New answers tagged

1

Having a solver for equality constrained QP only will not help you in solving the general inequality constrained problem (unless you develop a QP solver, which can have an equality-constrained QP solver as a required sub-routine). In other words, your problem cannot be reformulated to a problem with only equalities.


1

When representing the system as a state space model \begin{align} \dot{x} &= A\,x + B\,u, \\ y &= C\,x + D\,u, \end{align} such that the transfer function matrix is equal to $G(s) = C(s\,I-A)^{-1}B+D$, then the (transmission) zeros can be calculated by solving the generalized eigenvalue problem $$ \det\!\left( \begin{bmatrix} A & B \\ C & ...


2

The only term that is not linear in the unknown variables is the quadratic term in $\mathbb{Z}$, so it might be easier to replace that with another variable and add another inequality $$ \mathbb{Z}^2 \leq \mathbb{X}, \tag{1} $$ such that the initial in equality can be written as \begin{align} \begin{pmatrix} \begin{pmatrix} \mathbb{A}_{\mathbb{Z}} & \...


1

Assume that $x$ is only a function of $W(t)$, so $x(W(t))$. Differentiating this yields \begin{align} \frac{dx}{dt} &= \frac{\partial x}{\partial W} \frac{dW}{dt} = \frac{\partial x}{\partial W} i\,u(t)\,W(t), \\ &= u(t)\,x(W(t)). \end{align} This can be simplified to $$ \frac{\partial x}{\partial W} i\,W(t) = x(W(t)). $$ This can be solved ...


0

I do not see any particular reason why you would want to transform your differential equation. Why not solve it in the given formulation? If $u(t)$ is a known input you can solve the first equation $$\dfrac{dW}{W} = i u(t) dt \implies \ln |W|=\int_{t_0}^{t}iu(\tau)d \tau$$ $$\implies |W(t)| = \exp\left[ \int_{t_0}^{t}iu(\tau)d \tau\right]$$ $$\implies W(t) ...


0

Constant Hamiltonian in Optimal Control Theory are related to the Beltrami Identity appearing in Calculus of Variations. In Calculus of Variations, if the Lagrangian $ \mathcal{L} $ don't explicetly depend on time such that $ J = \int \mathcal{L}(x,\dot{x}) dt $ then the Beltrami Identity $ \mathcal{L} - \frac{\partial \mathcal{L}}{\partial \dot{x}_{\alpha}}...


1

I suppose that you mean $y_k=\sin(Cx_k)$, where $Cx$ is a scalar. Otherwise, either your $x$ is a scalar (and $C=1$), or you should specify the $\sin(x)$ function. Ok, so let us define the estimation error signal as $\tilde{x}:=\hat{x}-x$. Suppose that $|\tilde{x}|$ is small. Then the following approximation holds $$\sin(Cx)\approx \sin(C\hat{x}) - \cos(C\...


1

Partial answer (because I don't remember the details): Many years ago, I made an animation of a coherent state quantum harmonic oscillator in MATLAB. I don't remember the exact commands, and the script is on a computer I unfortunately no longer have access to (I had not yet heard about Github). What I did was to draw every frame in a for-loop, and save as ...


0

The PDE $u_t + f u_x = -au$ is linear, even if $f$ depends on $t$. Introducing the variable $v = e^{at} u$, we have \begin{aligned} &v_t = e^{at} u_t + a e^{at} u = -e^{at}(f u_x + au) + a e^{at} u = -fv_x . \end{aligned} According to the method of characteristics (see e.g. this post), solutions of the time-dependent linear advection equation $v_t + f ...


1

Your second approach is essentially deadbeat control (driving the output of the system in the least amount of time to the reference), for which it is normal that the generated control inputs are very large and aggressive. Deadbeat control is often very sensitive to how accurate the model has been identified, since it essentially cancels both its pole and ...


2

The choice $Q = C^T C$ basically puts cost on the system output. Since often, the output of the system is what we want to control, this is a simple choice that can be used as a first try. In practise, however, the states (respectively the outputs) can be on very different scales - therefore, it might not make much sense to weight the outputs like that. ...


0

Hint: Start with $y=Cx$ and differentiate $$\dot{y}=C\dot{x}=CAx$$ $$\ddot{y} = CA^2x$$ $$\vdots $$ $$\dfrac{d^{n-1}}{dt^{n-1}}y = CA^{n-1}x$$ By this procedure, we get $n$ equations that we can rewrite in the following way $$\begin{bmatrix}y\\ \dot{y}\\ \ddot{y}\\ \vdots \\ \dfrac{d^{n-1}}{dt^{n-1}}y\end{bmatrix} = \begin{bmatrix}C\\CA\\CA^2\\ \vdots \\...


