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# D-06 - am=1(Dm KtKb)=1(0.13(1x1)=1.32 JmRa0.255...

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am = 1  (Dm + KtKb  )  = 1( 0.13 + (1x1 )) = 1.32         Jm            Ra       0.25            5 m(s) θ  = Km      .  =     0.800  ………………………………………………………21 Ea(s)    s(s+a m )     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  θ m (s). For configuration #1: Kg =  θ 0 (s)  = N1  = 25  = 0.1 ………………………………………………………….22          θ m (s)    N2   250 For configuration #3: Kg =  θ 0 (s)  = N1  = 50  = 0.2 …………………………………………………………..23          θ m (s)    N2   250

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