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stability11

# stability11 - Stability QUESTION 11 For the system shown in...

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Stability Q UESTION 11 For the system shown in Figure 11 with ( 29 5 4 3 2 18 7 7 18 G s s s s s s = + - - - , create the Routh table and determine how many closed–loop poles are in the left s–plane, in the right s–plane, and on the imaginary axis. Without the use of the Routh table, what can be said about the stability of this closed–loop system? Explain. Numerically determine the location of the closed–loop poles. + - R(s) G(s) Y(s) E(s) Figure 11 The closed–loop transfer function is ( 29 ( 29 ( 29 5 4 3 2 18 1 7 7 18 18 G s T s G s s s s s s = = + + - - - + . The closed–loop characteristic polynomial is 5 4 3 2 7 7 18 18 s s s s s + - - - + . Since the signs of the coefficients are not the same, there is at least one unstable pole. Therefore, the closed–loop system is unstable. The Routh table is 5 4 3 2 2 1 0 1 7 18 1 7 18 0 36 0 36 7 18 0 18 36 0 0 36 7 18 0 0 s s s s s s ε + - - - + + - + + - + - - + - - - - - - + +

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Stability Since there are two sign changes in the first column, there are two poles in the right s–plane. Since there are five total poles, the other three poles are in the left s–plane.
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stability11 - Stability QUESTION 11 For the system shown in...

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