Confused about Pole Zero Compensation of Double Integrator

RRogers wrote:


I think that the more ways one can see a thing, the better one's grasp of it is likely to be. Each representation highlights certain aspects; together they evoke a more complete picture, even if any one of them has all the information.

And by me.
Jerry
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Engineering is the art of making what you want from things you can get.

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Haven't looked at Nichol's charts yet but they're on the list. Jerry has summed up my dilemma: A system with two integrators has 180 degrees of phase shift at all frequencies. Even if compensated out at the unity gain frequency, with a phase boost circuit, that leaves a phase shift of 179.9 degrees or whatever at some low frequency where the compensator has run out of puff, and where a double integrator has plenty of gain. A step input applied to this system ( in my imperfect understanding ) ought at the least to make the output ring with a very underdamped oscillation.
I have looked at it on an S-plane root-lous diagram and the same problem occurs ( as it should ). A compensated double integrator pulls the root locus away from the jw axis and into the left hand plane, but it still snakes back to graze the jw axis at low frequency ( near the origin ) as it connects up with the poles at the origin, so it still looks marginally stable at low frequencies. I can accept that it will not oscillate, but the settling times might be large.
The question occurs, since a normal system with say, three poles will easily manage 180 degrees of phase shift at one particular frequency, and if there is gain at that frequency it will oscillate at that frequency.
A double integrator in theory, uncompensated, could oscillate at any and all frequencies up the the unity gain frequency. What am I to make of that? A strange beast indeed. I haven't revisited the example given in Dorf yet as I want to take a fresh look at it. In the mean time I'm just working through some more straightforward examples of lead, lag, and PID compensation.
Thanks for all the comments,
Andy.
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On Fri, 30 Sep 2005 23:27:51 +0100, Andy wrote

Yes, it does behave in a very underdamped way, but only if you make the gain too high or too low. Suppose the compensator consists of a zero at -1 and a pole at -10, then with RootLocs you can easily see that for a minimum damping ratio of 0.7 say, the gain must be not less than 20 and not more than 50 (rounded values). Note: these are the values of K in the open loop transfer function when it is expressed with all coefficients of S as 1, i.e. K(S+A)/S^2(S+B). You can get the same results from the Bode plot by shifting it up/down until the phase margin (at the crossover frequency) drops below a certain value. But its not immediately evident what that value is for a closed loop damping ratio of 0.7
AAR
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Aah, good point. I remembered I have a graph of damping ratio versus phase margin for a second order system, 0.7 is achieved with a phase margin of 65degrees. I had forgotten that a root locus diagram shows all values of gain, so by picking that carefully one can keep to a well-damped part of the curve.
cheers,
Andy.
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