me304_hw5_sol - ME304/Spring 2005/Sections...

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Unformatted text preview: ME304/Spring 2005/Sections 02+03+04/HW5-Solutions Page 1 / 9 SOLUTIONS TO HOMEWORK 5 for Sections 02, 03 & 04 Prepared by Gökhan BAYAR (C-204), Atilla BAYRAM (A-148), Erdinç İ Y İ AY (C-209) Problem 1 a) Setting ) ( = s T d , the closed loop transfer function for the servo characteristics of the system can be obtained as ) ( 1 ) ( ) ( ) ( ) ( s G K K s G K s s s M a h a R + = Θ Θ = where 100 s 10 s 100 ) s ( G 2 + + = . Then 100 10 100 100 10 100 10 100 100 10 100 1 100 10 100 ) ( 2 2 2 2 2 + + + + + + + = + + + + + = s s K K s s s s K s s K K s s K s M h a a h a a h a a R K K s s K s s s M 100 100 10 100 ) ( ) ( ) ( 2 + + + = Θ Θ = (1) Using equation (1), we can write ) ( ) 100 100 ( ) ( 10 ) ( ) ( 100 2 s K K s s s s s K h a R a Θ + + Θ + Θ = Θ (2) MIDDLE EAST TECHNICAL UNIVERSITY MECHANICAL ENGINEERING DEPARTMENT ME 304 CONTROL SYSTEMS SPRING 2005 Θ (s) G(s) K a K h Θ R (s) T d (s) – + ME304/Spring 2005/Sections 02+03+04/HW5-Solutions Page 2 / 9 Taking the inverse Laplace transforms of both sides of equation (2), the corresponding differential equation to represent the servo characteristics of the system is obtained as { } { } ) ( ) 100 100 ( ) ( 10 ) ( ) ( 100 2 1 1 s K K s s s s L s K L h a R a Θ + + Θ + Θ = Θ − − ) ( ) 100 100 ( ) ( 10 ) ( ) ( 100 t K K t t t K h a R a θ θ θ θ + + + = & & & ) ( 100 ) ( ) 100 100 ( ) ( 10 ) ( = − + + + t K t K K t t R a h a θ θ θ θ & & & b) h a a R K K s s K s s s M 100 100 10 100 ) ( ) ( ) ( 2 + + + = Θ Θ = The sensitivity of M(s) with respect to K a and K h are a a M Ka K s M s M K S ∂ ∂ = ) ( ) ( ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + + − + + + ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ + + + = 2 2 2 2 ) 100 100 10 ( ) 100 ( 100 ) 100 100 10 ( 100 100 100 10 100 h a h a h a h a a a M Ka K K s s K K K K s s K K s s K K S Rearranging this expression h a M Ka K K s s s s S 100 100 10 100 10 2 2 + + + + + = h h M Kh K s M s M K S ∂ ∂ = ) ( ) ( ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + + − ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ + + + = 2 2 2 2 ) 100 100 10 ( ) 100 ( 100 100 100 10 100 h a a h a a h M Kh K K s s K K K s s K K S Rearranging this expression h a h a M Kh K K s s K K S 100 100 10 100 2 + + + − = c) Recall from part (a) that, the differential equation for the system is...
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This note was uploaded on 05/16/2011 for the course ME 304 taught by Professor Samim during the Spring '11 term at Middle East Technical University.

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me304_hw5_sol - ME304/Spring 2005/Sections...

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