homework3_solution.pdf

# homework3_solution.pdf - Solutions to Homework 3 ECE 460...

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Solutions to Homework 3 ECE 460: Automatic Control Due: Monday, February 1, at 2:10pm Note 1. Your solutions should be neat and legible; always show your work. Problem 1. Figure 1 shows a DC motor with R a = 2 Ω, J m = 4 kg-m 2 , D m = 12 kg-mm 2 /s, K b = 25 mV-sec, K t = 7 N-mm/A, and K m = 0. Figure 1: DC motor (a) (20 Points) Find the transfer function, G ( s ) = Ω m ( s ) E a ( s ) for L a = 1.0 mH and find the pole(s) and zero(s) of the system. Solution: The equations of motion with K m = 0 can be written as v b ( t ) = K b ω m ( t ) T m ( t ) = K t i a ( t ) e a ( t ) = R a i a ( t ) + L a di a ( t ) dt + v b ( t ) T m ( t ) - D m ω m ( t ) = J m ˙ ω m ( t ) Taking the Laplace transform, we have V b ( s ) = K b Ω m ( s ) T m ( s ) = K t I a ( s ) E a ( s ) = ( R a + sL a ) I a ( s ) + V b ( s ) T m ( s ) = ( sJ m + D m m ( s ) 1

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Then, we have E a ( s ) = ( R a + sL a ) I a ( s ) + K b Ω m ( s ) K t I a ( s ) = ( sJ m + D m m ( s ) Now, we have E a ( s ) = ( R a + sL a )( sJ m + D m ) K t Ω m ( s ) + K b Ω m ( s ) The transfer function of the system is given by G ( s ) = Ω m ( s ) E a ( s ) = 1 ( R a + sL a )( sJ m + D m ) K t + K b = K t ( R a + sL a )( sJ m + D m ) + K t K b = 0 . 007 (2 + 10 - 3 s )(4 s + 12) + 0 . 007 · 0 . 025 We have (2 + 10 - 3 s )(4 s + 12) + 0 . 007 · 0 . 025 (2 + 10 - 3 s )(4 s + 12) because 0 . 007 · 0 . 025 is too small compared to 2 · 12. Therefore, there are two poles at -3 and -2000, and there is no zero. (b) (10 points) Find the transfer function, G ( s ) = Ω m ( s ) E a ( s ) for L a = 0 and find the pole(s) and zero(s) of the system. How can you compare them with your solutions to (a)?
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