steadystateerror12

steadystateerror12 - Steady State Error QUESTION 12 For the...

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Steady State Error Q UESTION 12 For the system shown in Figure 12 with ( 29 ( 29 ( 29 ( 29 2 4 1 10 26 K s F s s s s s + = + + + , calculate K such that e (∞) = 0.1 for unit step inputs. + - R(s) F(s) Y(s) E(s) Figure 12 ( 29 ( 29 ( 29 1 1 E s R s F s = + The steady–state error for unit step inputs is ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 2 2 0 0 2 1 10 26 1 1 1 lim lim 0 4 1 10 26 4 1 1 10 26 s s s s s s e s s K s s s s s s s K s s s s s + + + ∞ = = = + + + + + + + + + + It would appear that the steady–state error is zero, regardless of the value of K , and that a steady– state error for ramp inputs of 0.1 cannot be achieved. However, the closed–loop system must be stable. The closed–loop characteristic polynomial is ( 29 ( 29 ( 29 ( 29 2 4 3 2 1 10 26 4 11 36 26 4 s s s s K s s s s K s K + + + + + = + + + + + . The Routh Table is
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Steady State Error
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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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steadystateerror12 - Steady State Error QUESTION 12 For the...

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