# 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
Commented 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. Commented Mar 24, 2021 at 6:40
• @kb314 Yes, you're correct. My mistake.
– Chu
Commented 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. Commented 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? Commented 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. Commented 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 $s_d,\bar{s}_d$. Commented Mar 24, 2021 at 6:55