 7 '11 at 15:46

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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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Yes, that is how I did it first. But in my exercise class we did this example. There I was told to make a polynomial division with $G_1$ because of this leapability. –  madmax Oct 7 '11 at 17:43
    
Added some the differential equations and $G_1$. –  madmax Oct 7 '11 at 17:50
    
@madmax What is leapability? This is the first time that I hear about it. –  user13838 Oct 7 '11 at 17:51
    
@madmax You will get a state space representation.Use $C(sI-A)^{-1}B$ to obtain the transfer function. You would get, $ \frac{s - 1}{s^2 + 2 s - 3}$ which can be simplified to $1/(s+3)$. –  user13838 Oct 7 '11 at 17:54
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@madmax I used the simple technique that I wrote in one of your earlier questions. –  user13838 Oct 7 '11 at 18:43

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