steadystateerror6 - Steady State Error QUESTION 6 For the...

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Steady State Error Q UESTION 6 For the system shown in Figure 6 calculate e (∞) for the following inputs: r ( t ) = 1, r ( t ) = t , and r ( t ) = 0.5 t 2 . 2 2 s ( 29 ( 29 ( 29 10 1 3 4 s s s s + + + R(s) E(s) Y(s) + - + - Figure 6 From the block diagram, ( 29 ( 29 ( 29 E s R s Y s = - and ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 10 2 2 1 3 4 Y s E s sY s s s s s = - + + + Combining equations, ( 29 ( 29 ( 29 3 2 4 3 2 8 19 32 8 19 32 20 s s s s E s R s s s s s + + + = + + + + The closed–loop characteristic polynomial is 4 3 2 8 19 32 20 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.
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Steady State Error
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steadystateerror6 - Steady State Error QUESTION 6 For the...

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