MAE 340 Lecture Notes - Lecture 6: Root Locus, Settling Time, Asymptote
11 Oct 2017
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2 + 2 + k ( + 3) = 0. The two branches begin at 0 and 2, move toward each other, meet at 1. 27, break out at that point to become a complex conjugate pair, then break back in at 4. 73. After they break in, they go in opposite directions along the real axis, with one branch ending at 3 and the other branch going to . Substituting = 1. 27 into the characteristic equation, we nd that k = 0. 54, and when = 4. 73, k = 7. 46: substituting = i into the original polynomial, we nd. One solution is k = 0, = 0, and the other solution is k = 2, = 3. Since we are only considering values of 0 k , there are no crossings of the imaginary axis other than k = 0, = 0. 3 2 + (2 + k)i = 0.
