Closed Loop Transfer Function - PD Control

Question

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.

Any working or help as to how I go about this would be greatly appreciated.

Answer 1

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)

Answer 2

I find it best not to do algebra until it is needed. Loop Gain here is G(s)=Kp(s+9){10/(s+a2)}{1/(s^2+6s+15)}, a gain factor 10*Kp, a zero and three poles. A Bode plot of this function shows that the unity frequency depends on Kp. You change the phase margin with this parameter. Find the PM needed for your overshoot and adjust Kp to get this PM. Doing the algebra too early confuses pole-zero information.