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One way of doing this is using the Kalman decomposition. For this you need the reachable and unobservable subspaces. These subspaces can be constructed using the image of the controllability matrix and the kernel of the observability matrix respectively. The controllability matrix and its image, the observability matrix and its kernel are, $$ \mathcal{C} = \...


7

Observability means that you can estimate the complete state using only the output, without knowing the initial state. In other words, you have to figure out where you are without knowing where you were initially. A more practical reason why this rarely works is that when you are limited by non-perfect sensors and non-zero sampling time, taking the ...


5

The solution is fairly straightforward. Short answer The system is controllable without any modifications. You made a small mistake calculating the new controllability matrix: you are missing the first column of $\bar{M}$, which should be $\begin{bmatrix}0 & I\end{bmatrix}^T$. Full proof Let's consider the second system, and define some augmented ...


2

Found the solution. Looks like there is a bug in Nastran where if it doesnt see enough space in your C: drive (even if your scratch directory is in your D: drive), it just hangs. Once I cleared out some space in my C: drive (AND restarted my computer) then it started working.


1

your question is ill-conditioned: If $A$ must have a higher degree than $B$, 2 things can happen: $A$ is a constant, meaning $B$ must be 0: which means you cannot solve the equation as there is no $s$ term in $A(s)(s^2-1)$. Or $A$ is of order one (atleast one $s$) and $B$ is a constant: Which means the function can not be solved as there will be a $s^3$ ...


1

Couldn't you just define $y(t)$ and be done? $$ y'(t)=u(t) \\ y(t) = \int{u(t)}dt \\ $$ As long as the system is controllable, as defined by the controllability matrix $M$ being full rank, it doesn't matter what the input signal is, does it? The only issue I can think of is maybe if $u(t)$ is not able to be integrated, but that seems beyond the scope of ...


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