Model of Proportinal SystemBias = 0Cu =0Ko = feedback gain (voltage divider in lab)Ko = Proportional controller gainx(t) = input functionr(t) = output response of the control systemPlant is modelled using RC circuitMethod of solutional flow alI solutionSignFinasformsn'tcrart:FKR.rtrul+ceinpt. (tgebra ato stepKpnd Laplachanget-L' e.* o)r(t, x o) :-9 et438a-5.pptK p'Kfe6 0.6.Eocooo.4Response to Step lnput0 0.2 0.4 0.6 0.8 't 1.2 1.4 1.6Tlme (s)- Step InputKP=1-- KP=4-- KP=10II,ir;ttlI| .,"| .'L"iitltAs Ko increases, steady-state error decreasesSystem responds faster than natural timeconstantMathGAD x(t) := o(t - 0.2)formulasused to | -r,-.2).o(t-.21.(r.r-rp.ra)generate K^ l. R-cplots '(t'K p) '= 1 + K;G'lt - "l0 et438a-5.pptError As a Function of ProportionalGainv"erro(2, r o)Computing residual errorsso-ssa.100%o/oAffOt=WheressoSSo = desired steady-state output (1 forunit stepSS. = actual output at steady-stateplot on a log axis then this graph becomeslf Ko islinearI I et438a-5.pptLaplace TransformsLaplace transform converts time domain problemsinto functions of a complex variable, s, that isrelated to the frequency response of the systemju= j?rflef t holfnlnneright holfnlnner'' '-Time domainComplex frequencyComplex Frequency combines transient responsewith sinusoidal steady-state response to get totalresponse of system to inputS=o r j .Crlo = exponential decay/increase constantrelated to time constants of systems transientresponse RC = L/R = o in circuit analysiso'te" ' exponentially increasing function over t-o't .. rr . . .e v \ exponentially decreasing function over tq\/qTpmnespones- jor j?rfet438a-6.MCD 1Sinusiodal response From Laplacejar = j2nf same frequency used in phasor analysisLaplace related to sine response through Euler'sldentity. Eule/s relates complex exponentials tosine and cosine time functionsej 'o-cos(r,r.t) . j .sin(ro.t)e-j 'o-cos(rrr.t) - j .sin(r,r.t)Adding and subtracting the above retationshipsgives the exponential forms of sine and cosineej 'o + e-j 'o-2.cos(ro.t)Icos(r,r't, - ej 'od.- e-i 't{2ej 'o+- e j 'tt=2.j.sin(or.t)sin(o't) - "i 'ai- t-i 'art2.jSince "t't = eo't."j .crrt Laplace can givecomplete responseDc, transient, steady-state sinusiodalet438a-6.MCD 2Basic Laplace Transform pairsTime function6(t) lmpulseu r(t) unitstep"- t'tLaplace function by constantExamples: time5.u s(t)3.sin ( 4.t)LaplaceS+a1"t'ts-a0)sin ( o.t),2 +(DScos(ar.t)22s-+0)Linear rampslope 1Note: time functions multiplied by constants giveFunction1Is121JLaplace!S43.,2 + 16et438a-6.MCD 3More Laplace Examplestimee-2'te5't10.1t.e- a't3.cos ( t)Laplace1S+21Theoremss-510^sz1 Laplace- ^ Table 3.2(s + a)t text.rS"' . crl:lSt+ 1Laplace of an unknown functionLinearity of transform - can multiply by constantI (t,' tt)) F 1(s)Examples!(, ,(t))= | 1(s) I (u ,' (t))= V 1(s)1(t))= r 1(s) I (r r(t))= F 2(s)("'f 1(t) + b'f z(t)) =a.'F 1(s) + b'F 2(s)lf I(fThen Iet438a-6.MCD 4Laplace Transforms of CalculusOperatorsinto multiplication by sThenSubtract any initial1(s) - f 1(o) conditionLaplace turns derivativerr r( 1(t))=r 1(s)rllf \\dt'1(t)i=s'FFor higher order derivall/l (t:rr(,))=, (.l\otz '' I \tivesF 1(s)ftr(o)fr(o))r(r,)]=12F1(s)\fr(t,l s (s F!wThenrf I(1(t))rf r1(t),,)lf initial conditions are all zera, formula reduces tod2at2Laplace turns integration into division by s= F 1(s)=: F 1(s)et438a-6.MCD 5Examples: Lumped circuit elementsInductor voltagevr(t)=Lfti(t)r v r( t) =! (r.