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I would like to have some help on this system:

System Diagram

I have to consider that all initial conditions at t = 0 are null. I've tried some things and I have some answers but I don't know if they are true. Could someone shed some light on this?

Here is what I've tried:

calculations

(Original image showing calculations)

Is this right? or wrong?

Also, is this system asymptotically stable if and only if a <0 ? Or does B effect on that?

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  • \$\begingroup\$ I can't see anything particularly wrong with it. Did you expect something else or more? Do you perhaps want to add another effort with a different answer and see where it went wrong? \$\endgroup\$ – Sven B Jun 17 '18 at 21:21
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    \$\begingroup\$ I just want to know if this equation is the right way of representing this diagram, because I wasn't sure if it was, all I want is the H(s) of the diagram, if it is wrong, another option would be good :D \$\endgroup\$ – Matheus Ribeiro Jun 17 '18 at 21:25
  • \$\begingroup\$ I think you're good. The input of the integrator is \$bx(t) - ay(t) \$ so the output is \$y(t) = \int_0^t (bx(\tau) - ay(\tau)) d\tau\$ which is a differential equation you then solve. The only comments I'd have are only minor. \$\endgroup\$ – Sven B Jun 17 '18 at 21:35
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    \$\begingroup\$ The block diagram represents the classical active first-order lowpass. \$\endgroup\$ – LvW Jun 18 '18 at 7:32
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    \$\begingroup\$ added a question to the main question \$\endgroup\$ – Matheus Ribeiro Jun 18 '18 at 16:09
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This is how I would do it, personally. The diagram at the bottom has a + on the top left to signify adding bX(s) and a negative at the bottom to signify subtracting KY(s).

This is probably what SeventyMTB had in mind when he said you could save yourself trouble by using the Laplace earlier on.

My approach

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  • \$\begingroup\$ I see, thanks, but is this system asymptotically stable if and only if a <0 ? \$\endgroup\$ – Matheus Ribeiro Jun 18 '18 at 16:09
  • \$\begingroup\$ Yes, for a > 0 (or K in my diagram), you get a pole on the left hand side of the s-plane, which is stable. For a = 0, it is marginally stable (the system will oscillate infinitely to any input), and for a < 0, the system has an unstable pole on the right hand side of the s-plane. I can draw this out if you wish. \$\endgroup\$ – Tri Jun 18 '18 at 20:13
  • \$\begingroup\$ Another fun fact is that b does effect the stability of the system (too large a b will make the system unstable), but this depends partially on the sampling frequency of the control system. \$\endgroup\$ – Tri Jun 18 '18 at 20:15
  • \$\begingroup\$ @Tri "sampling frequency of the control system" This is incorrect, the system is of continuous time. There is no sampling involved here, and the system is stable for any non-infinite \$b\$. You could argue that a discrete version of the system could be unstable, but then a discretization method (zoh, euler, tustin, ...) should be mentioned explicitly, as some will guarantee stability and some won't. \$\endgroup\$ – Vicente Cunha Jun 19 '18 at 12:11
  • \$\begingroup\$ Yes, you are correct. When I said "sampling frequency" I assumed that Matheus would understand that I was talking about what would happen if you implemented this in the real world with a computer, but I also didn't want to go into too much detail unless he asked more of me. \$\endgroup\$ – Tri Jun 19 '18 at 19:43

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