⚙ Root Locus in the Real World
- DC motor speed control — increasing K moves poles: underdamped → critically damped → unstable
- Antenna tracking system — gain tuning via root locus prevents tracking oscillation
- Op-amp feedback amplifier — root locus shows how bandwidth-gain tradeoff plays out
- Hard disk read head — positioning controller gain set using root locus to avoid servo buzz
- Radar platform stabilizer — root locus determines safe gain range for vibration-free operation
1. Root Locus — Concept
The root locus is the plot of closed-loop poles in the s-plane as gain K varies from 0 to ∞. It tells us exactly how pole locations — and thus system response — change with K.
For unity feedback with G(s) = KG₀(s), the closed-loop CE is:
1 + KG₀(s) = 0 ⟹ G₀(s) = −1/K
Angle condition: ∠G₀(s) = ±180°(2q+1), q = 0,1,2,...
Magnitude condition: |G₀(s)| = 1/K
Angle condition: ∠G₀(s) = ±180°(2q+1), q = 0,1,2,...
Magnitude condition: |G₀(s)| = 1/K
Physical interpretation: At K=0, closed-loop poles = open-loop poles (locus starts). As K→∞, closed-loop poles → open-loop zeros (finite ones) or → ∞ along asymptotes (for poles without matching zeros).
2. Seven Rules of Root Locus Construction
RULE 1
Starting and Ending Points
Root locus starts (K=0) at open-loop poles and ends (K→∞) at open-loop zeros (finite zeros) or goes to ∞ along asymptotes.
RULE 2
Number of Branches
Number of branches = max(P, Z) where P = number of open-loop poles, Z = number of open-loop zeros. Branches going to ∞ = P − Z.
RULE 3
Symmetry
Root locus is symmetric about the real axis (complex poles always appear in conjugate pairs).
RULE 4
Real-Axis Segments
A point on the real axis lies on the root locus if the total number of real poles and zeros to its right is odd.
RULE 5
Asymptote Angles
For (P−Z) branches going to ∞, asymptote angles:
φₐ = (2q+1)×180° / (P−Z), q = 0, 1, ..., P−Z−1
φₐ = (2q+1)×180° / (P−Z), q = 0, 1, ..., P−Z−1
RULE 6
Asymptote Centroid
All asymptotes originate from the centroid:
σₐ = (Σ poles − Σ zeros) / (P − Z)
σₐ = (Σ poles − Σ zeros) / (P − Z)
RULE 7
Breakaway / Break-in Points
Where multiple locus branches meet and split. Find from:
dK/ds = 0 where K = −1/G₀(s)H(s)
dK/ds = 0 where K = −1/G₀(s)H(s)
Additional Rules
Angle of departure from complex pole p:
φd = 180° − Σ∠(p − pᵢ) + Σ∠(p − zⱼ) [sum of angles from other poles, minus sum from zeros]
Gain at any point s* on root locus:
K = 1/|G₀(s*)| = (product of distances from s* to all poles) / (product of distances from s* to all zeros)
φd = 180° − Σ∠(p − pᵢ) + Σ∠(p − zⱼ) [sum of angles from other poles, minus sum from zeros]
Gain at any point s* on root locus:
K = 1/|G₀(s*)| = (product of distances from s* to all poles) / (product of distances from s* to all zeros)
3. Worked Example — Complete Root Locus
G(s) = K / [s(s+2)(s+4)] — Unity feedback
Poles: 0, −2, −4 (P=3) | Zeros: none (Z=0)
Branches: 3 | Branches to ∞: 3 − 0 = 3
Rule 4 (Real axis): RL exists on real axis where odd number of poles/zeros to right:
0 to −2: 1 pole to right → on RL ✓
−2 to −4: 2 poles to right → NOT on RL
−∞ to −4: 3 poles to right → on RL ✓
Centroid: σₐ = (0 + (−2) + (−4) − 0) / (3−0) = −6/3 = −2
Asymptote angles: q=0: 60°, q=1: 180°, q=2: 300° (= −60°)
Breakaway point: K = −s(s+2)(s+4) = −[s³+6s²+8s]
dK/ds = −[3s²+12s+8] = 0 → s = (−12±√(144−96))/6 = (−12±√48)/6 = −2 ± 2√3/3
s ≈ −0.845 (between 0 and −2, on RL ✓) → breakaway at s ≈ −0.845
jω-axis crossing: CE = s³+6s²+8s+K = 0. From Routh: K_marginal = 48 (Unit 3 example).
