Solved root locus exercises, with the loci drawn

Nine exercises with solutions: gain for a damping, asymptotes and breakaway, circular locus, departure angles, stabilising an unstable process, stabilising zero, locus with respect to a parameter, design for overshoot.

ielettronica 25 October 2026 8 min read 0 comments

Root locus exercises of the kind set in Control Systems courses, in three levels: basic, with second-order systems and gain computation; intermediate, with asymptotes, breakaway points, departure angles and imaginary-axis crossings; exam, with unstable processes, compensation with a zero, locus with respect to a parameter and design to a specification. The result is given straight away; the full solution, with the locus drawn, opens with a click. The proved rules and the MATLAB and Python code are in the in-depth study linked at the top.

Basic level

Exercise 1. The simplest locus

Sketch the locus of $L(s) = \dfrac{K}{s(s + 2)}$ and find the gain for which the closed-loop poles have damping $\zeta = 0.707$ .

Result: branches on the real axis between 0 and $-2$ , breakaway at $-1$ for $K = 1$ , then the vertical line $\text{Re}\,s = -1$ ; $K = 2$ .

Solution

Characteristic equation: $s^2 + 2s + K = 0$ . Real axis: the stretch between 0 and $-2$ (to the right of each of its points there is one pole). Two asymptotes at $\pm 90°$ from the centroid $\frac{0 - 2}{2} = -1$ .

Breakaway: $\frac{d}{ds}(s^2 + 2s) = 2s + 2 = 0$ , $s = -1$ , with $K = 1$ . For $K \gt 1$ the poles are $-1 \pm i\sqrt{K - 1}$ : the real part stays at $-1$ (the sum of the poles is constant, rule 7).

Damping: comparing with $s^2 + 2\zeta\omega_ns + \omega_n^2$ , $\omega_n = \sqrt K$ and $\zeta = \frac{1}{\sqrt K}$ . For $\zeta = 0.707$ : $K = 2$ , poles $-1 \pm i$ . Increasing the gain makes the response more oscillatory but not faster to settle: the settling time $4/\zeta\omega_n = 4$ s does not change.

Root locus of K over s times s plus 2 — The branches start from 0 and minus 2, meet at minus 1 and go up vertically; the dashed green line at 45 degrees corresponds to damping 0.707 and meets the locus at minus 1 plus or minus i.

Exercise 2. A gain for an overshoot

For $L(s) = \dfrac{K}{s(s + 2)(s + 4)}$ find the gain that gives the dominant poles a damping of 0.5, the position of the third pole and the critical gain.

Result: $K = 8.3$ ; dominant poles $-0.667 \pm 1.155i$ , third pole $-4.67$ ; $K_{crit} = 48$ .

Solution

The dominant poles lie on the line $s = r\,e^{\pm i120°}$ (damping 0.5, angle 60° from the negative half-axis). By rule 7, the sum of the three poles is always $-6$ : if the dominant ones are $-a \pm ia\sqrt 3$ , the third is $-6 + 2a$ .

Requiring $(s^2 + 2as + 4a^2)(s + 6 - 2a) = s^3 + 6s^2 + 8s + K$ , the coefficient of $s$ gives $4a^2 + 2a(6 - 2a) = 8$ , i.e. $12a = 8$ , $a = 0.667$ . Then the dominant poles are $-0.667 \pm 1.155i$ , the third $-4.67$ , and $K = 4a^2(6 - 2a) = 1.778\cdot 4.667 = 8.30$ .

Critical gain (Routh, or $s = i\omega$ ): $6\cdot 8 = K$ , $K = 48$ , $\omega = 2\sqrt 2$ . The gain margin with $K = 8.3$ is $48/8.3 = 5.8$ .

The third pole is seven times further from the axis than the dominant ones: its influence is small, and the actual overshoot (15.5 %) is close to that of the second-order system (16.3 %).

The locus crossed by the damping line 0.5 — The green dots are the three closed-loop poles for K equal to 8.3: the two dominant ones on the dashed line and the third on the real axis at minus 4.67.

Intermediate level

Exercise 3. Four poles: asymptotes, breakaway, stability

For $L(s) = \dfrac{K}{s(s + 1)(s + 3)(s + 4)}$ find the asymptotes, the breakaway points and the critical gain.

Result: centroid $-2$ , angles $\pm 45°$ and $\pm 135°$ ; breakaway at $-0.419$ and $-3.581$ (both with $K = 2.25$ ); $K_{crit} = 26.25$ with $\omega = 1.22$ .

Solution

Real axis: stretches $[-1, 0]$ and $[-4, -3]$ . Asymptotes: $n - m = 4$ , centroid $\frac{0 - 1 - 3 - 4}{4} = -2$ , angles $45°, 135°, 225°, 315°$ .

