For example, i have this differential equation:

$$\dot{y} = 1.8 + 0.58u \sqrt{y}$$

Is it possible to model this system in open loop?? My understanding is that since the term $y$ is in the output as well as one of the input terms, then it is only possible to model with a feedback loop.

  • $\begingroup$ You are aware that the term in the left doesn't appear on the right ? y_dot is time rate derivative most likely. $\endgroup$
    – ShadowMan
    Jun 12 '19 at 0:24
  • $\begingroup$ Yes, y \dot is dy/dt and it is the output in this equation. But how can there be an input term with the squareroot of y, if there is no feedback in an open loop. What i need to do is model this equation in open loop, so that's the problem. $\endgroup$
    – DryRun
    Jun 12 '19 at 0:53
  • $\begingroup$ There are a number of methods available to solve differential equations numerically have you considered any of these? $\endgroup$
    – ShadowMan
    Jun 12 '19 at 2:24

The answer really boils down to how you define your system. If you say "this is my plant" and give that equation, then as long as $u$ does not depend on $y$ then the system is operating in open loop.

Yes, there is some inherent feedback internal to your plant, but that's the case with just about anything that might be described with a differential equation.


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