Auxiliary: 6s²+48=0 → s = ±j2√2 → RL crosses jω at ±j2.83 when K=48.
Poles: 0, −2, −4 (P=3) | Zeros: none (Z=0)
Branches: 3 | Branches to ∞: 3 − 0 = 3
Rule 4 (Real axis): RL exists on real axis where odd number of poles/zeros to right:
0 to −2: 1 pole to right → on RL ✓
−2 to −4: 2 poles to right → NOT on RL
−∞ to −4: 3 poles to right → on RL ✓
Centroid: σₐ = (0 + (−2) + (−4) − 0) / (3−0) = −6/3 = −2
Asymptote angles: q=0: 60°, q=1: 180°, q=2: 300° (= −60°)
Breakaway point: K = −s(s+2)(s+4) = −[s³+6s²+8s]
dK/ds = −[3s²+12s+8] = 0 → s = (−12±√(144−96))/6 = (−12±√48)/6 = −2 ± 2√3/3
s ≈ −0.845 (between 0 and −2, on RL ✓) → breakaway at s ≈ −0.845
jω-axis crossing: CE = s³+6s²+8s+K = 0. From Routh: K_marginal = 48 (Unit 3 example).
Auxiliary: 6s²+48=0 → s = ±j2√2 → RL crosses jω at ±j2.83 when K=48.
4. Gain Calculation from Root Locus
The gain K at any point s* on the root locus is found from the magnitude condition:
K = |s* − p₁| × |s* − p₂| × ... / (|s* − z₁| × |s* − z₂| × ...)
Each factor |s* − pᵢ| is the geometric distance from the test point s* to the corresponding pole on the s-plane.
Example: For G(s) = K/[s(s+2)], find K when closed-loop poles are at s = −1 ± j1.
Test point s* = −1+j1.
Distance to pole at 0: |−1+j1−0| = √(1+1) = √2
Distance to pole at −2: |−1+j1−(−2)| = |1+j1| = √2
No zeros → K = √2 × √2 / 1 = 2
Test point s* = −1+j1.
Distance to pole at 0: |−1+j1−0| = √(1+1) = √2
Distance to pole at −2: |−1+j1−(−2)| = |1+j1| = √2
No zeros → K = √2 × √2 / 1 = 2
GATE Trap: The angle condition (±180°) determines WHETHER a point is on the root locus. The magnitude condition gives the GAIN at that point. Never use the magnitude condition to check if a point is on the locus — always use the angle condition first.
5. Effect of Adding Poles and Zeros
| Modification | Effect on Root Locus | Practical Impact |
|---|---|---|
| Add open-loop pole (LHP) | Locus pushed toward RHP; system harder to stabilize | Integrator (pole at 0) improves steady-state but can destabilize |
| Add open-loop zero (LHP) | Locus pulled toward LHP; more stable | Lead/derivative action improves stability margins |
| Add pole-zero pair (dipole) | Local effect only if pair close together | Notch filter, partial cancellation (risky if imperfect) |
⚙ Lab: Observing Root Locus Effects
- MATLAB: rlocus(G) plots root locus; rlocfind(G) gives K at clicked point
- Start with G=K/[s(s+1)]: stable for all K. Add pole at −0.1: observe how locus enters RHP
- Add a zero at −0.5: observe locus bending back into LHP
- Physical: op-amp circuit → increase RF (gain K) → measure when oscillation starts → that K is at jω crossing
- Verify: K_oscillation matches K from Routh-Hurwitz analysis in Unit 3
GATE Previous Year Questions
G(s) = K/[s(s+1)(s+3)]. The centroid of asymptotes is:
Answer: B — σₐ = −4/3
Poles: 0, −1, −3. Zeros: none. P=3, Z=0.
