# Lead Compensator for double integrator

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.

• With p=10 and z=0.9 you have a lag compensator, not lead: the break frequencies are 1/p and 1/z.
– Chu
Feb 6, 2016 at 19:33
• @Chu the definition of lead compensation implies that the transfer function $\frac{s+z}{s+p}$ with $0<z<p$ represents a lead compensator. Mar 24, 2021 at 6:40
• @kb314 Yes, you're correct. My mistake.
– Chu
Mar 24, 2021 at 12:40
• @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. Mar 24, 2021 at 15:16

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\$$.

• 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? Feb 7, 2016 at 10:23
• 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. Feb 7, 2016 at 15:57
• 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 $s_d,\bar{s}_d$. Mar 24, 2021 at 6:55