Timeline for How to get second derivative of state as a output of state-space model?
Current License: CC BY-SA 4.0
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| Jan 18 at 11:07 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Sep 8, 2023 at 10:08 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| May 6, 2023 at 12:03 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| May 5, 2021 at 2:14 | answer | added | AJN | timeline score: 1 | |
| May 5, 2021 at 2:12 | comment | added | jonk | @ChanwooAhn Well, don't assume I'm interested just yet. I may be when I feel I understand what might help you. I don't feel that way, just yet. It actually sounds more like what you want is Kalman (discrete time) or Kalman-Bucy (continuous time.) The moment you start talking about fusing information from multiple sensors, the subject has in my mind changed. | |
| May 5, 2021 at 2:07 | comment | added | Chanwoo Ahn | @jonk Thank you for your kind interest in my question. Could you please check the edited question? | |
| May 5, 2021 at 2:05 | history | edited | Chanwoo Ahn | CC BY-SA 4.0 |
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| May 5, 2021 at 1:58 | comment | added | jonk | @ChanwooAhn Like that??? (Admittedly, that's going to have increasing \$x\$ in the down direction.) You have to decide the output function(s), yet. The above is A and B, but to get C and D you need the rest. | |
| May 5, 2021 at 1:58 | comment | added | jonk | @Chanw Not sure what you want. But if:$$\ddot{x}=g$$Then two state vars:$$\left\{\begin{array}{l}x_1&=&x&&&&\dot{x_1}&=&\dot{x}&=&x_2\\x_2&=&\dot{x}&&&&\dot{x_2}&=&\ddot{x}\end{array}\right.$$ So:$$\left\{\begin{array}{l}\dot{x_1}&=&x_2\\\dot{x_2}&=&g\end{array}\right.$$State and input matrices:$$\begin{array}{c} \left[\begin{smallmatrix}\dot{x_1}\\\\\dot{x_2} \end{smallmatrix}\right]&=&\left[\begin{smallmatrix}0 & 1\\\\0 & 0 \end{smallmatrix}\right]\cdot\left[\begin{smallmatrix}x_1\\\\x_2 \end{smallmatrix}\right]+\left[\begin{smallmatrix}0\\\\g \end{smallmatrix}\right] \end{array}$$ | |
| May 5, 2021 at 1:55 | comment | added | Chanwoo Ahn | @AJN Could you explain a bit more about the way how "applying a double derivative" could be done on A, B, C, D state-space model? | |
| May 5, 2021 at 1:53 | comment | added | Chanwoo Ahn | @LetterSized I would like to stay on the time domain. So, If I want to attach an accelerometer on the free-falling ball with some disturbance force, what would be my C matrix? should I expand the state vector? Hard to understand "use the derivative operator on x dot" on A, B, C, D state-space model. | |
| May 5, 2021 at 1:49 | comment | added | Chanwoo Ahn | @jonk I am interested in both case, but the former seems more general. So, I would say the former. Does it make any difference? | |
| May 5, 2021 at 1:38 | comment | added | AJN | In a general case apply a double derivative on the position state or a single derivative on the velocity state as mentioned in the previous comment. | |
| May 5, 2021 at 1:36 | comment | added | AJN | In this particular example, the second derivative of position is given by the original differential equation itself. Adjust the C matrix and the D matrix so that one row of the output equation replicates the original differential equation. | |
| May 5, 2021 at 1:30 | comment | added | LetterSized | The derivative is a linear operator. If you take the Laplace transform, you need only multiply X(s) by s^2 to obtain the Laplace transform of x double dot. If you keep things in the time domain, then just use the derivative operator on x dot (d/dt). | |
| May 5, 2021 at 1:29 | comment | added | jonk | Are you just asking about \$F-m\ddot{x}=0\$ or \$g-\ddot{x}=0\$? | |
| May 5, 2021 at 1:08 | history | asked | Chanwoo Ahn | CC BY-SA 4.0 |