steadystateerror4 - ( 29 2 3 2 1 10 26 4 11 159.5 520 s s s...

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Steady State Error Q UESTION 4 For the system shown in Figure 4 with ( 29 ( 29 ( 29 ( 29 2 4 1 10 26 K s F s s s s + = + + + , calculate K such that e (∞) = 0.05 for a unit step input. + - R(s) F(s) Y(s) E(s) Figure 4 ( 29 ( 29 ( 29 1 1 E s R s F s = + If e (∞) = 0.05 for a unit step input, ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 2 2 0 0 2 1 10 26 1 1 1 26 lim lim 0.05 4 26 4 1 10 26 4 1 1 10 26 s s s s s e s s K s s s K s s s K s s s s + + + ∞ = = = = + + + + + + + + + + + Solving, K = 123.5 The closed–loop characteristic polynomial is ( 29 ( 29
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Unformatted text preview: ( 29 2 3 2 1 10 26 4 11 159.5 520 s s s K s s s s + + + + + = + + + . From the Routh Table below, it can be seen that the closedloop system is stable and, therefore, the steadystate error can be calculated. Steady State Error 3 2 1 1 159.5 11 520 112.7 520 s s s s 2...
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steadystateerror4 - ( 29 2 3 2 1 10 26 4 11 159.5 520 s s s...

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