steadystateerror10

steadystateerror10 - Steady State Error QUESTION 10 For the...

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Steady State Error Q UESTION 10 For the system shown in Figure 10 with ( 29 ( 29 ( 29 ( 29 ( 29 2 5 1 5 3 8 12 s s F s s s s s + + = + + + , calculate e (∞) for the following inputs: r ( t ) = 11, r ( t ) = 4 t , and r ( t ) = 2 t 2 . + - R(s) F(s) Y(s) E(s) Figure 10 The closed–loop transfer function is ( 29 ( 29 ( 29 ( 29 ( 29 2 4 3 2 5 6 5 1 11 41 66 25 s s Y s F s R s F s s s s s + + = = + + + + + . From the Routh Table below, it can be seen that the closed–loop system is stable and, therefore, the steady–state error can be calculated. 4 3 2 1 0 1 41 25 11 66 0 35 25 0 58.14 0 0 25 0 0 s s s s s ( 29 ( 29 ( 29 1 1 E s R s F s = +
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Steady State Error ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 2 2 0 0 2 3 8 12 1 lim lim 5 1 5 3 8 12 5 1 5 1 3 8 12 s s s s s s e s R s s R s s s s s s s s s s s s s + + + ∞ = = + + + + + + + + + + + + For r ( t ) = 11, ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 2 2 0 3 8
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This note was uploaded on 02/01/2012 for the course MECH ENG 279 taught by Professor Landers during the Spring '11 term at Missouri S&T.

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steadystateerror10 - Steady State Error QUESTION 10 For the...

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