52 States in 8 Months

Where does the integral action go?

Posted in University stuff by Ulf on April 1, 2009

I just had to refresh some of my memories from our first Control Theory lecture. I was thinking about how to eliminate a steady state control offset. Yes, everybody knows: We need an integrator! The next question was: Where does the integrator go? Do we have to put it into the controller K or can we rely on an integrator in the plant G, for example if we use state feedback?

First case, let’s assume there is a disturbance d at the output of our plant:
output_disturb
case1
If G contains an integrated signal, then G(0)=∞. That means that a constant input will be amplified … infinitely. Now, since the coefficient in front of d will then be 1/(1+∞·K)=0, we don’t have to worry about the disturbance in this case. Any stabilizing proportional controller K will achieve a vanishing steady state control offset.

But what if a disturbance acts on the input of our plant?
output_disturb
case2
Things are different here. Even if G(0)=∞, the coefficient of d will be nonzero. Specifically, it will be 1/(1/∞+K) = 1/K. This means that the steady control offset can be made arbitrarily small with sufficiently high controller gains, but it will never be zero. Unless of course you have an integrator in your controller, K(0)=∞ :-).

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One Response

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  1. Andreas said, on April 1, 2009 at 11:35 am

    toll


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