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stability1

# stability1 - Stability therefore there are three poles in...

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Stability Q UESTION 1 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 5 4 3 2 8 4 4 3 2 s T s s s s s s + = - + - + - . Without the use of the Routh table, 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 5 4 3 2 4 4 3 2 s s s s s - + - + - . Since the signs of the coefficients are not the same, there is at least one unstable pole. Therefore, the closed–loop system is unstable. The Routh table is 5 4 3 2 2 1 0 1 4 3 1 4 2 0 1 0 1 4 2 0 2 4 1 0 0 1 4 2 0 0 s s s s s s ε ε ε ε ε ε ε ε + - + + - - - - - + - - - + - - + + + - - - - . Since there is a zero in the first column of the s 3 row, it must be replaced by an arbitrarily small number denoted by ε . When the Routh table is complete, sign changes will be determined as ε goes to zero and is positive and as ε goes to zero and is negative. There are three sign changes;

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Unformatted text preview: Stability therefore, there are three poles in the right s–plane. Since there are five poles, the other two poles are in the left s–plane. An alternate method is to examine the stability of the characteristic polynomial of the reciprocal roots. The characteristic polynomial for the reciprocal roots is 5 4 3 2 2 3 4 4 1 1 s s s s s-+-+-+ . The Routh table is 5 4 3 2 1 2 4 1 3 4 1 1.333 2.333 1.251 1 3.398 1 s s s s s s-------. There are three sign changes; therefore, there are three poles in the right s–plane and two poles in the left s–plane. As expected, this method yields the same result as the previous method. The Matlab code for computing the closed–loop pole locations is roots([1 –1 4 –4 3 –2]) The closed–loop poles are located at 0.8423, 0.1309 ± 0.8600 j , and –0.0521 ± 1.771 j . 2...
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