# Placing a Compensating Pole

Discussion in 'Homework Help' started by Alex Panebianco, Mar 27, 2012.

1. ### Alex Panebianco Thread Starter New Member

Mar 21, 2012
2
0
I'm currently working with a transfer function and altering its compensator through use of the Matlab rltool command.
I've placed a compensating zero, which has led to a root locus showing an impossibility of instability, however, I now have to place a compensating pole.

This pole has to be placed far away from the jω axis, in an attempt to make the system including the compensating zero physically realisable, yet cause negligible effect to the overall stability of the system.

I'm unsure how to go about placing this pole, as I don't fully understand what a negligible effect, in this sense, would be.

Upon placing the pole in a random place, the root locus seems to reshape similarly to the original transfer function's root locus.

In this situation, is a negligible effect one that is unnoticeable on the whole root locus diagram, or one which has relatively little effect with relation to the transient specifications I'm working with?

My transfer function is

$G(s)=\frac{5}{s(\frac{s}{6}+1)(\frac{s}{2}+1)}$

and my compensating zero has been placed at s = -3.

My transient specifications are:

A steady state error of 0.2
A settling time < 4 seconds
A damping ratio of 0.6

2. ### daviddeakin Active Member

Aug 6, 2009
207
27
I assume you're using unity-negative feedback?

A damping ratio of 0.6 gives you a line of constant damping at arccos(0.6) = 53°.

For a settling time of 4sec the real part of the dominant closed-loop poles needs to be: -4/4sec= -1. The poles also need to lie on the line of constant damping, so the imaginary part must be:
ω = σ x tan(53) = 1.3
So you need the locus branches to pass through the points: -1 ± j1.3.

Next you need to find the angle contribution of all the poles and zeros you have so far (I assume you're familiar with this?) It's easy, if tedious, so I'm not doing this for you. There's probably a way to make Matlab do it, but I'm not that familiar with it.

The Angle criterion states that:
(sum of the zero angles - sum of pole angles) must be an odd multiple of 180 degrees.

If it isn't (and it won't be) then you need to make up the deficit with some extra phase lag, which comes from the pole you are trying to place. Since you know the required angle, some quick trigonometry will tell you where to put it (it will automatically end up being far away from the others, making it negligible). If all goes well you should find your locus branches now pass through the desired points, and Matlab will even tell you the DC gain at those points.