σₐ = (Σpoles − Σzeros)/(P−Z) = (0+(−1)+(−3) − 0)/(3−0) = −4/3 ≈ −1.33
Poles: 0, −1, −3. Zeros: none. P=3, Z=0.
σₐ = (Σpoles − Σzeros)/(P−Z) = (0+(−1)+(−3) − 0)/(3−0) = −4/3 ≈ −1.33
G(s) = K(s+2)/[s(s+1)(s+4)]. The number of root locus branches going to infinity is:
Answer: B — 2 branches to infinity
P = 3 (poles at 0, −1, −4), Z = 1 (zero at −2).
Branches to infinity = P − Z = 3 − 1 = 2.
One branch ends at the finite zero at s = −2.
P = 3 (poles at 0, −1, −4), Z = 1 (zero at −2).
Branches to infinity = P − Z = 3 − 1 = 2.
One branch ends at the finite zero at s = −2.
G(s) = K/[s(s+2)]. The breakaway point on the root locus lies at:
Answer: A — s = −1
K = −s(s+2) = −(s²+2s). dK/ds = −(2s+2) = 0 → s = −1.
Check: s=−1 is between poles at 0 and −2, and lies on the real-axis RL (1 pole to right) ✓.
K at breakaway = −(−1)(−1+2) = −(−1)(1) = 1.
K = −s(s+2) = −(s²+2s). dK/ds = −(2s+2) = 0 → s = −1.
Check: s=−1 is between poles at 0 and −2, and lies on the real-axis RL (1 pole to right) ✓.
K at breakaway = −(−1)(−1+2) = −(−1)(1) = 1.
G(s) = K/[s(s+4)(s+8)]. The asymptote angles are:
Answer: B — 60°, 180°, 300°
P=3, Z=0 → P−Z=3 → 3 asymptotes.
φₐ = (2q+1)×180°/3: q=0: 60°, q=1: 180°, q=2: 300° (= −60°).
Centroid = (0+(−4)+(−8)−0)/3 = −12/3 = −4.
P=3, Z=0 → P−Z=3 → 3 asymptotes.
φₐ = (2q+1)×180°/3: q=0: 60°, q=1: 180°, q=2: 300° (= −60°).
Centroid = (0+(−4)+(−8)−0)/3 = −12/3 = −4.
The root locus of a unity feedback system with G(s) = K/[s(s+2)] has how many branches, and where do they start?
Answer: B
G(s) has 2 poles (P=2) and 0 zeros (Z=0). Branches = max(P,Z) = 2.
At K=0: closed-loop poles = open-loop poles → locus starts at s=0 and s=−2.
As K→∞: both branches go to ∞ along asymptotes at 90° and 270°.
G(s) has 2 poles (P=2) and 0 zeros (Z=0). Branches = max(P,Z) = 2.
At K=0: closed-loop poles = open-loop poles → locus starts at s=0 and s=−2.
As K→∞: both branches go to ∞ along asymptotes at 90° and 270°.
A system has open-loop poles at s=0, s=−1, s=−2 and one open-loop zero at s=−3. Which segment of the real axis is NOT part of the root locus?
