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The following system is given and I'm asked to find the transfer function $$\frac{Y(s)}{U(s)}=G(s)$$ $$\bar {\dot x}=\begin{bmatrix} 0 & 1 & 0 & 0 &0 \\0 & 0 & 1 & 0 & 0 \\ 0 & -1 & -2 & 0 & 0 \\ 0 & 0 & 0 & -2 & 0\\ 0 & 0 & 0 & 0 & -3\end{bmatrix} \bar x + \begin{bmatrix} 0\\0\\1\\1\\0\end{bmatrix}u=A\bar x + Bu \\ $$

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

I haven't practised that much on state space models and I don't remember that much from matrix algebra but here's what I've thought.

I found the eigenvalues seeing the matrix as a 2x2 diagonal matrix with matrix elements therefore getting the denominator of my transfer function $$s(s+1)^2(s+2)(s+3)$$.

I also have my backdoor $$G(s)=C(sI-A)^{-1}B $$ but this includes many calculations and I guess that there is a faster solution.

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  • \$\begingroup\$ I'm trying to decide if we should migrate this somewhere like the Engineering StackExchange or the Theoretical Computer Science StackExchange... though the latter might be the best place to ask this. I honestly have never heard of state space in control systems. \$\endgroup\$ – KingDuken Jul 22 '17 at 22:09
  • \$\begingroup\$ You also haven't asked an actual question. \$\endgroup\$ – Transistor Jul 22 '17 at 22:30
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    \$\begingroup\$ @KingDuken, if you Google state space you will find that it is very much in the control systems subject area. \$\endgroup\$ – Chu Jul 22 '17 at 22:36
  • \$\begingroup\$ Your final equation is the TF, \$\frac{Y(s)}{X(s)}\$, not \$y(t)\$. What do you mean by 'faster solution'? \$\endgroup\$ – Chu Jul 22 '17 at 22:45
  • \$\begingroup\$ There is also one eigenvalue in the origin missing, i.e. \$(s+1)^2(s+2)(s+3)\$ needs to be multiplied with a factor \$s\$. \$\endgroup\$ – Koen Tiels Feb 17 '19 at 20:31
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I think the answer should be \$G(s) = \frac{1}{(s+1)^2}\$, but I haven't checked it with Matlab yet. That should be

A = [0 1 0 0 0; 0 0 1 0 0; 0 -1 -2 0 0; 0 0 0 -2 0; 0 0 0 0 -3];
B = [0; 0; 1; 1; 0];
C = [0 1 0 0 1];
D = 0;
tf(ss(A,B,C,D))

Since there are quite a lot of zeros in your matrices, the formula \$G(s) = C (sI-A)^{-1}B\$ is not as daunting to compute by hand. Since \$C = \begin{bmatrix} 0 & 1 & 0 & 0 & 1\end{bmatrix}\$, you only care about the second and the fifth row of \$(sI-A)^{-1}\$. Since \$B=\begin{bmatrix} 0 & 0 & 1 & 1 & 0 \end{bmatrix}^T\$, you only care about the third and the fourth column of \$(sI-A)^{-1}\$. The other values in \$(sI-A)^{-1}\$ will get multiplied by zero. This means that you only need to compute four elements of \$(sI-A)^{-1} = \frac{adj(sI-A)}{det(sI-A)}\$: $$ \begin{aligned} G(s) & = C \frac{adj(sI-A)}{det(sI-A)} B \\ & = \begin{bmatrix} 0 & 1 & 0 & 0 & 1\end{bmatrix} \frac{ \begin{bmatrix} \bullet & \bullet & \bullet & \bullet & \bullet \\ \bullet & \bullet & \theta_1& \theta_2& \bullet \\ \bullet & \bullet & \bullet & \bullet & \bullet \\ \bullet & \bullet & \bullet & \bullet & \bullet \\ \bullet & \bullet & \theta_3& \theta_4& \bullet \end{bmatrix}} {s(s+1)^2(s+2)(s+3)} \begin{bmatrix} 0 \\ 0 \\ 1 \\ 1 \\ 0\end{bmatrix} \end{aligned} $$ The four elements that you need to compute are elements of an adjugate matrix. The first element is $$ \theta_1 = (-1) det( \begin{bmatrix} s & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & s+2 & 0\\ 0 & 0 & 0 & s+3 \end{bmatrix} ) = s(s+2)(s+3) $$ The other \$\theta\$s are zero, since there is either a zero row or a zero column in the 4 by 4 matrix of which you need to compute the determinant.

The transfer function is then equal to $$ G(s) = \frac{s(s+2)(s+3)}{s(s+1)^2(s+2)(s+3)} = \frac{1}{(s+1)^2} $$

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