stability5

# stability5 - K for which the closedloop system is stable....

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Stability Q UESTION 5 For the system shown in Figure 5 with ( 29 ( 29 ( 29 ( 29 ( 29 2 2 4 5 3 K s s s G s s - + + = + , create the Routh table and determine for what range of K the closed–loop system is stable. Figure 5 The closed–loop transfer function is ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 3 2 2 4 5 1 7 1 2 3 40 G s K s s s T s G s Ks K s Ks K - + + = = + + + + + - . The closed–loop characteristic polynomial is ( 29 ( 29 3 2 7 1 2 3 40 Ks K s Ks K + + + + - . The Routh table is 3 2 2 1 0 2 7 1 3 40 54 0 7 1 3 40 0 s K K s K K K K s K s K + - - + - . If K is negative, the first entry in the s 0 row is always positive. Therefore, there is no range of negative

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Unformatted text preview: K for which the closedloop system is stable. If K is positive, the first entry in the s 2 row will always be positive, the first entry in the s 1 row will be negative if K < 1/54, and the first ME 279 Automatic Control of Mechanical Systems Professor Robert G. Landers entry in the s row will be negative if K > 3/40. Therefore, the closedloop system is stable for 1 3 54 40 K < < . 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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stability5 - K for which the closedloop system is stable....

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