Page - (000031) - in Control Theory Tutorial - Basic Concepts Illustrated by Software Examples

3.1 Open-LoopControl 21 L(s)= ω˜0 ω0 ( ω0 s+ω0 )3(s2+ω20 ω0 )( ω˜0 s2+ ω˜20 ) , (3.3) inwhich thefirst termadjusts thegain tobeoneat s=0. Thegold curves inFig.3.1 show theBodeplot for this open loop, usingω0 =1 and ω˜0 =1.2.Note theresonantpeakin theuppermagnitudeplot.Thatpeakoccurs when the input frequencymatches the natural frequency of the intrinsic oscillator, ω˜0.Near that resonant frequency, the system“blowsup,” because the denominator in the last term, s2+ ω˜20, goes tozeroas s= jω→ jω˜0 and s2 →−ω˜20. In summary, open-loop controlworkswellwhen one has accurate information. Successful open-loopcontrol is simple andhas relatively lowcost.However, small variations in the intrinsic process or themodulating controller can cause poor per- formanceor instabilities, leading to systemfailure. 3.2 FeedbackControl Feedback and feedforward have different properties. Feedforward action is obtained by matchingtwotransferfunctions,requiringpreciseknowledgeoftheprocessdynamics,while feedbackattempts tomake theerror smallbydividing it bya largequantity. —ÅströmandMurray (2008,p.320) Feedback often solves problems of uncertainty or noise.Human-designed systems andnaturalbiological systems frequentlyuse feedbackcontrol. Figure2.1cshowsacommonformofnegativefeedback.Theoutput,y, isreturned to the input.Theoutput is then subtracted fromtheenvironmental reference signal, r. The new system input becomes the error between the reference signal and the output,e= r− y. In closed-loop feedback, the system tracks its target reference signal by reduc- ing the error. Any perturbations or uncertainties can often be corrected by system dynamics that tend tomove the error toward zero.Bycontrast, a feedforwardopen loop has no opportunity for correction. Feedforward perturbations or uncertainties lead touncorrectederrors. In the simple negative feedbackofFig.2.1c, the key relationbetween the open- loopsystem,L(s)=C(s)P(s), and the full closed-loopsystem,G(s), is G(s)= L(s) 1+L(s). (3.4) ThisrelationcanbederivedfromFig.2.1cbynotingthat, fromtheerror input,E(s), to theoutput,Y(s),wehaveY = LE and thatE = R−Y. Substituting the second equation into thefirstyieldsY = L (R−Y).Solvingfor theoutputY relative to the inputR,which isG=Y/R, yieldsEq.3.4.