quiz2_sol

# quiz2_sol - 2.171 Quiz 2 Solutions Problem 1 2 a H s = s 2...

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Unformatted text preview: 2.171 Quiz 2 Solutions Problem 1: 2 a H ( s ) = s 2 − a 2 a) Calculate the zero order hold equivalent H eq ( z ). H eq ( z ) = z − z 1 Z { G s ( s ) } G ( s ) a 2 Z { s } = Z { s ( s 2 − a 2 ) } G ( s ) A B Z { s } = Z { s ( s + a ) + s ( s − a ) } G ( s ) 1 a 1 a Z { s } = Z {− 2 s ( s + a ) + 2 s ( s − a ) } Using Transform 12 in Table B.2 of FPW, we have: Z { G s ( s ) } = − ( z − z (1 1)( − z e − − aT e − ) aT ) − ( z − z (1 1)( − z e − aT e ) aT ) G ( s ) z (1 − e − aT )( z − e aT ) + z (1 − e aT )( z − e − aT ) Z { s } = − ( z − 1)( z − e − aT )( z − e aT ) z − 1 z (1 − e − aT )( z − e aT ) + z (1 − e aT )( z − e − aT ) H eq ( z ) = − z ( z − 1)( z − e − aT )( z − e aT ) (1 − e − aT )( z − e aT ) + (1 − e aT )( z − e − aT ) H eq ( z ) = − ( z − e − aT )( z − e aT ) aT − e − aT ) z − e aT − e − aT H eq ( z ) = − (2 − e ( z − e − aT )( z − e aT ) ( e aT + e − aT − 2) z + e aT + e − aT H eq ( z ) = ( z − e − aT )( z − e aT ) 1 b) Show that for 8 × 10 − 4 ( z + 1) a = 40 ,T = 0 . 001 sec,H eq ( z ) = z 2 − 2 . 0016 z + 1 . H eq ( z ) = ( e . 04 + e − . 04 − 2) z + e . 04 + e − . 04 ( z − e − . 04 )( z − e . 04 ) 8 × 10 − 4 ( z + 1) H eq ( z ) = z 2 − 2 . 0016 z + 1 The answer matches, as expected. The poles map closely to z = e sT due to the fact that the sampling frequency is significantly higher than the frequency of the poles. sT e 1 = 1 . 0408 = p 1 z sT e 2 = 0 . 9608 = p 2 z-1-0.5 0.5 1 1.5-1-0.8-0.6-0.4-0.2 0.2 0.4 0.6 0.8 1 Pole-Zero Map Real Axis Imaginary Axis Figure 1: Pole Zero Map of Plant 2 c) The DC value of the transfer function is-1,...
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## This note was uploaded on 02/27/2012 for the course MECHANICAL 2.171 taught by Professor Davidtrumper during the Fall '06 term at MIT.

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quiz2_sol - 2.171 Quiz 2 Solutions Problem 1 2 a H s = s 2...

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