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Steady State Error Q UESTION 16 For the system shown in Figure 16 calculate e (∞) for the following sets of inputs: r ( t ) = d ( t ) = 1, r ( t ) = d ( t ) = t , and r ( t ) = d ( t ) = 0.5 t 2 . 1 s 1 s + 10 2 s + + - R(s) + + D(s) Y(s) Figure 16 From the block diagram, ( 29 ( 29 ( 29 E s R s Y s = - and ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 1 10 1 2 Y s D s R s s Y s s s = + - + + Combining equations, ( 29 ( 29 ( 29 2 2 2 10 12 10 12 10 s Y s D s R s s s s s + = + + + + + and ( 29 ( 29 ( 29 2 2 2 12 2 12 10 12 10 s s s E s R s D s s s s s + + = - + + + + The closed–loop characteristic polynomial is 2 12 10 s s + + . Since the characteristic polynomial is 2 nd order and all the coefficients have the same sign, the closed–loop system is stable and, therefore, the steady–state error may be calculated.

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Steady State Error For r ( t ) = 1 and d ( t ) = 0, ( 29 2 2 0 12 1 lim 0 12 10 s s s e s s s s + ∞ = = + + For r ( t ) = t and d (
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