Stability: Routh–Hurwitz & root locus: practice problems with step-by-step solutions

Free Stability: Routh–Hurwitz & root locus practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: Routh array and pole counts, range of K for stability, root locus.

Free practice problemEasy
s⁴16−2s³416s²??s¹?s⁰?

The characteristic polynomial of a feedback system is Q(s) = s⁴ + 4s³ + 6s² + 16s − 2. Build the Routh array and find how many roots N of Q(s) lie in the right half-plane.

Q(s) = s⁴ + 4s³ + 6s² + 16s − 2

Exact answer: a whole number, no rounding.

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What you practise

  1. EasyRouth array and pole counts
  2. Mediumrange of K for stability
  3. Hardroot locus

How to solve Stability: Routh–Hurwitz & root locus problems

  1. Write the closed-loop characteristic polynomial: the denominator of 1 + K·G(s) = 0, multiplied out.
  2. Put the coefficients alternately into the first two rows of the Routh array and build the remaining rows.
  3. Count the sign changes in the first column: each one is a pole in the right half-plane. With K in the array, demand that every first-column entry stays positive.
  4. For a root locus, mark the poles and zeros, the real-axis segments, the asymptotes from the centroid and the break points where dK/ds = 0. Use K = 1/|G(s)| for the gain at a point.

Common mistakes

  • Concluding that a system is stable because all coefficients are positive; for a cubic or higher order that is not enough.
  • Forgetting the constant term of the plant when K is added: with three poles away from the origin the s⁰ entry is abc + K.
  • Taking the midpoint of two poles as the break-away point when a third pole or a zero is nearby.
  • Using pole locations without their signs in the centroid formula.

Key formulas

Stability
all roots of the characteristic polynomial in the left half-plane
Routh array, s¹ entry of a cubic
b1 = (a2a1 − a3a0)/a2
Right half-plane roots
number of sign changes in the first column
Cubic s³ + a₂s² + a₁s + a₀
stable when all a > 0 and a₂·a₁ > a₀
K/(s(s + a)(s + b))
Kmax = ab(a + b), ω = √(ab)
Characteristic equation
1 + K·G(s) = 0
Asymptotes
σa = (Σp − Σz)/(n − m), θ = (2q + 1)·180°/(n − m)
Break-away and break-in points
dK/ds = 0
Magnitude condition
K = 1/|G(s)|
Angle of departure
θd = 180° − Σ∠(from other poles) + Σ∠(from zeros)

Learn Stability: Routh–Hurwitz & root locus from scratch

Before this: Laplace transforms, Block diagrams

1What stability means

A system is stable when every bounded input gives a bounded output. After a disturbance a stable system settles down; an unstable one runs away or oscillates with growing amplitude.

The behaviour is set by the poles of the closed loop: the roots of its characteristic polynomial. A pole at s = p adds a term ept to the response.

ReIm−2−1 + j2−1 − j2+1Re < 0Re > 0
  • A pole with a negative real part (left half-plane) gives a term that dies out.
  • A pole with a positive real part (right half-plane) gives a term that grows without limit.
  • So the system is stable only when all poles lie in the left half-plane.
Self-check A closed loop has poles at −3, −1 + j2, −1 − j2 and +0.5. Is it stable?
  1. No, because complex poles always make a system unstable
  2. No, the pole at +0.5 makes it unstable
  3. Yes, because the complex poles have negative real parts
  4. Yes, because three of the four poles are in the left half-plane

Answer: No, the pole at +0.5 makes it unstable. One pole in the right half-plane is enough: its term e0.5t grows. Complex poles in the left half-plane are fine; they give a decaying oscillation.

2The Routh array

For a cubic or higher order it is hard to find the roots by hand. The Routh–Hurwitz criterion tells you how many roots lie in the right half-plane without solving for them.

