# Re: Basic Question

Pramit 'Jake' Sarma (psarma@che.iitb.ernet.in)
Thu, 1 Jul 1999 15:29:48 +0200 (MET DST)

On Tue, 22 Jun 1999, Feijun Song wrote:

>
> Generally, when the control objective is near the set point, one
> would say a FLC would function equally to a conventional P(I)D
> controller since the objective is fairly a linear system now.
>

The nearness to setpoint/equilibrium allows many nonlinear systems to be
well-defined by a linear one - but only the *system*, not the controllers.
If the FLC is non-adaptive (say), or fixed, then it will remain exactly
as nonlinear near or far from equilibrium. Much analysis shows simply
that the FLC-PI is like a gain-scheduled PI. Anyway, a good nonlinear
controller (FLC) will control even a linear process equal or better than a
a linear one. The emphasis on this resemblance is more theoretical, and
tells very little to nothing about the performance.

> Where the above statement is true for 2D systems, based on my limited
> experience on 4D systems, I would like to point out that even near
> the set point, a FLC could outperform a P(I)D controller for 4D systems.
> The reason is below:
>

See parts of above.

> We know for 4D system, a system trajectory oscillates a lot with a P(I)D
> controller, and a trajectory goes to the set point after several
> oscillations. This kind of trajectory is energy-comsuming, time-consuming.
> There are possibilities that a better trajectory exist within the
> objective's ability, and all the control commands along this better
> trajectory is a nonlinear function of system states. A FLC can approximate
> this nonlinear function fairly well whereas a P(I)D could never do.
> So even when system is near the set point, a P(I)D controller still
> cause lots of oscillation, by contrast, a FLC can directly drive the
> objective to the set point without much oscillation.
>

The nonlinear function mapped by the FLC is removed from the system
states. The FLC(e, \delta e) depends only on "output feedback" error,
not the directly the states of the system. The inverted pendulum has 4
states but *only* 2 outputs (theta, omega) hence only 2 SISO (TISO)
FLC's. The (e, \delta e) map is not a state phase-space map, but an
input/output map only. This is nominally independent of the no. of
states.

> I have designed a LQR and a TS type FLC for a 4D inverted pendulum, lots
> of trajecotries tell me that the FLC drive the pole and cart directly to
> the set point, whereas the LQR always drive the pole and cart in an
> oscillating way, which costs lots of time and energy.
>

LQR is a linear structure. FLC is nonlinear. The plant is strongly
nonlinear. Ergo.

> Since I am sitll quite a naive in this field, I may have wrong
> understanding and please excuse me if did make the mistake.
> I would thank you very much if you could point out any mistakes. Your time
> on this will be highly appreciated.
>
>
> Feijun Song
> Ph.D candidate
> Ocean Engineering Dept.
> Florida Atlantic Univ..
>

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Pramit "Jake" Sarma
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e-mail: jake_n_jazz@yahoo.com e-mail: psarma@che.iitb.ernet.in

Process Systems and Control
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