splane2

splane2 - Routh-Hurwitz Criterion Special Cases Zero only...

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Routh-Hurwitz Criterion: Special Cases Zero only in the first column When forming the Routh table, replace the zero with a small number ε and evaluate the fist column for positive or negative values of ε . Problem: Determine the stability of the closed- loop transfer function: T ( s )= 10 s 5 +2 s 4 +3 s 3 +6 s 2 +5 s (1) s 5 1 3 5 s 4 2 6 3 s 3 - 13 26 2 = 6 0 ε - 15 23 2 = 7 2 - 10 20 2 =0 s 2 ε 7 2 ε = 6 ε - 7 ε - ε 0 ε =3 - ε 0 ε s 1 ε 7 2 6 ε - 7 ε 3 6 ε - 7 ε = 42 ε - 49 - 6 ε 2 12 ε - 14 - ε 0 6 ε - 7 ε 0 6 ε - 7 ε - ε 0 6 ε - 7 ε 0 6 ε - 7 ε s 0 3 0 0 1
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Label First Column ε =+ ε = - s 5 1 + + s 4 2 + + s 3 ε + - s 2 6 ε - 7 ε - + s 1 42 ε - 49 - 6 ε 2 12 ε - 14 + + s 0 3 0 0 The changes in sign occur in both instances so it doesn’t matter which we choose. they both indicate the system is unstable with two poles in the rhp. 2
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Another, less computationally expensive method to use when a zero occurs in the first column is to create the Routh table using the polyno- mial that has the reciprocal roots of the origi- nal polynomial. i.e.
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splane2 - Routh-Hurwitz Criterion Special Cases Zero only...

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