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Unformatted text preview: ( 29 ( 29 ( 29 cos 4 sin 4 y t A B t C t = + + . Since the poles are purely imaginary, the time response is marginally stable. Time Response For the transfer function ( 29 2 5 2 100 T s s s = + + , the poles are at 1 99 j + and 1 99 j , and there are no finite zeros. Since the poles are complex, the time response is underdamped and the general time response to a step input is ( 29 ( 29 ( 29 ( 29 cos 99 sin 99 cos 99 t t t y t A Be t A Ce t De t φ= + + = + + . Since the real parts of the poles are negative, the time response is stable. For the transfer function ( 29 ( 29 2 5 1 s T s s + =, the poles are at 1 and 1, and there is a zero at –5. Since the poles are real and equal, the time response is critically damped and the general time response to a step input is ( 29 t t y t A Be Cte = + + . Since the real parts of the poles are positive, the time response is unstable. 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.
 Spring '11
 LANDERS

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