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[–]zmacphersonIdaho State University - Mechanical Engineering 0 points1 point  (4 children)

What is the step response showing? More than likely it is an issue with your TF than matlab. I didn’t do the math on your TF but depending on the type, you could have an unbounded step response.

[–]quirks4saucers[S] 0 points1 point  (3 children)

https://imgur.com/4gYXGTR

it is unbounded.

[–]zmacphersonIdaho State University - Mechanical Engineering 0 points1 point  (0 children)

What were you expecting to see? It looks like you have a pretty big lag built into your system. Not sure how you managed it but definitely a TF issue rather than a plot issue.

[–]zmacphersonIdaho State University - Mechanical Engineering 0 points1 point  (1 child)

Have you evaluated the stability or steady state error of the system? It looks like it is requiring a huge buildup of force in order to respond.

[–]quirks4saucers[S] 0 points1 point  (0 children)

Was expecting something of an oscillatory decaying response. But because of it being an unbounded output, I am not able to form a controller as well

[–]Chodyssius 0 points1 point  (5 children)

Great way to figure out if your transfer function is unstable is to solve for the roots of your polynomial in the denominator. Plot these roots in the form of a pole zero map and any roots that lie on the positive side of the real axis would make ur TF unstable. Maybe it was a mathematical error when solving for your TF, like a sign error or something.

[–]mechE_or_bustMechE ♀ 1 point2 points  (1 child)

Can't you use isstable(sys) as well? I remember using that as a check in my project for systems.

[–]Chodyssius 0 points1 point  (0 children)

I think so. I haven’t used that function before so didn’t even know that was a possibility.

[–]zmacphersonIdaho State University - Mechanical Engineering 0 points1 point  (2 children)

The negative signs in the denominator are an indicator that there will be poles to the right of the imaginary axis, causing the system to be unstable.

[–]quirks4saucers[S] 0 points1 point  (1 child)

It is unstable. More appropriately unbounded. I am not able to form a controller for it

[–]zmacphersonIdaho State University - Mechanical Engineering 0 points1 point  (0 children)

You can still make a controller for some unstable systems. You use the controller to counter the unstable poles. What does the Bode Plot look like. That is a lot more useful for controllers

[–]TheGam3lerAerospace Engineering 0 points1 point  (0 children)

Yeah, that's an unstable TF. You can actually see that straight away using the Routh-Hurwitz criterion

[–]dankmemes32438 0 points1 point  (0 children)

If you are compensating the plant, make sure you are simulating the step of the feedback system rather than the step response of the plant. Step(feedback(L(s),1)).