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I want to know how I can get from a set of differential equations to a set of transfer functions for a multi-input multi-output system. I can do this easily with Matlab or by computing $G(s) = C[sI - A]^{-1}B + D$. I have the following two equations: $ \ddot{y}_1 + 2\dot{y}_1 + \dot{y}_2 + u_1 = 0 \\ \dot{y}_2 - y_2 + u_2 - \dot{u}_1 = 0 $There are 2 inputs, $y_i$, and 2 outputs, $u_i$. At first I thought when I want to retrieve the transfer function from $y_1$ to $u_1$ that I had to set $y_2$ and $u_2$ equal to zero. Thus I would have been left with, $\ddot{y}_1 + 2\dot{y}_1 + u_1 = 0$ and $\dot{u}_1 = 0$. However this does not lead to the correct answer, $ y_1 \rightarrow u_1: \frac{-s^2 - s + 1}{s^3 + s^2 - 2 s} $ I also thought about substituting the two formulas in each other. So expressing $y_2$ and $u_2$ in terms of $y_1$ and $u_1$ however this also lead to nothing.

Can someone explain to me how to obtain the 4 transfer functions, $y_1 \rightarrow u_1$, $y_1 \rightarrow u_2$, $y_2 \rightarrow u_1$ and $y_2 \rightarrow u_2$?

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    First, write the system as a system of first-order equations. Then, use Laplace transforms.2012-09-17

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I am guessing that you are looking for the transfer function from $u$ to $y$, this would be consistent with current nomenclature.

Taking Laplace transforms gives $ (s^2+2s) \hat{y_1} + s\hat{y_2} + \hat{u_1} = 0\\ (s-1)\hat{y_2} + \hat{u_2}-s \hat{u_1} = 0 $ Solving algebraically gives $\hat{y_1} = \frac{1-s-s^2}{s(s+2)(s-1)} \hat{u_1} + \frac{1}{s(s+2)(s-1)}\hat{u_2} \\ \hat{y_2} = \frac{s}{s-1} \hat{u_1} -\frac{1}{s-1} \hat{u_2} $ from which all four transfer functions can be read off.

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    @saulus proposed some edits that were correct, but unfortunately rejected. I'm not sure how to include their edits.2016-10-23