Lecture 10: Stability Analysis: the Routh-Hurwitz Approach

A useful feedback loop needs a good transient response, stability, and small steady-state errors. The transient-response calculations in the preceding lectures assume that the motion eventually settles. We now examine that assumption directly. The Routh-Hurwitz criterion lets us count right-half-plane poles from a characteristic polynomial and determine a stable gain range without solving for every root.

Natural response and closed-loop poles

The total response of a linear system separates into a forced response and a natural response:

The natural response is the motion caused by initial conditions when the external input is zero. Its modes are determined by the poles of the closed-loop dynamics, or equivalently by the roots of the closed-loop characteristic polynomial $D_{CL}(s)$.

• Asymptotically stable: every natural mode decays. All poles lie in the open left half-plane (LHP).
• Marginally stable: every natural mode remains bounded, but some do not decay. In the scalar modal description, imaginary-axis poles are simple and all other poles lie in the LHP.
• Unstable: some natural mode grows without bound. A right-half-plane (RHP) pole produces exponential growth; a repeated imaginary-axis pole in a scalar mode produces a growing polynomial envelope.

An unstable oscillation need not tend to positive infinity: its amplitude can grow while its sign alternates. Repeated LHP poles, by contrast, still yield decaying terms such as $t^k e^{-at}$ with $a\gt 0$.

Use the characteristic polynomial of the full closed-loop dynamics. A pole-zero cancellation in a reference-to-output transfer function can hide an internal unstable mode. In state space, imaginary-axis eigenvalues must be semisimple, meaning their Jordan blocks have size one; repeated eigenvalues alone do not imply instability.

Why avoid computing every pole?

Computers are indispensable, but a root calculation can be sensitive when roots are repeated or nearly repeated. Consider

Every exact root is $-1$. If we expand the polynomial and pass its coefficients to a floating-point root finder, the computed roots can form a cluster around $-1$. The cluster reflects numerical sensitivity, not different physical poles.

# Author: Xu Chen, University of Washington
from math import comb
import numpy as np

for n in (12, 20):
    a = [comb(n, k) for k in range(n + 1)]
    p = np.roots(a)
    print(n, np.max(np.abs(p + 1)))

The sensitivity also has a simple analytical example. If a small constant perturbation changes the polynomial to

then every perturbed root satisfies $|s+1|=|\delta|^{1/12}$. Even $\delta=10^{-12}$ gives a displacement of $0.1$.

Python experiment: vary the polynomial order

Use the live repeated-pole Python demo and the repository setup instructions. From the repository root, run:

python demos/repeated_poles_app.py

Open the printed local URL and vary $n$ from 2 to 40. Compare the exact repeated root with the computed cluster. The browser runs Python/NumPy in a worker; its first load requires internet access to obtain Pyodide and NumPy. To compute directly with local Python instead, run python demos/repeated_poles_app.py --n 20.

Routh avoids explicit root finding when we need a pole count or a symbolic gain condition. It does not make floating-point arithmetic infallible: numerical Routh calculations also require care near zero pivots and stability boundaries.

Quick checks before building an array

Assume a real polynomial with positive leading coefficient. Strict LHP stability requires every coefficient to be positive.

1. Mixed nonzero coefficient signs rule out strict stability. For example, $s^2-4s+4=(s-2)^2$ has RHP poles at $+2$.
2. A missing power rules out strict stability. The examples $s^2-1$ and $s^2+4$ have roots $\pm1$ and $\pm2j$, respectively. The former is unstable; the latter has bounded, nondecaying natural modes.
3. Positive coefficients are necessary, not sufficient. In particular,

has roots $-3$ and $1\pm3j$, so it is unstable despite having positive coefficients.

The quadrantal example is

with roots $\pm1\pm2j$. An even polynomial has roots symmetric about the origin; real coefficients also give conjugate pairs. A missing coefficient alone does not guarantee this symmetry. For example, $s^3+s+1$ has a missing power but is not an even polynomial.

Constructing the Routh array

The Routh criterion counts RHP roots through sign changes down the first column. Start with a quartic:

Place alternating coefficients in the first two rows, padding with zeros as needed:

Each new coefficient uses two rows above it. Multiply the lower-left entry by the upper-right entry, subtract the upper-left entry times the lower-right entry, and divide by the lower-left pivot:

These formulas assume nonzero pivots. A positive scaling of an entire row does not change its contribution to the sign count. Resolve zero pivots and entire zero rows before drawing a conclusion. No RHP poles and no imaginary-axis poles means asymptotic stability; a zero sign-change count alone does not exclude marginal stability.

