Question 5.2: Plot the root locus of a closed-loop system having the follo...

This can be expressed as

G(s)H(s) = \frac{K}{(s + 1 + j)(s + 1 -j)(s + 2)}                 (5.172)

which has three poles at s = (- 1 + j ), s = (- 1 –  j) and s = – 2, and no zeros.
Following the rules above:
(1) Symmetry about the real axis will be assumed and is satisfied by the open-loop poles.
(2) There are three open-loop poles so there are three paths on the root locus plot.
(3) The paths start at the open-loop poles s = (- 1 + j ), s = (- 1 – j ) and s = – 2, which are marked on the diagram with X symbols (Figure 5.88(a))

(4) There are no zeros to be marked on the plot, so all of the paths end in asymptotes rather than at defined positions.
(5) n_{P} = 3, n_{Z} = 0, so there are three asymptotes at angles π/3, π and 5π/3 radians (or 60°, 180° and 300°) to the real axis (Figure 5.88(b))
(6) The asymptotes radiate outwards from the point:

s = \frac{(- 1 +j) + (- 1 -j) + (- 2)}{3} = – \frac{4}{3}           (5.173)

as shown in Figure 5.88(b).

(7) The only point at which a path can be present on the real axis is to the left of the root s = – 2, because only in this region is there an odd number of poles and zeros (in this case, only one pole and no zeros) to its right on the real axis. By enforcing this rule it is seen that this particular path starts at the pole s = – 2 and extends infinitely along the negative real axis (Figure 5.88(c)).

(8) There is no breakout or breakin as the paths in this case either lie on the real axis or away from it.

(9) The angle of departure of from the pole s = (- 1 + j ) (measured relative to the real axis), plus the total of the angles of that pole from the other poles, minus the total of the angles to that pole from the zeros, must add up to an odd multiple of π. It can be seen from Figure 5.88(c) how this is applied in the present case, with the total of the angles simply being π itself. The other angles are (in this example) simple to find geometrically, so the angle of departure is found as

By symmetry about the real axis, the angle of departure from the pole s = (- 1 – j)
must be – π/4.

(10) Two of the paths cross the imaginary axis after leaving the complex poles. At this stage it is noted that (by inserting the open-loop transfer function into (5.47),
expanding the bracketed terms and rearranging) the characteristic equation of the
closed-loop system can be expressed as
s³ + 4s² + 6s + 4 + K = 0                                         (5.174)

Making the substitution s = jω gives

– jω³  –  4ω² + 6jω + 4 + K = 0                                       (5.175)
Taking the real part of equation (5.175) and rearranging gives

6ω = ω³                                                                              (5.176)

ω is therefore either zero (not a valid solution) or ±√6 rad/s, so the loci intersect the imaginary axis at the points s = ±√6j = ±2.44j. Now taking the imaginary part of equation (5.175), rearranging and inserting this value of v gives the value of gain for limiting stability:

K = 4ω² – 4 = 4 × 6  –  4 = 20                                              (5.177)

At this stage, the root locus plot can be completed as shown in Figure 5.88(d),
with the two complex paths leaving the two complex poles at ±π/4, crossing the
imaginary axis at ±2.44j, and converging on the asymptotes at π/3 and 5π/3. A
control engineer may then choose (by eye and experience) a point on the plot for
which the gain can then be calculated from the distances of that point to the poles and zeros. For example, if point (– 0.5 + 1.65j ) on the locus is chosen by eye (Figure 5.89), K is calculated as

= 4.93                                                                       (5.178)

noting that (in this case) the denominator is simply 1 as there are no zeros.