steadystateerror2 - and, therefore, the steadystate error...

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Steady State Error Q UESTION 2 For the system shown in Figure 2 with ( 29 ( 29 ( 29 ( 29 ( 29 2 2 9 7 20 3 5 7 s s F s s s s s + + = + + + , calculate e (∞) for the following inputs: r ( t ) = 13, r ( t ) = 4 t , and r ( t ) = 5 t 2 . + - R(s) F(s) Y(s) E(s) Figure 2 The closed–loop transfer function is ( 29 ( 29 ( 29 ( 29 ( 29 2 5 4 3 2 9 7 20 1 15 71 114 63 180 s s Y s F s R s F s s s s s s + + = = + + + + + + . From the Routh Table below, it can be seen that the closed–loop system has two unstable poles
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Unformatted text preview: and, therefore, the steadystate error cannot be calculated. 5 4 3 2 1 1 71 63 15 114 180 63.4 51 101.9 180 61 180 s s s s s s-...
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