Consider the generic feedback loop, and the transfer function $G(s)$ shown by the following root locus plot.


Root Locus

Where $\mathbf{x}$ denotes the open-loop poles and $\square$ denotes the closed loop poles.

I want to determine if this root-locus produces any of the following output responses to a unit step reference signal:

output response


I attempted to build the open loop transfer function:

$$G_{OL} = \frac{K}{s^2+2s+2}$$ Find the value $K$ using the FVT for the step response and graph a) $$\lim_{s\to 0} = s\cdot\frac{1}{s}\frac{K}{s^2+2s+2} = 1$$ which gives $K = 2$, plug this back in and see if the step response matches any of the figures. It is possible none of the figures match, but I think I am doing something incorrect.

Root locus shows all the possibilities of K, but I am also given the closed loop poles. Can I use those to get K? Is the unit step response in relation to the Closed Loop Transfer Function? or the open loop? I can see that building the connection between the graphs is important, I seem to be missing something though. Thanks!


I believe the corresponding Step Response Graph is C) but I'm still uncertain on if I followed the right process: Finding the closed-loop transfer function I get $$G_{CL} = \frac{K}{s^2+2s + 2 + K}$$ Applying FVT to graph C I get $K = 3$, which makes $$G_{CL} = \frac{3}{s^2+2s + 5}$$ and plotting the unit step response with Matlab gives C). What if I didn't have access to Matlab? Is there an easier way? Is it sufficient to match the closed loop poles after I determine $K$?

  • 1
    $\begingroup$ Note that the denominator of $G(s)$ should be $s^2+2s+2$ instead of $s^2-2s+2$. $\endgroup$
    – Matt L.
    Commented Jan 19, 2021 at 20:26
  • $\begingroup$ @MattL. You're right, thanks for catching that. I updated the question $\endgroup$
    – Clark
    Commented Jan 19, 2021 at 20:36

1 Answer 1


The closed loop poles are the roots of the polynomial


and, according to the root locus plot, they are $s_{1,2}=-1\pm 2j$. Consequently, we get

$$2+K=|1+2j|^2=5\quad\Longrightarrow\quad K=3\tag{2}$$

With $K=3$ we obtain $$H(0)=\frac{K}{2+K}=\frac35\tag{3}$$

which leaves step response $C$ as the only option.

  • $\begingroup$ How did you know step 2? are you simply factoring $D(s)$? Then what is $H(0)$? is that simply the application of FVT to the closed-loop transfer function. Physical meaning, I suppose. I get you're just plugging in $0$ $\endgroup$
    – Clark
    Commented Jan 19, 2021 at 22:16
  • 1
    $\begingroup$ @Clark: Eq. $(2)$ is just a consequence of the quadratic polynomial $(1)$ having two complex conjugate zeros. So the constant part $2+K$ must equal the squared magnitude of these zeros. And $H(0)$ is the DC response of the filter, which is the value the step response converges to. $\endgroup$
    – Matt L.
    Commented Jan 20, 2021 at 8:09

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