# Tag Info

## Hot answers tagged nonlinear-control

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-...

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In nonlinear control theory, you will recognize most concepts such as controllability and observability where the linear case is often a special case of the nonlinear case. I would highly recommend digging into linear control theory first if you have not done so. Depending on the course you take, concepts such as Lyapunov stability are discussed here, which ...

3

A set is invariant with respect to its dynamics if $x(t_0)\in M \implies x(t)\in M$, this is not the case for the set $E$. The set $E = \left\lbrace x\in \Omega \mid \dot{V}(x) = 0 \right\rbrace$ does not need to be an invariant set, since it does not consider the solution of $x(t)$. I'll show this statement using the well known pendulum example. The ...

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It can be noted that the high-pass filter can also be written as $$\frac{s}{s + \omega} = 1 - \frac{\omega}{s + \omega},$$ such that $$\frac{s}{s + \omega} Y(s) = Y(s) - \underbrace{\frac{\omega}{s + \omega} Y(s)}_{\mathcal{L}\{\eta(t)\}(s)},$$ with $Y(s) = \mathcal{L}\{y(t)\}(s)$ denoting the Laplace transform of $y(t)$. In order to show that the second ...

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The sets $M$ and $E$ can be different. The set $E$ only considers $\dot{V}=0$ while $M$ also considers $f(x)$. Namely, invariant set means that for all $x(0) \in M$ the solution $x(t)\in M$ for all $t>0$, which can also be written as $x(0)+\delta\,f(x(0)) \in M$ for infinitesimally small $\delta$. For example consider the system \begin{align} \dot{x}_1 &...

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Input signal: ±U V. Scaled output signal: ±1 V. Required gain: $\frac 1 U$. Figure 1. Possible solutions. If U > 1 then a simple potential divider may suffice. $V_{OUT} = \frac {R_2}{R_1+R_2}$. Bear in mind that whatever follows this circuit may load it somewhat so keep the parallel combination value of R1 || R2 < 10% of the value of the ...

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Im going to step out of the comment section as it is fairly limited. The quickest answer to the question: Is it possible to somehow form a transient process (with given properties) if the steady state is not known in advance, but it is known that in the steady state f(x,t)=0? Given the proposed conditions is simple: no. And the explanation boils down to ...

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Taking the time derivative of $y$ yields: $$\dot{y}=\frac{\partial f}{\partial x}(f(x)+u)$$ We need $y(t)=y(0) e^{-\beta t}$ but this is possible if and only if $\dot{y}=-\beta y$ , if the hessian is invertible then this is possible if and only if $u=-f(x)-\left(\frac{\partial f}{\partial x}\right)^{-1}\beta y$. To get an intuition on why the hessian must ...

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You can show it with Laplace transform. Let's say Y = ℒ[y], H = ℒ[η], and ω_p is the filter pole. (Let's not simply call it ω -- when omega is used as a parameter rather than a variable, a subscript should be added for clarity) dη / dt = ω_p (y - η) ... transform ... s * H = ω_p ( Y - H ) s * H / ω_p = Y - H Y = H ( 1 + s/ω_p ) H/Y = 1 / ( 1 + s/ω_p ) which ...

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The problem is complicated by the fact that the function f(t) and the point at which its minimum/maximum x∗ is located, generally speaking, are not known to us. So if I understand it correctly: $f(t)$ is a unknown function, of which your controlled system should approach either the maximum or the minimum value $f(t)$ has over $t$. If I look at your system ...

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MPC finds the optimal input $u^*$ which is the input that minimizes the cost function $J$ or $c$. This means that regardless of what this actual value is, its proven to be its minimum. As such, multiplying the cost function with any constant value does not change this minimum, it just scales the value. Therefore, $\frac{t_{hor}}{N}$ does not affect the ...

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I think I have found an answer and please correct me if there are any other reasons apart from the following justification. Since $x^{*}(t)=\{x(t)\in\Omega|J(x)<\min J(y), \forall { y \in \Omega}\}$ is optimal between all admissible trajectories $x(t)\in\Omega$ for $t\in[t_0,t_1]$ with $x^{*}(t_0)=x(t_0)=x_0$ and $x^{*}(t_1)\neq{x(t_1)}$, it is still ...

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Being able to check the response of a system, particularly of a MIMO system, to a bunch of unknown external disturbances is very crucial when designing a controller. These external unknown disturbances (or even any unmodeled plant dynamics) may influence the system in these three different ways: Add only Gain Margin Add only Phase Margin Add both Phase and ...

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Based on your comment, it seems like you are trying to control a dynamical system to track some reference trajectory subject to Gaussian disturbances. Suppose you have some kind of model for the dynamics, then the discrete system can be written as \begin{align} x_{k+1} &= f_k(x_k) + a_k w_k\\ y_k &= x_k \end{align} where $w_k$ is a Gaussian ...

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You don't necessarily have to find exact constants $k_1,k_2,k_3$, only need to show that there exists some positive constants. In your example above, I can say that $$\dot{V} \leq -\bigg(\frac{x_1 + x_2}{\sqrt{2}}\bigg)^2 = -\frac{1}{2}\|x\|^2,$$ so $k_3 = 1/2$ would work. For the function $V(x)$ itself, we know that it is positive definite. Note that ...

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In control systems, the main focus is the design of a controller for machines or robots, here we mainly deal with linear systems application of linear control theory. While non-linear systems is an advanced topic, where we deal with advance and mathematically more complex systems. If its ur first course in control systems or automatic control. Go for the ...

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