Breakaway: $D = s^4 + 8s^3 + 19s^2 + 12s$ , $D' = 4s^3 + 24s^2 + 38s + 12 = 0$ : roots $-0.419$ , $-2$ , $-3.581$ . The point $-2$ is not on the locus (between $-1$ and $-3$ there are two poles to the right, an even number). At the other two $K = |D| = 2.25$ : the same $K$ by symmetry (the poles are symmetric about $-2$ ).

Routh on $s^4 + 8s^3 + 19s^2 + 12s + K$ : second row $8, 12$ ; third $\frac{8\cdot 19 - 12}{8} = 17.5$ and $K$ ; fourth $\frac{17.5\cdot 12 - 8K}{17.5}$ , positive for $K \lt 26.25$ . Auxiliary equation $17.5s^2 + 26.25 = 0$ : $\omega = 1.22$ .

Locus with four poles and its asymptotes — The pairs of branches leave the axis at minus 0.42 and minus 3.58 and approach the asymptotes at 45 and 135 degrees starting from minus 2; the right-hand branches cross the imaginary axis.

Exercise 4. The circular locus and maximum damping

For $L(s) = K\dfrac{s + 3}{s(s + 1)}$ find the breakaway points and the gain that gives maximum damping of the complex poles.

Result: breakaway at $-0.551$ and $-5.449$ ; maximum damping $\zeta = 0.577$ for $K = 1$ , with poles $-1 \pm i\sqrt 2$ .

Solution

Breakaway (in-depth study): $s^2 + 6s + 3 = 0$ , $s = -3 \pm\sqrt 6$ .

The complex part of the locus is the circle centred at $-3$ with radius $\sqrt 6$ . The damping of a pole is the cosine of the angle between the line to the origin and the negative half-axis: it is maximum where the line from the origin is tangent to the circle. There $\sin\theta = \frac{\sqrt 6}{3}$ , $\zeta = \cos\theta = \sqrt{1 - \frac{6}{9}} = 0.577$ , and the distance from the origin is $\sqrt{9 - 6} = \sqrt 3$ .

The pole is $\sqrt 3\,e^{i(180° - 54.7°)} = -1 + i\sqrt 2$ . From the equation $s^2 + (1 + K)s + 3K = 0$ , the real part is $-\frac{1 + K}{2} = -1$ , so $K = 1$ . Check: $s^2 + 2s + 3$ has roots $-1 \pm i\sqrt 2$ .

Circular locus around the zero — The circle centred at minus 3 with radius root 6 joins the two breakaway points; the point of maximum damping is where the line from the origin touches it.

Exercise 5. Departure angle and crossing

For $L(s) = \dfrac{K}{(s + 3)(s^2 + 2s + 2)}$ compute the departure angle from the pole $-1 + i$ , the asymptotes and the critical gain.

Result: $63.4°$ ; centroid $-1.67$ , angles $\pm 60°$ , $180°$ ; $K_{crit} = 34$ with $\omega = 2.83$ .

Solution

Departure (rule 6): $\theta = 180° - \arg(p + 3) - \arg(p - \bar p) = 180° - 26.6° - 90° = 63.4°$ .

Asymptotes: $\sigma_a = \frac{-3 - 1 - 1}{3} = -1.67$ .

Characteristic equation $s^3 + 5s^2 + 8s + 6 + K = 0$ . Routh: $5\cdot 8 \gt 6 + K$ , i.e. $K \lt 34$ . For $K = 34$ : $5s^2 + 40 = 0$ , $\omega = \sqrt 8 = 2.83$ .

Check with rule 7: for $K = 34$ the poles are $\pm 2.83i$ and the third must be $-5$ (sum $-5$ ): indeed $(s + 5)(s^2 + 8) = s^3 + 5s^2 + 8s + 40$ . ✓

Departure from the complex poles — The branch leaves the pole minus 1 plus i at an angle of about 63 degrees and crosses the imaginary axis for K equal to 34.

Exam level

Exercise 6. Stabilising an unstable process

The process $G(s) = \dfrac{s + 1}{s(s - 2)}$ is controlled with a gain $K$ in unity feedback. For which $K$ is the closed loop stable? For which $K$ are the poles real and coincident?

Result: stable for $K \gt 2$ ; coincident poles for $K = 4 + 2\sqrt 3 = 7.46$ , at $s = -2.73$ (and for $K = 4 - 2\sqrt 3 = 0.54$ , at $+0.73$ , unstable).

Solution

Characteristic equation: $s^2 + (K - 2)s + K = 0$ . Stable if all coefficients are positive: $K \gt 2$ .

Coincident poles: zero discriminant, $(K - 2)^2 = 4K$ , $K^2 - 8K + 4 = 0$ , $K = 4 \pm 2\sqrt 3$ . For $K = 7.46$ the double root is $-\frac{K - 2}{2} = -2.73$ : the breakaway on the left, on the stretch to the left of the zero. For $K = 0.54$ the root is $+0.73$ , between the two unstable poles: a breakaway in the right half-plane.

