D-04 - RaIa(s LasIa(s Vb(s = Ea(s)…………………………………….….10 We can now relate the input volage Ea with the motor position

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Unformatted text preview: RaIa(s) + LasIa(s) + Vb(s) = Ea(s)…………………………………….…...10 We can now relate the input volage Ea with the motor position θm. We begin by using Tm(s) = Kt Ia(s) …………… (8) to get rid of Ia in equation (10). (Ra + Las)Tm(s) + Vb(s) = Ea(s) …….………………………………………11 Kt We can now substitute Vb(s) = Kbs θm(s) …….. (9) (Ra + Las)Tm(s) + Kbs θm(s) = Ea(s) ……………………………………..12 Kt By analyzing the mechanical system shown below we are able to find an expression relating Tm(s) with θm(s). Assuming we have a single shaft with inertia Jm and damping Dm: Figure 2: A schematic of a typical equivalent mechanical loading system having a single shaft with inertia Jm and damping Dm Tm(s) = (Jms2 + Dms) θm(s) ……………………………………………………13 We substitute this mechanical equation (13) into equation (12) (Ra + Las)(Jm m θ s2 + D s) m(s) + Kb sθm(s) = Ea(s) ……………14 Kt if we assume Ra >>La; simplifying equation 14 θm(s) = Kt/( RaJm ) …………………………………………………15 Ea(s) s[s + 1 (Dm + KtKb ) ] Jm Ra Comparing equation 5 and 15, then Km = Kt/(RaJm) .............................................................16 ...
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This note was uploaded on 12/20/2009 for the course ECE 451 taught by Professor Staff during the Fall '09 term at Clarkson University .

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D-04 - RaIa(s LasIa(s Vb(s = Ea(s)…………………………………….….10 We can now relate the input volage Ea with the motor position

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