Write the coefficients alternately into the first two rows. For s³ + a₂s² + a₁s + a₀:

s³: 1, a1 s²: a2, a0

Each new row is built from the two rows above it. For the cubic, the s¹ entry is:

b1 = (a2·a1 − 1·a0)/a2, s⁰: a0

The number of sign changes in the first column equals the number of roots in the right half-plane.

Example: s³ + 2s² + 3s + 10. The first column is 1, 2, (2·3 − 10)/2 = −2, 10. The signs go +, +, −, +: two changes, so two poles lie in the right half-plane.

Self-check For s³ + 4s² + 5s + 2, what is the first entry of the s¹ row?
  1. 18
  2. 4.5
  3. 5
  4. −4.5

Answer: 4.5. b₁ = (4·5 − 1·2)/4 = 18/4 = 4.5. Forgetting to divide by 4 gives 18; swapping the products gives −4.5.

3The method step by step

  1. Write the closed-loop characteristic equation 1 + K·G(s) = 0 and multiply it out to a polynomial.
  2. Fill in the first two rows of the Routh array with the coefficients.
  3. Build the remaining rows; keep K as a symbol if it appears.
  4. Count the sign changes in the first column, or demand that every first-column entry is positive to find the allowed range of K.
  5. At the limiting gain the s¹ row is zero. The s² row then gives the auxiliary equation a·s² + b = 0, whose roots ±jω show the frequency of oscillation.

Quick test for a cubic s³ + a₂s² + a₁s + a₀: it is stable exactly when all coefficients are positive and a₂·a₁ > a₀.

4A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

The loop below has L(s) = K/(s(s + 2)(s + 4)). For which gains K is the closed loop stable, and at what frequency does it oscillate at the limit?

1

Look at the loop

R+K1/(s(s + 2)(s + 4))C−

The characteristic equation is 1 + L(s) = 0, so s(s + 2)(s + 4) + K = 0.

2

Multiply out

s³ + 6s² + 8s + K = 0
3

Build the Routh array

s³: 1, 8 s²: 6, K s¹: (48 − K)/6 s⁰: K

The s¹ entry is (6·8 − 1·K)/6.

4

Keep the first column positive

From the s⁰ row: K > 0. From the s¹ row: 48 − K > 0, so K < 48.

5

Frequency at the limit

At K = 48 the s¹ row is zero. The s² row gives 6s² + 48 = 0, so s² = −8.

ω = √8 = 2.83 rad/s
6

Check the answer

At K = 48 the polynomial is s³ + 6s² + 8s + 48 = (s + 6)(s² + 8). Its roots are −6 and ±j2.83, as expected.

Answer

The loop is stable for 0 < K < 48; at K = 48 it oscillates at ω = 2.83 rad/s.

5The root locus

The root locus shows how the closed-loop poles move as K goes from 0 to ∞. It starts at the open-loop poles (×) and ends at the open-loop zeros (○) or at infinity.

  • A point on the real axis is on the locus when the number of real poles and zeros to its right is odd.
  • The n − m branches that go to infinity follow asymptotes at the angles (2q + 1)·180°/(n − m).
  • The asymptotes start from the centroid σa = (Σ poles − Σ zeros)/(n − m).
  • Branches leave or join the real axis where dK/ds = 0, with K = −1/G(s) on the axis.
  • The gain at any point s on the locus is K = 1/|G(s)|, the product of the distances to the poles divided by that to the zeros.

For L(s) = K/(s(s + 2)(s + 4)) the centroid is (0 − 2 − 4)/3 = −2, and the asymptotes leave at 60°, 180° and 300°.

ReIm0−2−460°

The break-away point between 0 and −2 follows from dK/ds = 0: 3s² + 12s + 8 = 0, so s = −0.845.

Self-check For L(s) = K/((s + 1)(s + 5)), where do the two branches leave the real axis?
  1. s = −1
  2. s = −2.5
  3. s = −3
  4. s = −6

Answer: s = −3. With two real poles and no zeros, K = −(s + 1)(s + 5) is largest halfway between them: dK/ds = −(2s + 6) = 0 gives s = −3.

