Question

3. (25 marks) A system has a characteristic polynomial 45 +20s49 101s2 +2s +24 Using the Routh-Hurwitz criterion, determine the number of poles in the left half plane, the number on the jw-axis and the number in the right half plane.

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Answer 1

Given characteristic polynomial is CE = 4s^5 + 20s^4 - 9s^3 - 101s^2 + 2s + 24 rightarrow by routh-Hurwitz criterion, no. of right plane poles is equal to the no. of sign changes occurred in first column of routh Hurwitz criteria By routh Hurwitz criteria Auxiliary equation = -96 s^2 + 24 = 0 doublerightarrow 192 s + 0 = 0 doublerightarrow we got two sign changes in first column of the RH criteria. So, we have two right side poles in Right side of JW axis. \r\n doublerightarrow no. of sign changes below auxiliary equation = 1 total roots of Auxiliary equation = 2 So, each root will lie either side of the jW axis, bcz, one root is right side so another will be left side bcz, row of zero arises because of symmetrical roots of the Auxiliary equation around the origin. So, to satisfy the symmetry of roots, one of the roots of AE will be right side & one will be left side (So, no root will be on jW axis)