steadystateerror3 - and, therefore, the steady–state...

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Steady State Error Q UESTION 3 For the system shown in Figure 3 with ( 29 ( 29 ( 29 ( 29 ( 29 2 3 9 7 20 3 5 7 s s F s s s s s + + = + + + , calculate e (∞) for the input r ( t ) = 18 t 3 . + - R(s) F(s) Y(s) E(s) Figure 3 The closed–loop transfer function is ( 29 ( 29 ( 29 ( 29 ( 29 2 6 5 4 3 2 9 7 20 1 15 71 105 9 63 180 s s Y s F s R s F s s s s s s s + + = = + + + + + + + . From the Routh Table below, it can be seen that the closed–loop system has two unstable poles
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Unformatted text preview: and, therefore, the steady–state error cannot be calculated. 6 5 4 3 2 1 1 71 9 180 15 105 63 64 4.8 180 103.9 20.8 8.02 180 2351 180 s s s s s s s-...
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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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