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stability14

# stability14 - K< 0 or K> 19.69 Therefore the closed–...

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Stability Q UESTION 14 For the system shown in Figure 14 with ( 29 ( 29 ( 29 ( 29 1 2 5 K G s s s s s = + + + , create the Routh table and determine for what range of K the closed–loop system is stable. Determine the frequency of oscillation at marginal stability. + - R(s) G(s) Y(s) E(s) Figure 14 The closed–loop transfer function is ( 29 ( 29 ( 29 4 3 2 1 8 17 10 G s K T s G s s s s s K = = + + + + + . The closed– loop characteristic polynomial is 4 3 2 8 17 10 s s s s K + + + + . The Routh table is 4 3 2 1 0 1 17 8 10 0 126 0 8 32 10 0 0 63 0 0 s K s s K s K s K - + . There will only be a sign change in the first column if

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Unformatted text preview: K < 0 or K > 19.69. Therefore, the closed– loop system will be stable for 0 19.69 K < < . When K = 19.69, the s 1 row becomes a row of ME 279 – Automatic Control of Mechanical Systems Professor Robert G. Landers zeros. The even polynomial is in the s 2 row and is 2 126 19.69 8 s + = . Solving for s , the frequency of oscillation is 1.118 rad/s. 2...
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stability14 - K< 0 or K> 19.69 Therefore the closed–...

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