# Closed Loop Transfer Function - PD Control

For the following control system, I am trying to find the characteristics of the system. I have firstly found the closed loop transfer functions $$\T(s) =\frac{C(s)}{R(s)}\$$ , to be:

$$\G(s) =K_p(s+9)*\frac{10}{s+12}*\frac{1}{s^2+6s+15}\$$

$$\G(s) =\frac{10K_ps+90K_p}{s^3+18s^2+87s+180}\$$

Now as feedback is Uniity Feedback, therefore H = 1, I get;

$$\T(s) = \frac{G(s)}{1+G(s)H(s)}\$$

Which means $$\T(s) = \frac{10K_ps+90K_p}{s^3+18s^2+87s+180+10K_ps+90K_p}\$$

My first question for this problem is, is this Transfer Function correct, or have I miscalculated during a particular step up to this point.

Continuing on, I am now trying to find the $$\K_p\$$ value that leads to less than a 15% Overshoot for the system.

Unfortunately, I am unsure what to do next to find the $$\K_p\$$ value for this problem.

To find the $$\K_p\$$ value that leads to less than a 15% Overshoot for the system. we can use the root locus plot, by looking at the position of the dominant poles (the two closer to the imaginary line). I have also added the sgrid lines showing the $$\\zeta\$$ and $$\\omega\$$ positions. Using the formula that $$\zeta \geq \frac{-\ln(PO/100)}{\sqrt{ \pi^2 + \ln^2(PO/100) }},$$ we have that $$\zeta \geq \frac{-\ln(15/100)}{\sqrt{ \pi^2 + \ln^2(15/100) }}= 0.517.$$ So, by positioning the poles below that 0.55 damping line should get you the overshoot to meet your requirement. Therefore, for a gain in $$\K_p \in [-1,2]\$$ you would meet that requirement. (the image looks horrible, but if you click on it it gets slightly better) 