How do I confirm equivalency of two representation of the system

I have a system $$\dot x(t) = Ax(t)+Bu(t)$$ $$y(t) = Cx(t)+Du(t)$$

where $x$ is vector of states of the system, $y$ is output of the system, $u$ is input to the system and matrices $A$, $B$, $C$ and $D$ are following

$$A = \begin{bmatrix} 0&1&0&0\\3&0&0&2\\0&0&0&1\\0&-2&0&0\end{bmatrix}$$

$$B = \begin{bmatrix} 0&0\\1&0\\0&0\\0&1\end{bmatrix}$$

$$C = \begin{bmatrix} 1&0&1&0\end{bmatrix}$$

$$D = \begin{bmatrix} 0\end{bmatrix}$$

And I have another representation of the system $$\dot{\tilde{x}}=\tilde{A}\tilde{x}+\tilde{B}u$$ $$y=\tilde{C}\tilde{x}+\tilde{D}u$$

$$\tilde{A} = PAP^{-1}= \begin{bmatrix}0 &1 &0 &0\\0&0&0&0\\0&0&-i&0\\0&0&0&i\end{bmatrix}$$

$$\tilde{B} = PB = \begin{bmatrix}\frac23 &0\\0&1\\\frac12&i\\-\frac12&i&\end{bmatrix}$$

$$\tilde{C} = CP^{-1}=\begin{bmatrix}-3&2&2+i&-2+i\end{bmatrix}$$

$$\tilde{D} = D = \begin{bmatrix}0\end{bmatrix}$$

System representations are equivalent if their response to the same input is the same. So I should check the impulse response.

$$H(s) = C(sI-A)^{-1}B+D =\begin{bmatrix} \frac{1}{(s^2 + 1)} - \frac{2}{(s^3 + s)}& \frac{2}{(s^3 + s)} + \frac{(s^2 - 3)}{(s*(s^3 + s))}\end{bmatrix}$$ $$\tilde{H}(s) = \tilde{C}(sI-\tilde{A})^{-1}\tilde{B}+\tilde{D} = \begin{bmatrix}-\frac2{s}+\frac{1+\frac{i}{2}}{s+i}-\frac{\frac{i}2-1}{s-i}&-\frac3{s^2}+\frac{1}{s}+\frac{2i-1}{s+i}-\frac{2i+1}{s-i}\end{bmatrix}$$

MATLAB does confirm that transfer functions of both systems are equal, but I need to compute it and compare myself. I tried to play with the $\tilde{H}(s)$ but the $i$'s doesn't go away. Lecturer told us to use Euler's formula, but I can't figure out how.

How do I confirm equivalency of those two representation of the system?

If two state space models have the same transfer function matrix, then that does not mean that the two are equivalent. It only means that their minimal representation are equivalent. In this case both systems are both minimal, but you would first have to show this before you could use this.

If you do want to proof that these two transfer function matrices are equal, then it would be a good start to combine the complex terms first, since they should appear in complex conjugate pairs, such that the resulting coefficients are real. For example for the first entry,

\begin{align} \frac{1 + \frac12i}{s + i} - \frac{\frac12i - 1}{s - i} &= \frac{\left(s - i\right) \left(1 + \frac12i\right)}{\left(s + i\right)\left(s - i\right)} + \frac{\left(s + i\right) \left(1 - \frac12i\right)}{\left(s + i\right)\left(s - i\right)} \\ &= \frac{2s + 1}{s^2 + 1} \end{align}

Another option, which proofs that the two representations are equivalent, is to find $P$. This transformation can only exist if $A$ and $\tilde{A}$ have the same eigenvalues. In this case finding the eigenvalues relatively easy, since $\tilde{A}$ is in the Jordan canonical form. Checking if $A$ has the same eigenvalues is the same as checking if $\det(A-\lambda\,I)=0$, where $\lambda$ is each eigenvalue of $\tilde{A}$. If both systems would not be minimal, then you could check if they still could be equivalent by seeing if the Hautus test yields the same rank,

$$\text{rank}\left[\lambda\,I - A,\ B\right] = \text{rank}\left[\lambda\,I - \tilde{A},\ \tilde{B}\right],$$

$$\text{rank}\left[\lambda\,I - A^T,\ C^T\right] = \text{rank}\left[\lambda\,I - \tilde{A}^T,\ \tilde{C}^T\right].$$

