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A state space representation is given by: $$\dot{x}= \begin{bmatrix}0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 \\ 0&0&-2&-4\\0&0&1&0\end{bmatrix}x+\begin{bmatrix} 1\\0\\1\\0\end{bmatrix} u$$ $$y=\begin{bmatrix}0 & 1 & 0 & -1\end{bmatrix}x$$ I need to find a transfer function of this system. I don't know how, but I've tried it this way:

I wrote all equations: $$\dot x_1 = u$$ $$\dot x_2 = x_1$$ $$\dot x_3 = -x_3-4x_4+u$$ $$\dot x_4=x_3$$ $$y=x_2-x_4$$

And I've draw a scheme of such system: enter image description here

And then tried to find its transfer function. $$-4X_4-2sX_4+U = X_4s^2$$ $$U = X_4(s^2+4+2s)$$ $$X_4= \frac{U}{s^2+2s+4} $$ $$X_2 = \frac{U}{s^2}$$

$$Y = X_2 - X_4 = \frac{U}{s^2} -\frac{U}{s^2+2s+4}$$ $$G(s) = \frac{Y(s)}{U(s)} = \frac{1}{s^2} -\frac{1}{s^2+2s+4}=\frac{2(s+2)}{s^2(s^2+2s+4)}$$ I don't know, whether this is right answer nor solution.

What solution would you recomend for this task? Is the solution above correct?

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Use the fact that $\dot x_1 = s x_1$ and take it from there. It appears that your result is correct.

It should be noted that there is a closed form expression to map a state-space representation to a transfer function. However, it includes inverting a matrix with s variables and may hence be more challenging than just solving it as stated above

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