hw2 - E105 Homework 2 Solutions Stanford University Winter...

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E105 - Homework 2 Solutions Stanford University - Winter 2006 Stephen Russell January 25, 2006 Problem 1 - FPE 3.14 For the system in Fig. 2.45, compute the transfer function from the motor voltage to position θ 2 . Starting with the equations of motion: v a = L a di a dt + R a i a + K e ˙ θ 1 K t i a + k ( θ 2 - θ 1 ) + b ( ˙ θ 2 - ˙ θ 1 ) - B ˙ θ 1 = J 1 ¨ θ 1 - k ( θ 2 - θ 1 ) - b ( ˙ θ 2 - ˙ θ 1 ) = J 2 ¨ θ 2 Taking the Laplace transform of each equation and assuming zero initial conditions V a ( s ) = sL a I a ( s ) + R a I a ( s ) + sK e Θ 1 ( s ) (1) K t I a ( s ) + k 2 ( s ) - Θ 1 ( s )) + sb 2 ( s ) - Θ 1 ( s )) - sB Θ 1 ( s ) = s 2 J 1 Θ 1 ( s ) (2) - k 2 ( s ) - Θ 1 ( s )) - sb 2 ( s ) - Θ 1 ( s )) = s 2 J 2 Θ 2 ( s ) (3) Using equation (3) ( J 2 s 2 + bs + k ) Θ 2 ( s ) = ( bs + k 1 ( s ) Θ 1 ( s ) Θ 2 ( s ) = J 2 s 2 + bs + k bs + k (4) 1
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Now, rearranging equation (2) and plugging in the relation we’ve just derived for Θ 1 ( s ) ( J 1 s 2 + ( B + b ) s + k ) Θ 1 ( s ) - ( bs + k 2 ( s ) = K t I a ( s ) ˆ ( J 1 s 2 + ( B + b ) s + k ) ( J 2 s 2 + bs + k ) - ( bs + k ) 2 bs + k ! Θ 2 ( s ) = K t I a ( s ) I a ( s ) Θ 2 ( s ) = ( J 1 s 2 + ( B + b ) s + k ) ( J 2 s 2 + bs + k ) - ( bs + k ) 2 K t ( bs + k ) (5) Finally, plugging both equations (4) and (5) into equation (1) V a ( s ) = ( L a s + R a ) I a ( s ) + sK e Θ 1 ( s ) V a ( s ) = ( L a s + R a ) h ( J 1 s 2 + ( B + b ) s + k ) ( J 2 s 2 + bs + k ) - ( bs + k ) 2 i K t ( bs + k ) Θ 2 ( s )+ sK e ± J 2 s 2 + bs + k bs + k Θ 2 ( s ) V a ( s ) = ( L a s + R a ) h ( J 1 s 2 + ( B + b ) s + k ) ( J 2 s 2 + bs + k ) - ( bs + k ) 2 i + sK e K t ( J 2 s 2 + bs + k ) K t ( bs + k ) Θ 2 ( s ) Θ 2 ( s ) V a ( s ) = K
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hw2 - E105 Homework 2 Solutions Stanford University Winter...

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