# steadystateerror15 - Steady State Error For r t = 1 29 29...

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Steady State Error Q UESTION 15 For the system shown in Figure 15 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 + + + 5 20 10 R(s) Y(s) + - Figure 15 Let E 1 be the output from the summer and ( 29 ( 29 ( 29 4 5 8 s G s s s + = + + . From the block diagram, ( 29 ( 29 ( 29 ( 29 1 1 20 5 E s R s G s E s = - and ( 29 ( 29 ( 29 1 10 Y s G s E s = . Combining equations ( 29 ( 29 ( 29 ( 29 200 1 5 Y s G s R s G s = + . Since ( 29 ( 29 ( 29 E s R s Y s = - , ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 5 8 195 4 5 8 5 4 s s s E s R s s s s + + - + = + + + + The closed–loop characteristic polynomial is ( 29 ( 29 ( 29 2 5 8 5 4 18 60 s s s 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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Unformatted text preview: Steady State Error For r ( t ) = 1, ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 5 8 195 4 5 8 195 4 1 lim 12.33 5 8 5 4 5 8 5 4 s s s s e s s s s s → + +-+-∞ = = = -+ + + + + For r ( t ) = t , ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 2 5 8 195 4 1 lim 5 8 5 4 s s s s e s s s s s → + +-+ ∞ = = ∞ + + + + For r ( t ) = 0.5 t 2 , ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 3 5 8 195 4 1 lim 5 8 5 4 s s s s e s s s s s → + +-+ ∞ = = ∞ + + + + 2...
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steadystateerror15 - Steady State Error For r t = 1 29 29...

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