# Lead compensator for double integrator

1. May 2, 2015

### zoom1

I'm trying to design a lead compensator for a double integrator plant, which is plant = tf([1],[1 0 0])

I started with the angle condition. So, poles contribute (+) angle and zeros contribute (-) angle and that sums to 180

Desired pole location is 4.4 +- 4.4j

So, 45 + 45 = 90 degrees comes from the poles at the origin, assume that zero of the lead compensator is placed at -5, then atand(4.4/1) = 77.19 contribution from zero of the lead compensator.

45 + 45 - 77.19 + Lp = 180

Where Lp is the pole of the lead compensator. Lp should contribute +167.19 degrees, so tand(167.19) = -0.2274

At this point I wouldn't expect to get a negative number. Anyways, I proceeded and found the location of the Lp as 4.4/0.2274 = 19.3492, so Lp should be located at -19.3492 - 4.4 = 23.75

When I close the loop with unity feedback closedLoop = feedback(plant*lead,1) and check the gain with sisotool(closedLoop), I see that my desired pole location is not reached. at -4.4 I got -+6.84i

What am I doing wrong ?

2. May 2, 2015

### milesyoung

It sums to $r180^\circ,r = \pm1,\pm3,\pm5,\dots$ if the test point is on the root locus.

Should be $s = -4.4 \pm j4.4$?

$\arg(s) = \pm135^\circ$, and $\arg(s - z) \approx \pm82.2^\circ,z = -5$.

If $s = -4.4 + j4.4$, then the angle sum is $135^\circ + 135^\circ - 82.2^\circ = 187.8^\circ$. You want to remove those $7.8^\circ$, but that's not possible with a single pole (it adds to the angle sum).

The problem is your zero. You want it to subtract more than $90^\circ$, so the angle sum dips below $180^\circ$, and the pole makes up the difference, i.e. you should have $z > -4.4$.

3. May 2, 2015

### zoom1

Thank you so much. I was sure that I was missing something.