The locus is a circle centred at $-1$ with radius $\sqrt 3$ : $-1 \pm\sqrt 3$ are exactly $0.73$ and $-2.73$ . ✓

Locus of the unstable process — The branches start in the right half-plane, go round the zero at minus 1 and become stable for K greater than 2.

Exercise 7. A stabilising zero

To the system $\dfrac{K}{s(s + 2)(s + 4)}$ , unstable for $K \gt 48$ , a zero at $-2.5$ is added: $L = \dfrac{K(s + 2.5)}{s(s + 2)(s + 4)}$ . Show that the closed loop is now stable for every $K \gt 0$ .

Result: vertical asymptotes with centroid $-1.75$ ; Routh always satisfied.

Solution

Asymptotes: $n - m = 2$ , angles $\pm 90°$ , centroid $\frac{(0 - 2 - 4) - (-2.5)}{2} = -1.75$ . The branches going to infinity stay on the line $\text{Re}\,s = -1.75$ , in the left half-plane.

Routh on $s^3 + 6s^2 + (8 + K)s + 2.5K$ : all coefficients positive, and the third-order condition $6(8 + K) \gt 2.5K$ holds for every $K \gt 0$ (it becomes $48 + 3.5K \gt 0$ ).

The zero is obtained with a PD controller, $C = K(s + 2.5)$ , or in practice with a lead network (zero followed by a pole much further to the left, which leaves the picture almost unchanged).

Locus with the added zero — The pole at minus 4 goes towards the zero at minus 2.5; the other two branches meet and rise along the vertical line minus 1.75, never crossing the imaginary axis.

Exercise 8. The locus with respect to damping

The mass-spring-damper system $\ddot y + a\dot y + 4y = u$ has adjustable damping $a$ . Sketch the path of the poles as $a \ge 0$ varies and find $a$ for critical damping and for $\zeta = 0.5$ .

Result: the poles move on the circle $|s| = 2$ ; critical damping for $a = 4$ ; $\zeta = 0.5$ for $a = 2$ .

Solution

Characteristic equation $s^2 + as + 4 = 0$ , rewritten as $1 + a\frac{s}{s^2 + 4} = 0$ : a locus with poles $\pm 2i$ and a zero at 0 (in-depth study).

As long as the poles are complex, their product is 4, so $|s|^2 = 4$ : they lie on the circle of radius $\omega_n = 2$ . With $2\zeta\omega_n = a$ : $\zeta = a/4$ . Critical damping $\zeta = 1$ for $a = 4$ (poles coincident at $-2$ ); $\zeta = 0.5$ for $a = 2$ , poles $-1 \pm i\sqrt 3$ .

Poles as the damping varies — As a increases the poles go down the circle of radius 2 from plus and minus 2 i to minus 2, where they meet for a equal to 4.

Exercise 9. Design to an overshoot specification

The process $G(s) = \dfrac{1}{s(s + 5)}$ is controlled with a gain $K$ . Find $K$ for a 10 % overshoot and the corresponding settling time. Write the MATLAB and Python commands to check it.

Result: $\zeta = 0.591$ , $K = 17.9$ ; $t_s \approx 1.6$ s.

Solution

From the overshoot: $\zeta = \frac{-\ln 0.1}{\sqrt{\pi^2 + \ln^2 0.1}} = \frac{2.303}{3.896} = 0.591$ .

Characteristic equation $s^2 + 5s + K$ : $\omega_n = \sqrt K$ , $2\zeta\omega_n = 5$ , so $\sqrt K = \frac{5}{2\cdot 0.591} = 4.23$ and $K = 17.9$ . Poles $-2.5 \pm 3.42i$ .

2 % settling time: $\frac{4}{\zeta\omega_n} = \frac{4}{2.5} = 1.6$ s, fixed by the real part $-2.5$ , which a gain alone cannot move. For a faster system a zero is needed (lead network) to bring the locus further left.

Check in MATLAB: G = tf(1,[1 5 0]); stepinfo(feedback(17.9*G,1)). In Python: ct.step_info(ct.feedback(17.9*ct.tf([1],[1,5,0]),1)), which gives an overshoot of 10.0 % and a settling time of 1.42 s: the formula $4/\zeta\omega_n$ is a conservative estimate, obtained from the exponential envelope.

The gain for a 10 per cent overshoot — The dashed green line corresponds to damping 0.591 and meets the vertical branch of the locus for K equal to 17.9.

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Basic level

Exercise 1. The simplest locus

Exercise 2. A gain for an overshoot

Intermediate level

Exercise 3. Four poles: asymptotes, breakaway, stability

Exercise 4. The circular locus and maximum damping

Exercise 5. Departure angle and crossing

Exam level

Exercise 6. Stabilising an unstable process

Exercise 7. A stabilising zero

Exercise 8. The locus with respect to damping

Exercise 9. Design to an overshoot specification

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