D-06 - am = 1(Dm KtKb = 1(0.13(1x1 = 1.32 Jm Ra 0.25 5 θm(s = Km = 0.800 ………………………………………………………21 Ea(s

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Unformatted text preview: am = 1 (Dm + KtKb ) = 1(0.13 + (1x1)) = 1.32 Jm Ra 0.25 5 θm(s) = Km . = 0.800 ………………………………………………………21 Ea(s) s(s+am) s(s+1.32) Transfer function of the Gear The transfer function of the gear is given by the ratio of the load angle θ 0(s) to the motor load angle θm(s). For configuration #1: Kg = θ0(s) = N1 = 25 = 0.1 ………………………………………………………….22 θm(s) N2 250 For configuration #3: Kg = θ0(s) = N1 = 50 = 0.2 …………………………………………………………..23 θm(s) N2 250 Transfer function of output potentiometer The dynamics of the output potentiometer is neglected. The transfer function of the output is given by the ratio of the load angle θ0(s) to the load voltage Vi(s). Since the input potentiometer and the output potentiometer are configured in the same way, their transfer function will be the same. For configuration #1: Kpot = θ0(s) = 0.3183 ………………………………………………………………...24 Vi(s) For configuration #3: Kpot = θ0(s) = 3.183 ……………………………………………………………………25 Vi(s) ...
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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-06 - am = 1(Dm KtKb = 1(0.13(1x1 = 1.32 Jm Ra 0.25 5 θm(s = Km = 0.800 ………………………………………………………21 Ea(s

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