2
$\begingroup$

The question asks me to solve the positioning problem where:

$$ \dot{x_1} = x_2 $$ $$ \dot{x_2} = u_1 \in U_{bb} $$ $$ x_1(0) = - \text{X} (<0) $$ $$ x_2 (0) = 0 $$ $$ x_1(t_1) = 0$$ $$ x_2(t_1) = 0$$

I'm having trouble making the A matrix that I understand is required for the problem to be solved. i.e. the $$ \dot{x} = Ax + Bu$$ part of the control problem.

I think I may also need a bit of an explanation, too. When the lecturer teaches this I have a little trouble grasping the movements that lead to the answer.

Some help would be great. All the literature I've found on the topic turns out to be a little denser than I can handle.

Edit:

When I try this is how I see the formula:

$$ \begin{pmatrix} 1 \\ 1 \\ \end{pmatrix} \dot{x} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \\ \end{pmatrix} x + \begin{pmatrix} 0 \\ 1 \\ \end{pmatrix} u $$

Is this correct? If so, what happens from here?

$\endgroup$
1
$\begingroup$

The equation you've written down for the state space form is almost correct. On the left-hand side, typically the vector $x$ is defined as \begin{equation} x = \left(\begin{array}{c} x_1 \\ x_2 \end{array}\right) \end{equation} so that \begin{equation} \dot{x} = \left(\begin{array}{c} \dot{x}_1 \\ \dot{x}_2 \end{array}\right) \end{equation} which means you don't need the extra $(1\,\,\,\, 1)^T$ vector on the left-hand side (and indeed, virtually always in linear systems the left-hand side only contains $\dot{x}$ -- I can't recall seeing any other form of this equation). Then the corrected form of this equation is \begin{equation} \dot{x} = \left(\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right)x + \left(\begin{array}{c} 0 \\ 1\end{array}\right)u. \end{equation} To see that this makes sense, multiplying $Ax$ through lets us recover the original differential equation that we started with.

$\endgroup$
1
$\begingroup$

I'm not sure if you are still interested in the answer. Short answer: your answer is correct for $A$. How $A$ has been acquired, well I think everything set up for you. You are not only given the differential equation(s) that describes your system but also given the state space parameters. In other words, your task here is to put the differential equations in a compact (matrix) form. After that you need to check if the matrix $A$ time-invariant or time-varying. $A$ is time-invariant (constant), so the solution of your system is given by $$ x(t) = e^{At} x_{0} + e^{At} \int_{t_{0}}^{t} e^{-A\tau} B u(\tau) d \tau $$

Now you need to compute $e^{At}$, and it is ($A$ is 2x2) $$ e^{At} = a_{0}I + a_{1}A \\ a_{0} + a_{1} \lambda_{1} = e^{\lambda_{1}t} \\ a_{0} + a_{1} \lambda_{2} = e^{\lambda_{2}t} \\ $$ and to compute $\lambda_{1} \ \ \& \ \ \lambda_{2} $ $$ det[ A- \lambda I] = 0 $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.