Consider the continuous-time state-space representation

$\frac{d}{dt}x(t)=Ax(t)+Bu(t), \quad y(t)=Cx(t), \quad t \in R^+,$

with the matrices given by

$A =\begin{bmatrix}-4&-5&0&5\\-4&-4&0&4\\0&0&-2&0\\-4&-5&0&5\end{bmatrix}, \quad B=\begin{bmatrix}2\\0\\0\\1\end{bmatrix}, \quad C=\begin{bmatrix}1&-1&0&0 \end{bmatrix}$.

Note that the matrix $A$ satisfies $AV=V\Lambda$ with

$\Lambda=\begin{bmatrix}-4&0&0&0\\0&-2&0&0\\0&0&0&0\\0&0&0&1\end{bmatrix}\quad$ and $\quad V=\begin{bmatrix}1&0&0&1\\1&0&-1&0\\0&1&0&0\\1&0&-1&1\end{bmatrix}\quad$ and $V^{-1}=\begin{bmatrix}1&1&0&-1\\0&0&1&0\\1&0&0&-1\\0&-1&0&1\end{bmatrix}$

The transfer function $H(s)$ of the state-space model (with all poe/zero cancellations removed) corresponding to the system is given by:

A) $\quad H(s)=\frac{2s^2+6s-8}{s(s+4)(s-1)}$

B) $\quad H(s)=\frac{2s-2}{s(s-1)}$

C) $\quad H(s)=\frac{2s^3+10s^2+4s-16}{s(s+4)(s+2)(s-1)}$

D) $\quad H(s)=\frac{2}{s}$

E) $\quad$ None of the above

I know that $\Lambda$ is a diagonal matrix with the eiegenvalues of $A$ on its diagonal. and that $V$ is a matrix of eigenvectors of $A$.

Is there a way to quickly determine this? because all I know is that the transfer function can be determined using:


But this seems like a lot of work for a 4x4 matrix and during an exam.

Just for completeness. Here's a photo of the answers, showing that the correct transfer function $\frac{2s-1}{s(s-1)}$ is not listed among the possible choices.

possible answers

  • $\begingroup$ Could it be that you have made a typo, because when evaluating the minimal realization in Matlab I get $\frac{2\,s-1}{s(s-1)}$ instead of $\frac{2\,s-2}{s(s-1)}$? $\endgroup$ – Kwin van der Veen Aug 13 '18 at 19:57

What you basically have to determine is which modes are controllable and observable. Checking the rank of the controllability and observability matrices won't tell you which modes are controllable and observable, only whether the entire system is minimal (and to some extend how many modes are not present). A better method in this case would be the Hautus lemma. However a direct application of this would this require you to check the rank of eight matrices (all four eigenvalues with $B$ for controllability and $C$ for observability). This can be reduced significantly by using the similarity transformation $\hat{x} = V^{-1}\,x$, which gives

\begin{align} \hat{A} &= V^{-1}\,A\,V = \Lambda, \\ \hat{B} &= V^{-1}\,B, \\ \hat{C} &= C\,V. \end{align}

Here $\hat{B}$ is twice the first column plus the last column of $V^{-1}$ and $\hat{C}$ is the first minus the second row of $V$, which gives

$$ \hat{B} = \begin{bmatrix} 1 & 0 & 1 & 1 \end{bmatrix}^\top, \quad \hat{C} = \begin{bmatrix} 0 & 0 & 1 & 1 \end{bmatrix}. $$

From this result it is relatively easy to see, without actually applying the Hautus test, which modes are controllable and observable since $\hat{A}$ is diagonal, so the system can be seen as four decoupled first order systems. The first component of the state $\hat{x}$ has the associated eigenvalue $-4$ and its dynamics does get an input passed through by $\hat{B}$ (its first component is nonzero), but does not appear in the output (the first component of $\hat{C}$ is zero), so this state component is controllable but not observable. Similarly it can be shown that the state with eigenvalue $-2$ not controllable and not observable, and the states with eigenvalues $0$ and $1$ are both controllable and observable.

So a minimal representation of the system would only contain the states which are both controllable and observable. This would be the states with eigenvalues $0$ and $1$, therefore answer B) would be the only option. However B) is not minimal, since by factoring out the 2 it is easy to see that it has also a zero at $1$, which would mean you could do pole zero cancelation. This would imply the system was not minimal, but earlier we showed that it is minimal. So all that remains is E) none of the above.

The minimal transfer function can be calculated relatively easily by considering only the controllable and observable parts. A minimal state space model would then become

$$ \left[\begin{array}{c|c} A_m & B_m \\ \hline C_m \end{array}\right] = \left[\begin{array}{cc|c} 0 & 0 & 1 \\ 0 & 1 & 1 \\ \hline 1 & 1 \end{array}\right] $$

Its corresponding transfer function can be calculated with $C_m\,(s\,I-A_m)^{-1}B_m$, where $s\,I-A_m$ is diagonal so its inverse is just taking the inverse of each diagonal element. Using this inverse gives the following expression for the transfer function

$$ H(s) = \frac{1}{s} + \frac{1}{s-1} = \frac{2\,s-1}{s(s-1)}. $$

  • $\begingroup$ Thanks for your answer, it really saves me a lot of time. If you enter $C(sI-A)^{-1}B$ in Matlab you indeed get $\frac{2s-1}{s(s-1)}$ but the answers on the exam paper are exactly the same as what I've written down. No typos. Could it then be answer E, none of the above? because $\hat{C}=\begin{bmatrix}0&0&1&1\end{bmatrix}$ which would give $s+1$ in the numerator of the transfer function. $\endgroup$ – user463102 Aug 13 '18 at 21:21
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    $\begingroup$ @user463102 E) would indeed then be the only option (I do wonder if the person who made the exam maybe made a typo then). You are incorrect that $\hat{C}$ implies $s+1$ in the numerator. Instead you can use that $\hat{A}$ is diagonal, thus the transfer function is a sum of first order transfer function $$H(s)=\frac{1}{s}+\frac{1}{s-1}=\frac{2\,s-1}{s(s-1)}.$$ $\endgroup$ – Kwin van der Veen Aug 14 '18 at 4:02
  • $\begingroup$ Just for completeness, I uploaded the question form with the answers to show that, strangely enough, the correct answer is not listed. $\endgroup$ – user463102 Sep 27 '18 at 10:40
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    $\begingroup$ @user463102 You are correct, the numerator of B) is not correct. I will update my answer to show why. $\endgroup$ – Kwin van der Veen Sep 27 '18 at 11:07

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