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.

  • \$\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
  • \$\begingroup\$ @Chu the definition of lead compensation implies that the transfer function \$\frac{s+z}{s+p}\$ with \$0<z<p\$ represents a lead compensator. \$\endgroup\$ – kbakshi314 Mar 24 at 6:40
  • 1
    \$\begingroup\$ @kb314 Yes, you're correct. My mistake. \$\endgroup\$ – Chu Mar 24 at 12:40
  • \$\begingroup\$ @Chu I request you to hazard a guess at the OP's reasoning in computing the position of the 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\$ and the reason as to why \$\phi=65.21^{\circ}\$. The OP seems inactive and I was hoping you could help improve my understanding. \$\endgroup\$ – kbakshi314 Mar 24 at 15:16

I got the position of the pole to be \$8.873\$ and the gain to be \$8.873\$.


enter image description here

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

enter image description here

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\$.

  • \$\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
  • \$\begingroup\$ This comment clarifies the method of designing the zero \$z\$ of the lead compensation.The open-loop transfer function of the plant with the compensator \$G_cP=\frac{s+z}{s^2 (s+10z)}\$ has the associated closed-loop (CL) transfer function \$\frac{G_cP}{1+G_cP}=\frac{num(G_cP)}{num(G_cP)+den(G_cP)}\$ and the equation \$G_cP|_{s_d}=-1 \equiv num(G_cP)+den(G_cP) = 0\$, where \$s_d\$ indicates one of the desired poles of the CL system, was used to determine the value of \$z\$. The designed \$z\$ would thus result in CL poles identical to the required \$s_d,\bar{s}_d\$. \$\endgroup\$ – kbakshi314 Mar 24 at 6:55

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