steadystateerror5

steadystateerror5 - s 2 + 2 s + 2. Equating like...

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Steady State Error Q UESTION 5 For the system shown in Figure 5 with ( 29 ( 29 ( 29 K s F s s s α β + = + , calculate the parameters K , α , and β such that the closed–loop poles are located at –1 ± j and e (∞) = 0.01 for a unit ramp input. + - R(s) F(s) Y(s) E(s) Figure 5 ( 29 ( 29 ( 29 1 1 E s R s F s = + If e (∞) = 0.01 for a unit ramp input, ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 2 2 0 0 1 1 1 lim lim 0.01 1 s s s s e s s K s s s s K s s K s s + ∞ = = = = + + + + + + The characteristic equation is ( 29 ( 29 ( 29 2 0 s s K s s K s K + + + = + + + = . For closed–loop poles located at –1 ± j , the closed–loop characteristic equation is

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Unformatted text preview: s 2 + 2 s + 2. Equating like coefficients, K + β = 2 and K α = 2. Solving the three equations simultaneously, K = 1.98, β = 0.02, and α = 1.01. ME 279 – Automatic Control of Mechanical Systems Professor Robert G. Landers Since the closed–loop system is designed to be stable, the steady–state error can be calculated. 2...
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This note was uploaded on 02/01/2012 for the course MECH ENG 279 taught by Professor Landers during the Spring '11 term at Missouri S&T.

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steadystateerror5 - s 2 + 2 s + 2. Equating like...

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