I am given the response of a step of magnitude of 3 and the root locus and I have to find the transfer function of the system. The function I find gives me the step response(magnitude of 3 again) of the last diagram.

enter image description here

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 $\frac{-20K}{180-20K}=-4/3\implies K=5.14$. I divided 4 by 3 because of the step magnitude. With this K the step response is the one in the third diagram.It's close to the first one but it's not the one I'm looking for.

What am I missing here?

  • $\begingroup$ That is true when K is in the feedback path. Can you explain why my understanding is wrong? I'm very positive that I've done the root locus part right , except for the fact that G(s) might have a multiplying constant. $\endgroup$ – John Katsantas May 7 '17 at 19:16
  • $\begingroup$ Seeing the root locus , though, you can find the poles and zeros of the open-loop transfer function. The way I thought it (which , G(s) happened to be my open-loop transfer function. I wasn't aware of the fact that K is in the feedback in Matlab, it's gonna help. Thank you. And yes , I meant step of magnitude 3.My bad. $\endgroup$ – John Katsantas May 7 '17 at 19:30

The transfer function is

$$H(s)=\frac{K(s - 4)(s + 5)}{(s + 3)(s + 6)(s + 10)}=\frac{K(s^2 + s - 20)}{s^3 + 19s^2 + 108s + 180}$$ So we need to find $K$.

The LT of input step is $\frac{3}{s}$. Using the FVT for the step response $g(t)$:

$$\displaystyle\lim_{t\to \infty}g(t)=\displaystyle\lim_{s\to 0}s\cdot \frac{3}{s}\cdot\frac{K(s^2 + s - 20)}{s^3 + 19s^2 + 108s + 180}=-4$$ or $$\frac{3\cdot K\cdot (-20)}{180}=-4$$ which gives $K=12$. Then use the following code to plot the step response

K= 12;
sys = tf(K*[1 1 -20],[1 19 108 180]);

enter image description here

So there is no feedback.

  • $\begingroup$ So my mistake was assuming that we have feedback from the beginning and using the transfer function formula for the closed-loop. But how do I check if there is feedback? $\endgroup$ – John Katsantas May 8 '17 at 9:26
  • 1
    $\begingroup$ You should assume closed-loop if it is explicitly stated in the question. $\endgroup$ – msm May 8 '17 at 9:34

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