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.
Best Answer
I got the position of the pole to be \$8.873\$ and the gain to be \$8.873\$.
First,
yields \$z = 0.8873\$. Next plug in \$z\$ into the transfer function above and you get \$-0.1127\$. The gain is then \$\frac{1}{0.1127} = 8.873\$.
Step response is as follows
I also had issues with the overshoot (about \$33 \%\$) but at least it meets the pole-zero relationship. I'm guessing it's an issue with the two poles at \$0\$.