A closed-loop system has open-loop poles at s = 0, s = –2, and s = –5, and no finite zeros. As the gain K increases from 0 to ∞, where do the three root locus branches start and terminate?
AStart at the closed-loop poles and terminate at open-loop poles as K→∞
BStart at the open-loop zeros and terminate at the open-loop poles as K→∞
CStart at the open-loop poles (s = 0, –2, –5) at K=0 and travel to infinity along asymptotes as K→∞
DStart at the open-loop poles at K=0 and converge to the centroid of all poles as K→∞
Root locus branches always start at the open-loop poles (K=0) and terminate at open-loop zeros or at infinity (K→∞). Since this system has no finite zeros (n=3, m=0), all three branches escape to infinity along asymptotes. The asymptote angles are 180°(2k+1)/(n−m) = 60°, 180°, and 300°. The starting condition (branches begin at open-loop poles) follows from the fact that at K=0, the loop is barely closed and closed-loop poles approach the open-loop poles.
Question 2 Multiple Choice
A system has open-loop poles at s = 0, –1, –3, –4 and open-loop zeros at s = –2. A student claims the root locus exists on the real-axis segment between s = –1 and s = –2, because this segment lies between adjacent poles. Is this correct, and why?
AYes — locus always exists between adjacent poles on the real axis
BNo — the locus rule counts all poles AND zeros combined to the right of the test point; the segment –1 to –2 has exactly 2 real singularities to its right (pole at 0 and... depends on count)
CNo — the real-axis locus exists to the LEFT of an odd count of all open-loop poles and zeros combined; test each segment by counting poles+zeros to its right
DYes — any segment between a pole and an adjacent zero is automatically on the locus
The real-axis rule: a point on the real axis is on the locus if and only if the total count of open-loop poles PLUS zeros to its RIGHT is odd. For the segment –1 to –2: to the right lie only the pole at s=0, giving a count of 1 (odd) → this segment IS on the locus. The key misconception is thinking the rule only applies to poles, or that it counts from the nearest singularity. You must count ALL poles and zeros to the right of the test point.
Question 3 True / False
A root locus branch crossing the imaginary axis at a particular gain value means the closed-loop system becomes marginally stable at that gain.
TTrue
FFalse
Answer: True
Stability requires all closed-loop poles to be in the left half of the s-plane (negative real parts). When a locus branch crosses the imaginary axis, the corresponding closed-loop pole(s) have zero real part — the system is marginally stable (sustained oscillations with no decay or growth). Below that gain, those poles are in the left half-plane (stable); above it, they are in the right half-plane (unstable). The crossing gain can be found from Routh-Hurwitz, connecting the two methods.
Question 4 True / False
Adding a compensator pole or zero to the open-loop system shifts the root locus branches slightly but preserves the overall shape of the original locus.
TTrue
FFalse
Answer: False
Adding a pole or zero to the open-loop transfer function changes the angle condition for every point in the s-plane and fundamentally reshapes the entire locus — it does not merely shift existing branches. A new zero attracts locus branches toward it; a new pole repels them. The entire topology of the locus changes: the number of branches, the asymptote angles and centroid, the real-axis segments, and the breakaway points all change. This is precisely why compensator design (lead/lag) is so powerful — you can pull the locus into entirely new regions of the s-plane.
Question 5 Short Answer
Explain why the angle condition ∠G(s) = ±180°(2k+1) is the fundamental test for whether a point s₀ lies on the root locus.
Think about your answer, then reveal below.
Model answer: A closed-loop pole must satisfy 1 + KG(s) = 0, or G(s) = –1/K. For real positive K, this requires G(s) to be a negative real number, which means its phase angle must equal an odd multiple of 180°. So the angle condition ∠G(s₀) = ±180°(2k+1) is simply the requirement that K is positive and real at that point. If a point satisfies the angle condition, you can always find a positive K (namely K = 1/|G(s₀)|) that places a closed-loop pole there.
This is why the construction rules (real-axis rule, asymptotes, breakaway points) are all consequences of the angle condition: they are specialized forms of ∠G(s) = 180° applied to particular geometries. Understanding the angle condition as the definition means you can test any point in the s-plane directly, without memorizing rules. The magnitude condition |G(s₀)| = 1/K then gives you the specific gain value.