Answer: B — Segment −1 to −2 is NOT on RL
Count poles+zeros to the RIGHT of each segment:
0 to −1: 1 pole (at 0) → odd → ON RL
−1 to −2: 2 (poles at 0,−1) → even → NOT on RL
−2 to −3: 3 (poles at 0,−1,−2) → odd → ON RL
−3 to −∞: 4 (poles+zero) → even → NOT on RL
Count poles+zeros to the RIGHT of each segment:
0 to −1: 1 pole (at 0) → odd → ON RL
−1 to −2: 2 (poles at 0,−1) → even → NOT on RL
−2 to −3: 3 (poles at 0,−1,−2) → odd → ON RL
−3 to −∞: 4 (poles+zero) → even → NOT on RL
G(s) = K/[s(s+1)]. The gain K at which the closed-loop poles are at s = −0.5 ± j0.5 is:
Answer: B — K = 0.5
s* = −0.5+j0.5. Distance to pole at 0: |−0.5+j0.5| = √(0.25+0.25) = √0.5 = 1/√2
Distance to pole at −1: |−0.5+j0.5−(−1)| = |0.5+j0.5| = 1/√2
K = (1/√2)×(1/√2) / 1 = 1/2 = 0.5
s* = −0.5+j0.5. Distance to pole at 0: |−0.5+j0.5| = √(0.25+0.25) = √0.5 = 1/√2
Distance to pole at −1: |−0.5+j0.5−(−1)| = |0.5+j0.5| = 1/√2
K = (1/√2)×(1/√2) / 1 = 1/2 = 0.5
G(s) = K(s+1)/[s²(s+4)]. The angle of asymptotes and centroid for the root locus are:
Answer: A/B — 90° & 270°, centroid = −1.5
Poles: 0 (×2), −4 → P=3. Zeros: −1 → Z=1. Branches to ∞: 3−1=2 asymptotes.
φₐ = (2q+1)×180°/2: q=0: 90°, q=1: 270°
σₐ = (0+0+(−4)−(−1))/(3−1) = (−4+1)/2 = −3/2 = −1.5
Poles: 0 (×2), −4 → P=3. Zeros: −1 → Z=1. Branches to ∞: 3−1=2 asymptotes.
φₐ = (2q+1)×180°/2: q=0: 90°, q=1: 270°
σₐ = (0+0+(−4)−(−1))/(3−1) = (−4+1)/2 = −3/2 = −1.5
G(s) = K/[(s+1)(s+2)(s+3)]. The root locus has a break-away point between s=−1 and s=−2. Approximate location is:
Answer: B — ≈ −1.47
K = −(s+1)(s+2)(s+3). dK/ds = −[(s+2)(s+3)+(s+1)(s+3)+(s+1)(s+2)] = 0
Expanding: 3s²+12s+11=0 → s = (−12±√(144−132))/6 = (−12±√12)/6 = −2 ± √3/3
s = −2 + 0.577 ≈ −1.42 or s ≈ −2.577 (not between −1 and −2).
Valid breakaway ≈ −1.42 to −1.47 depending on rounding.
K = −(s+1)(s+2)(s+3). dK/ds = −[(s+2)(s+3)+(s+1)(s+3)+(s+1)(s+2)] = 0
Expanding: 3s²+12s+11=0 → s = (−12±√(144−132))/6 = (−12±√12)/6 = −2 ± √3/3
s = −2 + 0.577 ≈ −1.42 or s ≈ −2.577 (not between −1 and −2).
Valid breakaway ≈ −1.42 to −1.47 depending on rounding.
Adding a zero to a closed-loop system generally:
Answer: B
Adding a zero in the LHP attracts the root locus branches toward it, pulling them away from the RHP. This increases the range of K for which the system is stable and typically improves damping. This is the principle behind derivative (D) control action and lead compensation — the added zero improves stability margins. Adding a pole has the opposite effect (pushes locus toward RHP).
Adding a zero in the LHP attracts the root locus branches toward it, pulling them away from the RHP. This increases the range of K for which the system is stable and typically improves damping. This is the principle behind derivative (D) control action and lead compensation — the added zero improves stability margins. Adding a pole has the opposite effect (pushes locus toward RHP).