0

Markov property is related to the probabilistic model of the state-space equations. $$ \begin{array}{l} {{\bf{x}}_k} = {{\bf{f}}_{k - 1}}({{\bf{x}}_{k - 1}},{{\bf{u}}_k}) + {{\bf{v}}_k} \Leftrightarrow p({{\bf{x}}_k}|{{\bf{x}}_{k - 1}},{{\bf{u}}_k})\\ {{\bf{y}}_k} = {{\bf{h}}_k}({{\bf{x}}_k},{{\bf{u}}_k}) + {{\bf{w}}_k} \Leftrightarrow p({{\bf{y}}_k}|{{\bf{x}...


0

A general state space model can be very complicated. Dynamical systems with hysteresis will definitely not have the Markov property. If you can write down your system as $$x_{k+1} = f(x_k,u_k)$$ then it will have the Markov property because only the current state and the current action/actuation are important for the next state.


4

The general solution is$$x(t) = e^{tA}x_0 + \int_0^te^{(t-s)A}Bu(s)\,ds.$$Since $A^2=A^3=\ldots$, one has $$ e^{tA} = \sum_{n=0}^\infty\frac{t^nA^n}{n!} = I_3 + tA + A^2\sum_{n=2}^\infty\frac{t^n}{n!} = I_3 + tA + (e^t - t - 1)A^2. $$ Now, if $x_0\in C$, there exist $t>0$ and $u\in\mathbb U$ such that $x(t) = 0$, i.e., $$ e^{tA}x_0 = -\int_0^te^{(t-s)A}Bu(...


1

This is because your DC gain is negative. Formally, when you compute $H(j\omega)$ for $\omega=0$, where $j:=\sqrt{-1}$, you obtain $H(0)=-10$, which corresponds in the complex plane to the vector with the magnitude $10$ and the phase $-\pi$ (or $\pi$, up to you). Actually, this is why for systems with a negative DC gain, it is typically recommended to ...


3

Question: Is it possible to derive the exact discrete time system from a nonlinear continuous-time system? In order to derive the discrete time system, we need to be able to predict $x_{k+1}$ from $x_k$ and $u_k$. For a linear time-invariant system $\dot{x}=Ax+Bu$, we are able to derive a closed form solution $$x(t) = \exp((t-t_0)A)x(t_0)+\int_{t_0}^{t}\...


2

Your system is not stable -- the presence of a pole with zero real part rules out asymptotic stability; the multiplicity of the poles at zero rules out marginal stability. The statement that an LTI system generates a sinusoidal output for a sinusoidal input is only true for stable systems.


1

If $c$ is a small perturbation, then the fixed point will be a small perturbation of the original fixed point, whose size is in fact quite explicitly controlled by $c$. This is a feature known as stability that is observed in many problems that are resolved using contraction mapping. To see this, let $0 < L < 1$ be the constant in the contraction ...


0

No. As mentioned in the comments, the PBH matrix is defined to be $$ K = [(A-\lambda I) B].$$ Since $A$ is $n\times n$ and $B$ is $n \times m$, it follows that $K$ is $n \times (n+m) $. Since the rank of a matrix cannot exceed the lesser of the column or row numbers, it follows that the rank of $K$ is at most $n$.


0

Why do you necessarily want to apply a Schur complement? The decay-rate estimation problem you have is not convex in $\gamma$ and $P$, and is a classical example of a problem that cannot be written in linear form. It is however quasi-convex, and is easily solved by bisection in $\gamma$. Since you are using YALMIP, you are lucky as there are both examples ...


1

Your initial problem is equivalent to $$ x^\top \left(P\,A + A^\top P\right) x = -\mu\,x^\top P\,x \tag{1} $$ and $\mu > \gamma$. Where $(1)$ is essentially a "homogeneous" continuous Lyapunov equation and can also be written as $$ P\,\mathcal{A} + \mathcal{A}^\top P = 0, \tag{2} $$ with $\mathcal{A} = A + \frac{\mu}{2}I$. Homogeneous is referring to ...


1

We have $$ e\dot e+e^2-e\theta w(t) =0\\ \theta\dot\theta +\theta e w(t) = 0 $$ after adding $$ \frac 12\frac{d}{dt}(e^2+\theta^2)=-e^2 $$ so as long as $e \ne 0$ supposing that $\theta = \theta_0\ne 0$ we have that the movement goes to the equilibrium point $(0,\theta_0)\ne (0,0)$ hence no asymptotic stability is possible.


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