Physical System Modelling, Time Response And Stability - 275 Words

Physical System Modelling, Time Response And Stability (Math Problem Sample) Content: PHYSICAL SYSTEM MODELING, TIME RESPONSE AND STABILITYStudents NameCourseProfessors NameUniversityCity (State)DatePhysical System Modeling, Time Response and Stability 1 A differential equation that relates ea(t) and L(t)We know that eat=Raia(t)+Kbdm(t)dt and mt=N2N1L(t)dm(t)dt =ddt(mt)=N2N1dL(t)dtSubstituting the value of dm(t)dt in eat gives:eat=KbN2N1dLtdt+Raia(t)Therefore, eat=Raia(t)+(KbN2N1dL(t)dt) 2 GS=L(S)Ea(S) of the systemeat=Raia(t)+Kbdm(t)dt in the s-Domain becomes Eas=RaIa(s)+KbSm(s)Also, TmS=KtIa(s), IaS=TmSKtEaS=RaTmSKt+Kbsms..(i)We also know that TmS=(Jms2+Dms)ms..(ii)Substituting equation (ii) in equation (i)EaS=RaKt(Jms2+Dms)ms+KbsmsEaS=smsRaKt(Jms+Dm+Kb)After simplification, msEaS is found to be:msEaS=Kt(RaJm)s+1Jm(Dm+KtKbRa)sBut from mt=N2N1L(t) we find ms=N2N1L(s) ,Hence LsEaS=KtRaN1N2Jms+1Jm(Dm+KtKbRa)s 3 Time response of the system given that the input is a unit impulse inputJm=Ja+JL(N1N2)2=5+700(1001000)2=12Dm=Da+DL(N1N2)2=2+800(100 1000)2=10To get the electrical constant KtRa we use the torque-speed curve.Figure 1: Torque Speed CurveKtRa=Tstallea=500100=5LsEaS=500100100100012s+112(10+5002100)s=0.0417s(s+1.667) 4 The response of the system to a unit-step inputFor a unit-step input, Rs=1sFor a gain of K=40, the transfer function becomes 400.0417s(s+1.667)=1.668s(s+1.667)...