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I have the following 2 diagrams:

enter image description here

enter image description here

The step response is of magnitude of 3.

I'm a beginner at this so I've done something stupid probably but I have trouble finding answers regarding control engineering on the internet. This is what I tried doing: I found the poles and the zeros from the root locus. z=-5,+4 p=-6,-10,-3 I think my transfer function is given from this formula but I'm not sure if we have an H(s) in the feedback and it is not stated : $$ T(s)= \frac{KG(s)}{1+KG(s)} $$ From the poles and the zeros my open-loop transfer function G(s) is : $$ G(s)= \frac{(s+5)(s-4)}{(s+10)(s+6)(s+3)}$$ Doing the calculations I find : $$ T(s)= \frac{Ks^2+Ks-20K}{s^3+(K+19)s^2+(108+K)+180-20K}$$ From the step response(final value is 4) and the final value theorem I find $\frac{-20K}{180-20K}=-4/3=>K=5.14$ I divided 4 by 3 because the diagram above is for a step of magnitude of 3. Now testing the step response of the function I found I get this: enter image description here

It's similar to the first response but not identical.

What am I missing here?

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  • $\begingroup$ I use step(T) to generate the last diagram. The others are given by the exercise. $\endgroup$ – John Katsantas May 7 '17 at 12:44
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You are thinking too complicated. Namely the input to the root locus and step response diagram is the open loop transfer function, so the question is asking you to find this open loop transfer function. The step response diagram tells you the steady state gain, while the root locus diagram the poles and zeros of the transfer function. So the overal transfer function will look like,

$$ G(s) = \frac{K\, (s+5)\, (s-4)\, }{(s+10)\, (s+6)\, (s+3)}, $$

with $3\, G(0) = -4$. Substituting in the equation for $G$ and solving for $K$ yields $K = 12$.

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  • $\begingroup$ From what I've learnt the root locus is always for the open loop transfer function of a system.Even when we have a system with feedback we search for the open loop transfer function in order to draw the root locus diagram. How do I know that in this case the system has no feedback? $\endgroup$ – John Katsantas May 8 '17 at 9:32

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