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# steadystateerror7 - error may be calculated For r t = 1 29...

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Steady State Error Q UESTION 7 For the system shown in Figure 7 calculate e (∞) for the following inputs: r ( t ) = 1, r ( t ) = t , and r ( t ) = 0.5 t 2 . ( 29 ( 29 4 5 8 s s s + + + 10 5 R(s) Y(s) + - Figure 7 From the block diagram, ( 29 ( 29 ( 29 E s R s Y s = - and ( 29 ( 29 ( 29 ( 29 ( 29 4 5 10 5 8 s Y s R s Y s s s + = - + + Combining equations, ( 29 ( 29 2 2 18 60 23 80 s s E s R s s s + + = + + The characteristic polynomial is 2 23 80 s s + + . Since the characteristic polynomial is 2 nd order and the coefficients all have the same sign, the closed–loop system is stable and the steady–state

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Unformatted text preview: error may be calculated. For r ( t ) = 1, ( 29 2 2 18 60 1 60 lim 0.75 23 80 80 s s s e s s s s + + = = = + + Steady State Error For r ( t ) = t , ( 29 2 2 2 18 60 1 lim 23 80 s s s e s s s s + + = = + + For r ( t ) = 0.5 t 2 , ( 29 2 2 3 18 60 1 lim 23 80 s s s e s s s s + + = = + + 2...
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steadystateerror7 - error may be calculated For r t = 1 29...

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