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For the transfer function $G(s)$, I tried to design a lead compensator for the function to have a response to the step with the following specifications:

$$\%OS = 10\% \ \rightarrow \ Overshoot $$ $$ T_s = 5sec \ \rightarrow \ 2\% \ Criterion$$

When I simulate the function already with the lead compensator by the rlocus() command the system shows the desired behavior. However when I simulate with the step() function the parameters do not match.

Behavior of the compensated function with rlocus() and step() functions Matlab code used Steps used to design the lead compensator

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    $\begingroup$ Why are you using the closed loop transfer function in rlocus()? $\endgroup$ – fibonatic Jan 22 '19 at 7:34
  • $\begingroup$ I just noticed that you look at the overshoot of the rootlocus plot with a gain of zero, which should yield just the closed loop. However I can't find any documentation about how Matlab calculates the overshoot in rlocus, but it looks like it it is just based on the damping coefficient thus only take into consideration one complex conjugate pole pair. So the overshoot given by rlocus might not hold if there are more than two poles or any zeros. $\endgroup$ – fibonatic Jan 22 '19 at 14:23
  • $\begingroup$ @fibonatic To see if the poles at the origin went to the calculated location and to see the overshoot in closed loop. I calculated a new pole and zero for the compensator because I noticed that the pole is very far from the dominant poles which I think affects the performance of the controller, but even one pole closer to the desired dominant poles, the performance of the function continues going wrong. $\endgroup$ – Max Jan 25 '19 at 22:31
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The open loop transfer function of such a system with two poles at the origin

$$ G(s) = \frac{-0.6982}{s^2} $$

is, someone can say, "very" unstable despite the fact that the poles are at the origin. You can check this by first of all obtaining the step response of the open loop system which is:

enter image description here

It can be very challenging to reguate such a system. In order to use the root locus method to design the lead compensator the root locus of the feedforward path of the the system has to be drawn (same goes if we want to use the bode plot for the design process). In this particular system the feedforward path includes only the transer function $G(s)$. It is shown below (again it does not look like something easily manageable):

enter image description here

Next step of the design process is to add to the root locus graph the necessary requirements for the overshoot and the settling time which are shown by drawing the appropriate diagonal lines for overshoot and the appropriate vertical line for the settling time. Doing so produces the following graph. Notice that in order to achieve the requirements you have to move the poles of the closed loop system inside the white area of the graph (red x is for the pole of the lead compensator, red o is the pole of the lead compensator and the filled pink circles are the poles of the closed loop system which need to be moved in the white area).

enter image description here

Now, you can achieve the requirements by adjusting the pole and zero of the compeansator and moving the closed loop poles (which means you basically adjust some gain). I came up with the following compensator which gives the required specifications. However, it could be really challenging to deploy such compensator in real systems. Transfer function of the compensator is:

$$ G_c(s) = \frac{-66.62(s+0.45)}{s+15}$$

enter image description here

Let me note that the poles of the closed loop are not quite exactly placed in the white area of the root locus graph but I did that in order to meet the design requirements since I belive it is more important to achieve the specifications. Now, the step response of the system complies with the design from the root locus. More adjustments can always be made in order to improve the quality of the system.

enter image description here

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