If all these properties hold, then both systems might be equivalent. For a controllable SIMO system you can always find the transformation with,

$$\mathcal{C} = \left[B,\ A\,B,\ A^2B,\cdots ,\ A^{n-1}B\right],$$

$$\tilde{\mathcal{C}} = \left[\tilde{B},\ \tilde{A}\,\tilde{B},\ \tilde{A}^2\tilde{B},\cdots ,\ \tilde{A}^{n-1}\tilde{B}\right] = P\, \mathcal{C},$$

$$P = \tilde{\mathcal{C}} \, \mathcal{C}^{-1}.$$

If $\tilde{C}$ and $C\,P^{-1}$ are equal then both systems are equivalent. For an observable MISO system you can always find the transformation with,

$$\mathcal{O}^T = \left[C^T,\ A^T\,C^T,\ \left(A^T\right)^2C^T,\cdots ,\ \left(A^T\right)^{n-1}C^T\right],$$

$$\tilde{\mathcal{O}}^T = \left[\tilde{C}^T,\ \tilde{A}^T\,\tilde{C}^T,\ \left(\tilde{A}^T\right)^2\tilde{C}^T,\cdots ,\ \left(\tilde{A}^T\right)^{n-1}\tilde{C}^T\right] = P^{-T}\,\mathcal{O}^T,$$

$$P = \tilde{\mathcal{O}}^{-1} \mathcal{O}.$$

If $\tilde{B}$ and $P\,B$ are equal then both systems are equivalent.

This should be sufficient to solve your problem, but if you would like to generalize this to a MIMO system is might be harder to find $P$. Because is this case neither $\mathcal{C}$, $\tilde{\mathcal{C}}$, $\mathcal{O}$ and $\tilde{\mathcal{O}}$ would be square matrices, so their inverse would not exist. Since each $i$th column of $\tilde{B}$ would be only a function of $P$ and the $i$th column of $B$, so you might be able to "guess" a linear combinations of the columns of $B$ and $\tilde{B}$, which results into full rank square "controllability matrices" (using these linear combinations of the columns of $B$ and $\tilde{B}$). Or similarly use a linear combination of the rows of $C$ and $\tilde{C}$, which results into full rank square "observability matrices".

If the systems are not minimal, then there might be no unique $P$. For example when the systems are both completely not controllable and observable, then for a given $P$ another transformation could be found by multiplying each column of $P$ by a non-zero gain (this basically changes the length of the eigenvectors of $A$), however finding any $P$ would already show that both systems are equivalent.

Well, instead of carrying out the calculations you can prove it using matrix manipulations as follows $$\tilde{H}(s)=\tilde{C}(sI-\tilde{A})^{-1}\tilde{B}+\tilde{D}=CP^{-1}(sI-PAP^{-1})^{-1}PB+D\\ =CP^{-1}[P(sI-A)P^{-1}]^{-1}PB+D=CP^{-1}(P^{-1})^{-1}(sI-A)^{-1}P^{-1}PB+D\\=C(sI-A)^{-1}B+D=H(s)$$

• I think this is redundant. We probably got those formulas for $\tilde{A}$ etc. from this, what you wrote here. I computed the tilde representation and I need to compare it. But thank you. I appreciate your response. – user50222 Oct 16 '16 at 8:45
• @user50222 Sorry, I misunderstood the question. In this case there is not much here for us to help, you should just do the calculations correct. Note that the $(1,1)$ term in $\tilde{H}(s)$ is certainly not correct since you must also have an $s-i$ denominator, not an $s-1$. This is due to fact, that $1$ is not an eigenvalue of $A$ but $i$ is. – RTJ Oct 16 '16 at 9:18
• thank you! That was a mistake. I haven't noticed that before. – user50222 Oct 16 '16 at 9:24