stability9

# stability9 - to obtain 6 s , place this coefficient in the...

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Stability Q UESTION 9 Create the Routh table and determine how many poles of the following closed–loop transfer function are in the left s–plane, in the right s–plane, and on the imaginary axis: ( 29 2 4 3 2 4 3 4 8 20 15 s s T s s s s s + - = + + + + . By inspection, what can be said about the stability of this closed–loop system? Explain. Numerically determine the location of the closed–loop poles. The closed–loop characteristic polynomial is 4 3 2 4 8 20 15 s s s s + + + + . Since the signs of the coefficients are the same and there are no missing powers of s, the poles may all be in the left s– plane. Therefore, the closed–loop system may be stable. The Routh table is 4 3 2 1 0 1 8 15 4 20 0 3 15 0 06 0 0 15 0 0 s s s s s . Since there is a row of zeros in the s 1 row, the characteristic equation contains a 2 nd order even polynomial. The even polynomial is in the s 2 row and is 3 s 2 +15 = 0. Differentiate this polynomial

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Unformatted text preview: to obtain 6 s , place this coefficient in the s 1 row, and continue. Investigate sign changes from the s 1 row to the s row to determine the stability of the 2 nd order even polynomial. Since there are no sign changes and the roots of an even polynomial are symmetric about the origin, both poles are on the imaginary axis. To determine the location of the other poles, investigate sign changes from the s 2 row to the s 4 row. Since there are no sign changes, the remaining poles are in the left s–plane. Therefore, there are two poles on the imaginary axis and two poles in the left s–plane. Stability The Matlab code for computing the closed–loop pole locations is roots([1 4 8 20 15]) The closed–loop poles are located at –3, –1, and ± 2.236 j . 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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stability9 - to obtain 6 s , place this coefficient in the...

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