Worked quartic: table first, then arithmetic

For $D(s)=s^4+6s^3+11s^2+6s+2$, the completed array is

The three calculations are

The first column is $1,6,10,24/5,2$. Every entry is positive, and no exceptional row occurs, so all four poles lie in the LHP.

Stable open loop does not guarantee stable feedback

Consider unity negative feedback around

The open-loop poles $-2,-3,-5$ are stable. To analyze the feedback loop, first form its transfer function:

Expanding the denominator gives

The Routh array and the calculation of its negative entry are

The signs $+,+,-,+$ change twice. The closed loop has two RHP poles and is unstable. This is why stability must be checked after closing the loop.

Special case 1: a zero first entry

A zero first entry with other nonzero entries in the row prevents the next division. Replace only that first entry by $\epsilon\gt 0$, continue symbolically, and take $\epsilon\to0^+$ after finding the signs.

For

the first two rows are $[1,3,5]$ and $[2,6,3]$. The next row is $[0,7/2,0]$: this is not an entire row of zeros. Replacing the first zero gives

Keep the small pivot symbolic until its sign is known. The limits are

The first-column signs are $+,+,+,-,+,+$. Two sign changes mean two RHP poles, so the system is unstable.

Special case 2: an entire row of zeros

An entire zero row signals an origin-symmetric factor. Its roots may be real pairs, imaginary pairs, or quadrantal groups; the zero row does not by itself prove marginal stability. Use the row immediately above it to form an auxiliary polynomial $A(s)$, differentiate, and replace the zero row with the coefficients of $A'(s)$.

Consider

The zero row appears at $s^3$:

Both computed entries vanish:

The row above gives

Replace the $s^3$ row by $[28,84,0]$. Divide it by 28 and the $s^4$ row by 7, both positive scalings. The completed array is

Here $s^2$ starts with $(1\cdot6-1\cdot3)/1=3$, and $s^1$ starts with $(3\cdot3-1\cdot8)/3=1/3$. There are no RHP poles. To locate the symmetric roots, factor

The poles are $\pm j\sqrt2$, $\pm2j$, and $-7$. The imaginary-axis poles are simple, so the natural response is bounded but does not decay: the system is marginally stable.

Finding a stabilizing gain range

Now use the same unity-negative-feedback structure with

The closed-loop transfer function is

Its Routh array is

Every first-column entry is positive precisely when $K\gt 0$ and $1386-K\gt 0$. Thus

is the asymptotically stable range. For $K\gt 1386$, two sign changes give two RHP poles. For $K\lt 0$, one sign change gives one RHP pole. At $K=0$, the full characteristic polynomial has a pole at zero; the zero reference-to-output transfer function obtained by setting the numerator to zero must not be used to infer internal asymptotic stability.

The upper boundary and oscillation frequency

At $K=1386$, the $s^1$ row vanishes. Keep the array in view:

The auxiliary polynomial and replacement row are

The final row is $[1386,0]$. Factoring the characteristic polynomial gives

The simple poles $\pm j\sqrt{77}$ produce sustained oscillations. Their imaginary part gives the oscillation frequency:

For reference, with $\omega=\sqrt{77}$ the exact unit-step response is

Python experiment: vary the feedback gain

Run the gain laboratory from the repository root:

python demos/routh_gain_app.py

Compare $K=1000$, $1386$, and $1800$. Read the first-column signs, locate the poles, and explain the decaying, sustained, or growing response. Vary the displayed time window to distinguish a slow decay from a sustained oscillation.

This app uses Python/SciPy unit-step simulations with zero initial conditions. Its slider selects a precomputed gain grid from $-100$ to $2400$ in steps of 20, plus the exact boundary $1386$; it does not run Python again when the slider moves. The exported app works offline. These are illustrative mathematical simulations, not measured experimental data.

Summary

Stability belongs to the closed-loop dynamics. The Routh first column counts RHP poles, while exceptional rows require an epsilon limit or an auxiliary polynomial. A zero RHP count must be combined with an imaginary-axis check before declaring asymptotic stability. In a parameterized problem, the first-column inequalities give the stable gain range, and the auxiliary polynomial at a boundary gives the oscillation frequency.

For a state-space model, apply the same test to $\det(sI-A)$. Numerical root finding remains useful, but repeated poles and stability boundaries are reasons to check its output against symbolic structure.

References

• N. S. Nise, Control Systems Engineering, 6th ed., §6.1–6.5.
• X. Chen and M. Tomizuka, Introduction to Modern Controls, with Illustrations in MATLAB and Python, §8.2.1–8.2.2.
• Xu Chen, Introduction to Controls with Python: course code and setup.