stability6

# stability6 - s s K K s s K There will only be a sign change...

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Root Locus Q UESTION 6 For the system shown in Figure 6 with ( 29 ( 29 ( 29 2 10 4 5 K G 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. Figure 6 The closed–loop transfer function is ( 29 ( 29 ( 29 ( 29 3 2 1 14 45 50 G s K T s G s s s s K = = + + + + + . The closed– loop characteristic polynomial is ( 29 3 2 14 45 50 s s s K + + + + . The Routh table is 3 2 1 0 1 45 14 50 580 0 14 50
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Unformatted text preview: s s K K s s K +-+ . There will only be a sign change in the first column if K &amp;lt; 50 or K &amp;gt; 580. Therefore, the closedloop system will be stable for 50 580 K-&amp;lt; &amp;lt; . When K = 580, the s 1 row becomes a row of zeros. The even polynomial is in the s 2 row and is 14 s 2 + 630 = 0. Solving for s , the frequency of oscillation is 6.71 rad/s....
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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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