Physical System Modelling, Time Response And Stability (Math Problem Sample)

PHYSICAL SYSTEM MODELING, TIME RESPONSE AND STABILITY
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Physical System Modeling, Time Response and Stability
1 A differential equation that relates ea(t) and θL(t)
We know that eat=Raia(t)+Kbdθm(t)dt and θmt=N2N1θL(t)
dθm(t)dt =ddt(θmt)=N2N1dθL(t)dt
Substituting the value of dθm(t)dt in eat gives:
eat=KbN2N1dθLtdt+Raia(t)
Therefore, eat=Raia(t)+(KbN2N1dθL(t)dt)
2 GS=θL(S)Ea(S) of the system
eat=Raia(t)+Kbdθm(t)dt in the s-Domain becomes Eas=RaIa(s)+KbSθm(s)
Also, TmS=KtIa(s), IaS=TmSKt
EaS=RaTmSKt+Kbsθms………..(i)
We also know that TmS=(Jms2+Dms)θms………..(ii)
Substituting equation (ii) in equation (i)
EaS=RaKt(Jms2+Dms)θms+Kbsθms
EaS=sθmsRaKt(Jms+Dm+Kb)
After simplification, θmsEaS is found to be:
θmsEaS=Kt(RaJm)s+1Jm(Dm+KtKbRa)s
But from θmt=N2N1θL(t) we find θms=N2N1θL(s) ,
Hence θLsEaS=KtRaN1N2Jms+1Jm(Dm+KtKbRa)s
3 Time response of the system given that the input is a unit impulse input
Jm=Ja+JL(N1N2)2=5+700(1001000)2=12
Dm=Da+DL(N1N2)2=2+800(1001000)2=10
To get the electrical constant KtRa we use the torque-speed curve.
Figure 1: Torque Speed Curve
KtRa=Tstallea=500100=5
θLsEaS=500100×1001000×12s+112(10+500×2100)s=0.0417s(s+1.667)
4 The response of the system to a unit-step input
For a unit-step input, Rs=1s
For a gain of K=40, the transfer function becomes 40×0.0417s(s+1.667)=1.668s(s+1.667)
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