* i(t) ) u L(s) = L.s.r(s)\ dt ' 'lv6(t)=* Ir(uc(t) )=!(: I ic(t, o.)v p(t) = R.i(t) I(un(t))=/(n.i(t))V R( s) = R'l( s)Capactior Voltagei c(t) dt1V C( s) = C*'l( s)Resistor VoltageSimilar methods can be used on lumped elementsfor translational and rotational mechancial systemset438a-6.MCD 6Laplace and lmpedanceRemember phasor analysis, only valid for sinusoidalsteady-state. Turns ac analysis into an analysis similarto the dc. (Ohm's law)x,^, j 'al'GX L= j .rrl.La = 2'n'fI = 90 Phase shift-j =-90 PhaseshiftLaplace Ohm's LawlmpedanceV;:r, =J .O.LrL\,vc 1La =j -C=Lssince Laplace variable represents frequency, it'spossible to replace jrrr with s and s with jo. lf s isreplaced with jcD,_analysis becomes reverts to phasorswe can find the frequency response of a dynamicsystem by converting differential equation ihto Laplacedomian and replacing s with jrrr. sweeping frequencyproduces Bode pfot of system\/ /^\v cts) 1l( s) C.su*=*tRV ;-( s)l(s)V p(s)l(s)et438aS.MCD 71.)2.)3.)4.)5.)Solving Circuits and Systems Using theLaplace Transform MethodDevelop the differential equation model of thesystemTransform equation using Laplace tables. Includevoltage and current sources. (forces and torques)Solve resulting algebraic equations for variable(s) ofinterest. (usually a ratio of polynomials)For time response, take inverse Laplace transform.Results in form of exponential, sines and cosines.To find frequency response, replace s with jw andsweep frequency (note: take magnitude of complexquantity.Step 4 gives total response to system: transient andsteady-state, regardless of type of system excitationExample: Find current through capacitor, i"(t)et438a-6.MCD Iequation Using KVLi c(t) drWrite integro-differentiaV=R.ic(t) + : f'l\r, J Ofr|" Laplace transform/o.ftage source taken asf V volts.V:= R.tc(s) + *ofve for l.(s)Y=;"(rl l* + +\ c.s1.) \2.) T,Vof3.) ScvSuationnitudeof both sides of the eoa unit step with " rnroI c(s)= I c(s)(-= | c(s) simpfiffStarts to fookphasor analysisvs(-+C.sC.s1\f .-lc's i1\c's /.To* make the terms insidethe parenthesis toot tiiiJ'"-expression in Laptace tabfe)= , .(s)R.C.s + 1'u(et438a€.MCD 9(,tsrnr g LAPLA Ce % Rr PRe srnJTOP ArnP CtQCc.,-rT5-lrur5G pan(1rrsrnl oP AnnP THtop.yho f, f nterS tnve-f r^? no JeVit) "\-*(+) = oV.(*) Kcc o.l InvERTl^JG rvotn/til\+ re(t)=o? ,t,^(t) =-4(t)I ''-l "-9 ro fe too+ h s r des{t',-tni..J,, -r 1 n\rers€- o {*o 1e-+ % (' )AdtJ*RSCA u qse G rr.rFAlil l] fb GA rn,lnvtffr-lNG ArnP 4ruS Loplc*.cRel4zeo- a I c-( ar^re".ls I n -f err,.s o{ n o J .. v* /tc1es€,.^,1.1 (t) = -'Vc+y-u-Gtr(= c +(v,ot- v-(+))V-(+) = Y+6q .o Sc)tr'' /-t'\- -V; (t) Ii(t)=# Lr(t)=C#V'6)Sr( b.i,Ju*e rr.Jrs KCtVc- .Jff = - c:rv'ct)lr-! {,..,,.Jf : -C Vu(t)B,; J v;(t)-l a IRJ J v;c) Jr = V. cr; t"k* Lop l.,c.-q+qG) = V<'>r f ,, to,,*poQnnLA o{o{ Bo" rq €"[o*o^st'g-F/nd T^g /n Pwal1-)'L'rP,.-rV,G),-trrcLil0i\J^ rft roc' t.hv c,l/ecu-F{ Bglo,,^_:G.ne*l, aed lnvenlrnr ?q,rn.JV. =-{€*V;^ 3,".Usr AnPlnc^t Innprl ANCr ,RrtAzianlsHrps To frnr5 6ArnlLsQa) + ,S'lJ4aldt/l Ls =V;^(s\Z,'^