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This is the block diagram that I'd like to transform into a state-space representation, where u1 and u2 are inputs and y1 and y2 are the outputs of the system

enter image description here

I tried to place state variables on the diagram and go from there (is there a cleaner way to do this?): enter image description here

I am not sure how to go about from here. The $\frac{1}{s+1}$ block is confusing me. I know I can write the output signal from the $\frac{1}{s+1}$ block like this:

But I don't see if that's even useful and if it is, I don't know how to proceed.

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This is one odd diagram. And you are frankly in the right direction. Now you only need to express each state as the equation from before in the diagram.

Diagram with states

So lets start with $x_1$: \begin{equation} \begin{aligned} x_1 &= \frac{1}{s+1}u_1 & \rightarrow x_1 &= u_1-\dot{x}_1 \\ x_2 &= x_1 - u_1 + u_2 & \rightarrow x_2 &= u_2 - \dot{x}_1 \\ \dot{x}_3 &= x_2 \\ \dot{x}_4 &= x_3 \end{aligned} \end{equation} And the other side: \begin{equation} \begin{aligned} x_5 &= \frac{1}{s+1}u_2 &\rightarrow x_5 = u_2-\dot{x}_5 \\ x_6 &= x_5 + x_3 - x_7 \\ \dot{x}_7 &= x_6 \\ \dot{x}_8 &= x_7 \\ \end{aligned} \end{equation} And now substitute where ever needed to write everything as an equation of a state derivative: \begin{equation} \begin{aligned} \dot{x}_1 &=u_1 - x_1 \\ \dot{x}_3 &= u_2 - u_1 - x_1 \\ \dot{x}_4 &= x_3 \\ \dot{x}_5 &= u_2-x_5 \\ \dot{x}_7 &= x_5 + x_3 - x_7 \\ \dot{x}_8 &= x_7 \\ \end{aligned} \end{equation} As there are 6 integrators, having 6 states does make sense. Despite that, I do think it can be simplified, but lets first implement it into a state space $$\dot{x} = \begin{bmatrix}-1 & 0 & 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 &0 \\ 0 & 0 & 0 & -1 & 0 & 0 \\ 0 & 1 & 0 & 1 & -1 & 0 \\ 0 &0 &0&0&1&0\end{bmatrix}\begin{bmatrix}x_1 \\ x_3 \\ x_4 \\ x_5 \\ x_7 \\ x_8\end{bmatrix} + \begin{bmatrix}1 & 0 \\ -1 & 1 \\ 0&0 \\0&1\\0&0\\0&0\end{bmatrix}\begin{bmatrix}u_1 \\ u_2\end{bmatrix}$$

From this getting the C matrix is pretty self-explanatory. If there are any more questions, I am happy to hear.

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  • $\begingroup$ Thanks! Much appreciated $\endgroup$ Sep 2 '21 at 18:45

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