6Common mistakes

  • Calling a system stable because all its coefficients are positive. For a cubic or higher order you also need the Routh test.
  • Forgetting the constant term of the plant: for K/((s + a)(s + b)(s + c)) the s⁰ coefficient is abc + K, not K.
  • Taking the midpoint of two poles as the break-away point when a third pole or a zero pulls the locus.
  • Dropping the signs in the centroid: a factor (s + 4) is a pole at s = −4.
Self-check Is s³ + s² + 2s + 8 = 0 a stable characteristic equation?
  1. No: a cubic is never stable
  2. No: a₂·a₁ = 2 is less than a₀ = 8
  3. Yes: all coefficients are positive
  4. Yes: the coefficients do not change sign

Answer: No: a₂·a₁ = 2 is less than a₀ = 8. The s¹ entry is (1·2 − 8)/1 = −6, so the first column 1, 1, −6, 8 changes sign twice: two poles lie in the right half-plane. Positive coefficients are not enough.

7How exam questions look

  • Easy: the Routh array and pole counts. You count the right-half-plane poles of a cubic or quartic, find one entry of the array, find a closed-loop pole for a first-order plant, or the gain that puts it at a given point.
  • Medium: the range of K. You find the largest stable gain, the frequency of oscillation at that gain, the smallest gain that stabilises an unstable plant, or how far the gain can rise.
  • Hard: the root locus. You find the centroid of the asymptotes, the imaginary-axis crossing, break-away and break-in points, the angle of departure from a complex pole, or the gain at a given point or damping ratio.

For a gain at a damping ratio, write the characteristic polynomial and compare it with s² + 2ζωns + ωn².

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
R(s)+K1s + 10C(s)−

A proportional gain K controls the plant G(s) = 1/(s + 10) in a unity-feedback loop. Find the gain K that places the closed-loop pole at s = −10.5.

G(s) = 1/(s + 10)p = −10.5

Find K

Show hints
  1. The closed-loop pole is the root of s + 10 + 1K = 0.
  2. Put s = −10.5 into that equation.
  3. 1K = 10.5 − 10.
Show the answer and the step-by-step solution

Routh–Hurwitz criterion: a linear system is stable when every root of its characteristic polynomial lies in the left half of the s-plane. Write the coefficients alternately into the first two rows of the Routh array. Each entry of a new row comes from the two rows above it; for a cubic a₃s³ + a₂s² + a₁s + a₀ the s¹ entry is b₁ = (a₂·a₁ − a₃·a₀)/a₂. The number of sign changes in the first column equals the number of roots in the right half-plane. With a gain K in the coefficients, every first-column entry must stay positive: the gain that makes the s¹ entry zero is the stability limit, and the s² row then gives the frequency of the sustained oscillation.

Characteristic equation

s + 10 + 1K = 0

Required gain

−10.5 + 10 + 1K = 0 → K = (10.5 − 10)/1 = 0.5

Answer

K = 0.5
Example 2Medium
R(s)+Ks + 2(s − 4)(s + 1)C(s)−

The plant G(s) = (s + 2)/((s − 4)(s + 1)) has an unstable pole at s = 4. A proportional gain K closes a unity-feedback loop around it. Find the smallest gain Kmin that makes the closed loop stable.

G(s) = (s + 2)/((s − 4)(s + 1))

Find Kmin

Show hints
  1. The characteristic polynomial is (s − 4)(s + 1) + K(s + 2).
  2. s² + (-3 + K)s + (2K − 4).
  3. A second-order polynomial is stable when both coefficients are positive.
Show the answer and the step-by-step solution

Routh–Hurwitz criterion: a linear system is stable when every root of its characteristic polynomial lies in the left half of the s-plane. Write the coefficients alternately into the first two rows of the Routh array. Each entry of a new row comes from the two rows above it; for a cubic a₃s³ + a₂s² + a₁s + a₀ the s¹ entry is b₁ = (a₂·a₁ − a₃·a₀)/a₂. The number of sign changes in the first column equals the number of roots in the right half-plane. With a gain K in the coefficients, every first-column entry must stay positive: the gain that makes the s¹ entry zero is the stability limit, and the s² row then gives the frequency of the sustained oscillation.

