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

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

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