Question 8.9: Sketch the root locus as a function of negative gain, K, for...
Sketch the root locus as a function of negative gain, K, for the system shown in Figure 8.11.
The equivalent positive-feedback system found by pushing −1, associated with K, to the right past the pickoff point is shown in Figure 8.27(a). Therefore, as the gain of the equivalent system goes through positive values of K, the root locus will be equivalent to that generated by the gain, K, of the original system in Figure 8.11 as it goes through negative values.
The root locus exists on the real axis to the left of an even number of real, finite open-loop poles and/or zeros. Therefore, the locus exists on the entire positive extension of the real axis, between −1 and −2 and between −3 and −4. Using Eq. (8.27), the σa intercept is found to be
σ_{a} = \frac{\sum finite poles − \sum finite zeros }{\#finite poles – \#finite zeros} (8.27)
σ_{a} = \frac{(−1 − 2 − 4) − (−3) }{4 − 1} = – \frac{4}{3} (8.67)
The angles of the lines that intersect at −4/3 are given by
θ_{a} = \frac{k2π}{\# finite poles − \# finite zeros} (8.68a)
= 0 for k = 0 (8.68b)
= 2π/3 for k = 1 (8.68c)
= 4π/3 for k = 2 (8.68d)
The final root locus sketch is shown in Figure 8.27(b).
