# steadystateerror9 - 2 2 2 15 12 lim 2 15 20 3 4 8 s s s s e...

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Steady State Error Q UESTION 9 For the system shown in Figure 9 with ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 2 20 3 4 8 2 15 s s s F s s s s + + + = + + , calculate e (∞) for the following inputs: r ( t ) = 12, r ( t ) = 3 t , and r ( t ) = 8 t 2 . + - R(s) F(s) Y(s) E(s) Figure 9 The closed–loop transfer function is ( 29 ( 29 ( 29 ( 29 ( 29 3 2 4 3 2 20 15 68 96 1 37 330 1360 1920 s 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 1 330 1920 37 1360 0 293.2 1920 0 1118 0 0 1920 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 ( 29 ( 29 2 2 0 0 2 2 15 1 lim lim 20 3 4 8 2 15 20 3 4 8 1 2 15 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 ) = 12, ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29
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Unformatted text preview: 2 2 2 15 12 lim 2 15 20 3 4 8 s s s s e s s s s s s s s + + = = + + + + + + For r ( t ) = 3 t , ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 2 2 2 2 15 3 lim 2 15 20 3 4 8 s s s s e s s s s s s s s + + = = + + + + + + For r ( t ) = 8 t 2 , ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 2 2 3 2 15 8 2 2 15 8 2 lim 0.25 2 15 20 3 4 8 20 3 4 8 s s s s e s s s s s s s s + + = = = + + + + + + 2...
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steadystateerror9 - 2 2 2 15 12 lim 2 15 20 3 4 8 s s s s e...

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