105 105105 1052009-09-29 13:11:30 UTC / rev b19331f50bbd8Proportional and derivativecontrol8.1 Why derivative control 958.2 Mixing the two methods of control 968.3 Optimizing the combination 988.4 Handling inertia 998.5 Summary 103The goals of this chapter are:• to introduce derivative control; and• to study the combination of proportional and derivative controlfor taming systems with integration or inertia.The controllers in the previous chapter had the same form: The controlsignal was a multiple of the error signal. This method cannot easily controlan integrating system, such as the motor positioning a rod even withoutinertia. If the system has inertia, the limits of proportional control becomeeven more apparent. This chapter introduces an alternative: derivativecontrol.8.1 Why derivative controlAn alternative to proportional control is derivative control. It is motivatedby the integration inherent in the motor system. We would like the feed-back system to make the actual position be the desired position. In other106 106106 10696 8.2 Mixing the two methods of control2009-09-29 13:11:30 UTC / rev b19331f50bbdwords, it should copy the input signal to the output signal. We would evensettle for a bit of delay on top of the copying. This arrangement is shownin the following block diagram:+C(R) =?M(R) =R1 − RS(R) = R−1controllermotorsensorSince the motor has the functional R/(1 − R), let’s put a discrete-time de-rivative 1 − R into the controller to remove the 1 − R in the motor’s de-nominator. With this derivative control, the forward-path cascade of thecontroller and motor contains only powers of R. Although this method istoo fragile to use alone, it is a useful idea. Pure derivative control is fragilebecause it uses pole–zero cancellation. This cancellation is mathematicallyplausible but, for the reasons explained in lecture, it produces unwantedoffsets in the output. However, derivative control is still useful. As wewill find, in combination with proportional control, it helps to stabilize in-tegrating systems.8.2 Mixing the two methods of controlProportional control uses β as the controller. Derivative control uses γ(1 −R) as the controller. The linear mixture of the two methods isC(R) = β + γ(1 − R ).+C(R) = β + γ(1 − R)M(R) =R1 − RS(R) = R−1controllermotorsensorLet F(R) be the functional for the entire feedback system. Its numerator isthe forward path C(R)M(R). Its denominator is 1 − L(R), where L(R) isthe loop functional or loop gain that results from going once around thefeedback loop. Here the loop functional is107 107107 1078 Proportional and derivative control 972009-09-29 13:11:30 UTC / rev b19331f50bbdL(R) = −C(R)M(R)S(R).Don’t forget the contribution of the inverting (gain= −1) element! So theoverall system functional isF(R) =(β + γ(1 − R))R1−R1 +(β + γ(1 − R))R1−RR.Clear the fractions to getF(R) =whatever1 − R + (β + γ(1 − R))R2.The whatever indicates that we don’t care what is in the numerator. It cancontribute only zeros, whereas what we worry about are the poles. Thepoles arise from the denominator, so to avoid doing irrelevant algebra andto avoid cluttering up the expressions, we do not even compute the nu-merator as long as we know that the fractions are cleared.The denominator is1 − R + (β + γ)R2− γR3.This cubic polynomial produces three poles. Before studying their loca-tions – a daunting task with a cubic – do an extreme-cases check: Take thelimit γ → 0 to turn off derivative control. The system should turn into thepure proportional-control system from the previous chapter. It does: Thedenominator becomes 1 − R + βR2, which is the denominator from Sec-tion 7.2. As the proportional gain β increases from 0 to ∞, the poles, whichbegin at 0 and 1, move inward; collide at 1/2 when β = 1/4; then split up-ward and downward to infinity. Here is the root locus of this limiting caseof γ → 0, with only proportional control:108 108108 10898 8.3 Optimizing the combination2009-09-29 13:11:30 UTC / rev b19331f50bbd8.3 Optimizing the combinationWe would like to make the whole system as stable as possible, in the sensethat the least stable pole is as close to the origin as possible. The rootlocus for the general combination has three branches, one for each pole,whereas the limiting case of proportional control has only two poles andtwo branches. Worse, the root locus for the general combination is gener-ated by two parameters – the gains of the proportional and the derivativeportions – whereas in the limiting case it is generated by only one parame-ter. The general analysis seems difficult.Surprisingly, the extra parameter rescues us from painful mathematics. Tosee how, look at the coefficients in the cubic:1 − R + (β + γ)R2− γR3.The factored form is(1−p1R)(1−p2R)(1−p3R) = 1−(p1+ p2+ p3)|{z }1R+(p1p2+ p1p3+ p2p3)| {z }β+γR2−p1p2p3|{z }γR3.So the first constraint isp1+ p2+ p3= 1,showing that the center of gravity of the poles is 1/3. That condition isindependent of β and γ. So the most stable system has a triple pole at 1/3,if that arrangement is possible. To see why that arrangement is the moststable, imagine starting from it. Now move one pole inward along the realaxis to increase its stability. To preserve the invariant p1+ p2+ p3= 1,at least one of the other poles must move outward and become less stable.Thus it is best not to move any pole away from the triple cluster, so it is themost stable arrangement.Exercise 42. Where does the preceding argument require thatthe center of gravity be independent of β and γ?If the triple-pole arrangement is impossible, then the preceding argument,which assumed its existence, does not work. And we need lots of work tofind the best arrangement of poles.109 109109 1098 Proportional and derivative control 992009-09-29 13:11:30 UTC / rev b19331f50bbdFortunately, the triple pole is possible thanks to the extra parameter γ.Having freedom to choose β and γ, we can set the R2coefficient β + γindependently from the R3coefficient, which is −γ. So, using β and γ asseparate dials, we can make any cubic whose poles are centered on 1/3.Let’s set those dials by propagating constraints. With p1= p2= p3= 1/3,the product p1p2p3= 1/27. So the gain of the derivative controller isγ =127.The last constraint is that p1p2+ p1p3+ p2p3= 3/9 = 1/3. So β + γ = 1/3.With γ = 1/27, this equation requires that the gain of the
View Full Document
Unlocking...