# Find transfer function from step response and root locus?

I am given a step response 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.

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 first step response is of magnitude of 3. 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?

I think you did mix this up with proportional feedback. The transfer function is given by

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

in which $K$ is a parameter that needs to be determined.

Because $G(s)$ is a stable plant (all poles are in the left half plane) we can determine the DC gain by the final value theorem.

$$G(s=0)=K\frac{5\cdot (-4)}{10\cdot 6 \cdot 3}=-\frac{K}{9}$$ And $G(s=0)=y(s=0)/u(s=0)=\frac{-4}{3}\implies a=12.$ Hence,

$$G(s)=12\frac{(s+5)(s-4)}{(s+10)(s+6)(s+3)}.$$

MATLAB testing:

s = tf('s');
G = 12*(s+5)*(s-4)/((s+10)*(s+6)*(s+3));
step(3*G); % 3 to scale the unit step response