The following system is given : $$\dot{\bar x}=\begin{bmatrix}0&1&0&0\\-1&-2&0&0\\0&0&0&1\\0&0&0&-3 \end{bmatrix}\bar x+\begin{bmatrix}0&0\\-1&0\\0&0\\0&3\end{bmatrix}\begin{bmatrix}u_1\\u_2\end{bmatrix}=A\bar x+B\bar u\\$$

$$y=\begin{bmatrix}1&1&0&1\end{bmatrix}\bar x$$

After calculations I found $$Y(s)=\frac{-U_1(s)}{(s+1)}+\frac{3U_2(s)}{s+3}$$

The problem follows: Let $$\bar u=\begin{bmatrix}1&1\end{bmatrix}\begin{bmatrix}u_1\\u_2\end{bmatrix}+\bar r$$

Now how do I find $$\frac{Y(s)}{R(s)}$$ How does my transfer function change? Will it still depend on U1 and U2?

  • 2
    $\begingroup$ probably you are wrong in describing the problem. In the first equation you are saying that $\overline{u}$ is a $2$-dimensional vector, while in the fourth equation you are saying that it is a scalar. $\endgroup$ – trying Jul 23 '17 at 18:57
  • $\begingroup$ I noticed the u in the forth equation has a 'wavy' line above . That's the only difference between my description and the problem. $\endgroup$ – John Katsantas Jul 23 '17 at 20:26
  • $\begingroup$ The transfer function is the matrix $$\mathrm G (s) = \mathrm C (s \mathrm I - \mathrm A)^{-1} \mathrm B$$ $\endgroup$ – Rodrigo de Azevedo Jul 23 '17 at 21:58
  • $\begingroup$ @RodrigodeAzevedo That's what I used to get $Y(s)$. I can't follow the same process for $\frac{Y(s)}{R(s)} $. I'm not sure this formula stands if $u$ changes. $\endgroup$ – John Katsantas Jul 23 '17 at 22:06
  • $\begingroup$ Is $\bar u$ a scalar or a $2$-dimensional vector? $\endgroup$ – Rodrigo de Azevedo Jul 23 '17 at 22:08

Based on your comment and consequently on the fact that $\overline{r}$ is the control signal, probably that fourth equation is wrong and must be replaced with:

$$\overline{u} = \begin{bmatrix}u_1\\u_2\end{bmatrix} = \begin{bmatrix}1\\1\end{bmatrix} \overline{r}$$

This kind of equation is a needed one in order to know, using only one compensator (proportional compensator with gain $k$), how the control signal must be fed into the two inputs $u_1$ and $u_2$.

From now on, you can determine the transfer function $\frac{Y(s)}{\overline{R}(s)}$.


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