What Is Linearization?
Ғылым және технология
Why go through the trouble of linearizing a model? To paraphrase Richard Feynman, it’s because we know how to solve linear systems. With a linear model we can more easily design a controller, assess stability, and understand the system dynamics.
This video introduces the concept of linearization and covers some of the topics that will help you understand how linearization is used and why it’s helpful.
This video also describes operating points and the process of trimming your system to make an operating point an equilibrium. To end, we walk through an example of Jacobian linearization by looking at the first order partial derivatives of a system.
Check out our video to get a more practical understanding of how linearization is accomplished in Simulink: • Linearizing Simulink M...
Related resources:
- Learn about linearization for model analysis and control design: bit.ly/2E76uJe
- Watch more controls Systems Tech Talks: bit.ly/2rTc8Yp
- Trimming and Linearizing Simulink Models Documentations: bit.ly/3OSQgGp
- Linearize Nonlinear Models Documentation: bit.ly/2DXWy4G
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Пікірлер: 20
Every time he says tech talk i hear "TED talk", thanks Brian.
Hello Brian. Very well taught. Thanks a million times.
Hello. Thanks for interesting video. Can you give more detail process about step: x-xbar->x in state space equation and how simulink model and controller connect with it will look like. Thanks you so much.
great video, really helped me out
Hello Brian! in the final equation you have gotten and when I tried to substitute in value of the operating point(H_bar, V_bar), does the left side (H_dot) has to equal to zero? I tired to substitute but I have gotten nonzero value, why?
Thanks Brian !!
Thank you so much.
I have trigonometric dynamic equation. Is it highly nonlinear? I want to check the controllability and observability of my model.
thank you
I did not get the "re-label where we call zero" part. It looked like you made Hbar = 0, but then you took the square root of it n the slope term, making it result in -a/4A. It looks mathematically inconsistent, to consider Hbar to be zero only at one part. Could you explain? (Or someone that got this)
Hello, Brian if you are linearizing at H(bar) = 4 then how does H-H(bar) = H? is not H(bar) at 0
Thank you!
çok yararlı bi video
Two of the links link to AEM pages within MathWorks. Please edit those. Thanks
@BrianBDouglas
5 жыл бұрын
Samvith V Rao I asked Mathworks to update the links (I don’t post these so I can’t do it myself). Thanks for pointing it out!
@MATLAB
5 жыл бұрын
Hi Samvith, thanks for catching that. Links are updated now.
Fantastic
how pendulum will be at equilibrium pointing straight up ?
@louco2
Жыл бұрын
With friction forces helping/triming a a bit? Hahaha
@jayaramjonnada5855
8 күн бұрын
can someone explain this please?