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I am an autodidact in control engineering and I have been trying to design a lead compensator for double integrator system so that the closed loop poles have damping $$\zeta=0.5$$ and natural frequency $$\omega_n=1\,rad\cdot s^{-1}$$ while pole/zero ratio of the compensator would not be greater than 10.

Desired location of closed loop poles is $$s_d=-\zeta\cdot\omega_n\pm i\cdot\omega_n\cdot\sqrt{1-\zeta^2}=-0.5\pm i\cdot 0.866.$$ Transfer function of lead compensator is $$D(s)=K\cdot\frac{s+z}{s+p}.$$ I started with a pole at -10. Then I computed position of zero as $$\zeta\cdot\omega_n+\frac{wn\cdot\sqrt{1-\zeta^2}}{\tan(\phi)}=0.5\cdot 1+\frac{1\cdot\sqrt{1-0.5^2}}{\tan(65.21^{\circ})}=0.9.$$ According to the root locus I set the gain of the compensator $$K=10.$$ My problem is that this design doesn't fulfill the constraint of pole/zero ratio and also the percent overshoot is greater than I would expected according to $$\exp\left(\frac{-\pi\cdot\zeta}{\sqrt{1-\zeta^2}}\right)\cdot 100=16.3\%.$$ When I tried different pole locations I got approximately same results. How can I solve this problem? Thanks for any ideas.

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  • \$\begingroup\$ With p=10 and z=0.9 you have a lag compensator, not lead: the break frequencies are 1/p and 1/z. \$\endgroup\$ – Chu Feb 6 '16 at 19:33
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I got the position of the pole to be 8.873 and the gain to be 8.873.

First:

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Yields z = .8873

Next plug in z into the transfer function above and you get -.1127. The gain is then 1 / .1127 = 8.873

Step response:

enter image description here

I also had issues with over shoot (about 33 %) but at least it meets the pole / zero relationship. I'm guessing it's an issue with the 2 poles at 0.

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  • \$\begingroup\$ Thank you for your answer Mr. Barkely. Can you tell me how did you get pole at 8.873? Did you proceed by trial and error? \$\endgroup\$ – Steve Feb 7 '16 at 10:23
  • \$\begingroup\$ Set the combined transfer function -- (s+z) / (s^2 * (s+10z)) -- equal to -1 at the desired pole with the pole / zero ratio already set to 10 (hence (s+z) / (s+10z)). You could repeat this with other pole / zero ratios. \$\endgroup\$ – Swarles Barkely Feb 7 '16 at 15:57

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