# steadystateerror14 - Professor Robert G Landers The...

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Steady State Error Q UESTION 14 For the system shown in Figure 14 calculate e (∞) for the following inputs: r ( t ) = 1, r ( t ) = t , and r ( t ) = 0.5 t 2 . ( 29 2 1 1 s s + 2 1 s ( 29 2 1 3 s s + R(s) E(s) Y(s) + - + - Figure 14 Let ( 29 ( 29 1 2 1 1 G s s s = + and ( 29 ( 29 2 2 1 3 G s s s = + . Denote the output from the G 1 block as V ( s ). From the block diagram, ( 29 ( 29 ( 29 E s R s Y s = - , ( 29 ( 29 ( 29 ( 29 1 2 1 V s G s E s V s s = - , and ( 29 ( 29 ( 29 1 Y s G s V s = Combining equations, ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 4 1 2 4 1 1 2 2 1 1 1 3 3 1 1 3 3 1 1 G s s s s s s E s R s R s s s s s G s G s G s s + + + + + = = + + + + + + +

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ME 279 – Automatic Control of Mechanical Systems
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Unformatted text preview: Professor Robert G. Landers The closed–loop characteristic polynomial is ( 29 ( 29 ( 29 4 6 5 4 1 3 3 4 3 4 s s s s s s s s + + + + = + + + + . Since there are missing powers of s , the closed–loop system is not stable and, therefore, the steady–state error cannot 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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steadystateerror14 - Professor Robert G Landers The...

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