Characteristic polynomial

(s − 4)(s + 1) + K(s + 2) = s² + (-3 + K)s + (2K − 4)

Conditions

For s² + c₁s + c₀ all roots lie in the left half-plane exactly when c₁ > 0 and c₀ > 0.

-3 + K > 0 → K > 3
2K − 4 > 0 → K > 2
Kmin = 3

Answer

Kmin = 3
Example 3Hard
ReIm−2−4−6−8−10−12j2

The root locus of L(s) = K/(s(s + 4)(s + 10)) leaves the real axis between s = 0 and s = −4. Find the break-away point sb.

L(s) = K/(s(s + 4)(s + 10))

Find sb

Show hints
  1. On the real axis K = −s(s + 4)(s + 10) = −(s³ + 14s² + 40s).
  2. Set dK/ds = 0: 3s² + 28s + 40 = 0.
  3. Take the root that lies between 0 and −4.
Show the answer and the step-by-step solution

Root locus: as the gain K goes from 0 to ∞, the closed-loop poles, the roots of 1 + K·G(s) = 0, move from the open-loop poles to the open-loop zeros or to infinity. A point on the real axis belongs to the locus when the number of real poles and zeros to its right is odd. The n − m branches that go to infinity follow asymptotes at the angles (2q + 1)·180°/(n − m), which start from the centroid σₐ = (Σ poles − Σ zeros)/(n − m). Branches leave or join the real axis where dK/ds = 0. The angle condition ∠G(s) = 180° fixes the shape, and the magnitude condition K = 1/|G(s)| gives the gain at any point of the locus.

Gain along the real axis

K(s) = −s(s + 4)(s + 10) = −(s³ + 14s² + 40s)

Stationary points

dK/ds = −(3s² + 28s + 40) = 0
s = (−28 ± √(784 − 480))/6 = −1.761, −7.573

Pick the point on the locus

The locus lies on the real axis between 0 and −4; the other root is not on the locus.

sb = −1.761

Answer

sb = −1.761

Questions and answers

What is Stability: Routh–Hurwitz & root locus?

A system is stable when every bounded input gives a bounded output. After a disturbance a stable system settles down; an unstable one runs away or oscillates with growing amplitude.

The behaviour is set by the poles of the closed loop: the roots of its characteristic polynomial. A pole at s = p adds a term ept to the response.

How do I solve Stability: Routh–Hurwitz & root locus problems step by step?
  1. Write the closed-loop characteristic polynomial: the denominator of 1 + K·G(s) = 0, multiplied out.
  2. Put the coefficients alternately into the first two rows of the Routh array and build the remaining rows.
  3. Count the sign changes in the first column: each one is a pole in the right half-plane. With K in the array, demand that every first-column entry stays positive.
  4. For a root locus, mark the poles and zeros, the real-axis segments, the asymptotes from the centroid and the break points where dK/ds = 0. Use K = 1/|G(s)| for the gain at a point.
What are common mistakes in Stability: Routh–Hurwitz & root locus?
  • Concluding that a system is stable because all coefficients are positive; for a cubic or higher order that is not enough.
  • Forgetting the constant term of the plant when K is added: with three poles away from the origin the s⁰ entry is abc + K.
  • Taking the midpoint of two poles as the break-away point when a third pole or a zero is nearby.
  • Using pole locations without their signs in the centroid formula.
What do the Stability: Routh–Hurwitz & root locus exercises cover?
  • Easy: Routh array and pole counts
  • Medium: range of K for stability
  • Hard: root locus

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

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Stability: Routh–Hurwitz & root locus worksheet with answers
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