8

"Linear" imposes a set of restrictions. "Non-linear" simply means there are no restrictions. Many non-linear control schemes can be faster than linear ones. Linear control schemes are restricted to "smoothly" transitioning. Non-linear control can be implemented by suddenly slamming a digital value, for example. A good example of a fast-responding non-...


8

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 ...


3

If you do not meet both requirements, you designed the wrong compensator. If we take a look to the root locus figure of the compensator you designed we see that the poles (pink dots) are not in the white area, but there is still one in the yellow area. I came up with a different compensator, which clearly has all its poles in the white region. We can ...


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

You can design an asymptotic output tracker based on feedback linearization if the residual dynamics are stable. The theory for this can be found in the book 'Nonlinear Control Systems' by Isidori [Springer]. You can find examples worked out using Mathematica here and here. Another way is to develop a LQR tracking controller. See, for example, Chapter 4 of ...


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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