FOUNDATION · CHAPTER 6 OF 8

Stability, and where it is lost

A system is stable when its impulse response is absolutely integrable, which puts every pole in the open left half-plane. You see how stability is lost as a root reaches the imaginary axis, and read the Routh array as that event, to find the gain at which a loop goes unstable.

Updated 13 Oct 2026

What you will be able to do

You will show that a system is bounded-input, bounded-output (BIBO) stable exactly when its impulse response is absolutely integrable, so every pole lies in the open left half-plane. You will see why same-sign coefficients are needed but not enough. You will find the gain at which a cubic's roots reach the imaginary ( $j ω$ ) axis, and read the Routh array as that same event.

Before you start

Chapter 5, How pole positions set the time response
Laplace transforms, transfer functions and the closed loop $G / (1 + G H)$
Convolution with the impulse response
Partial fractions, poles and residues
Integrating $e^{a t}$ from $0$ to $\infty$ ; $∣ e^{j ω t} ∣ = 1$
Roots of a quadratic; the cube roots of $- 1$
Complex roots in conjugate pairs; real and imaginary parts of an equation
Polynomial roots move continuously with the coefficients

Stable means an integrable impulse response

A system is bounded-input, bounded-output (BIBO) stable if a bounded input always gives a bounded output: one constant $B$ makes $∣ y (t) ∣ \leq B M$ for every input with $∣ u (t) ∣ \leq M$ . Here $u$ is the input, $y$ the output and $g (t)$ the impulse response, all real and zero for $t < 0$ . The output is the input convolved with $g$ , so the question is whether $∣ g ∣$ has a finite area.

Conventions: negative feedback, zero initial conditions, inputs applied at $t = 0$ . $G (s) = Y (s) / U (s)$ is the one-sided (unilateral) Laplace transform of $g (t)$ , $\int_{0^{-}}^{\infty} g (t) e^{- s t} d t$ ; for a signal that is zero before $t = 0$ , the two-sided (bilateral) integral gives the same result. $s = σ + j ω$ , with $j$ the imaginary unit. LHP and RHP are the open half-planes $σ < 0$ and $σ > 0$ : the $j ω$ axis belongs to neither.

OUTPUT BOUND FROM THE IMPULSE RESPONSE

∣ y (t) ∣ = \int_{0}^{t} g (τ) u (t - τ) d τ \leq M \int_{0}^{\infty} ∣ g (τ) ∣ d τ

Linear time-invariant system, $∣ u (t) ∣ \leq M$ ; $τ$ is a dummy time (s). Use $∣ \int f ∣ \leq \int ∣ f ∣$ and $∣ u ∣ \leq M$ , then extend the upper limit: $B = \int_{0}^{\infty} ∣ g ∣ d τ$ works when it is finite. If it diverges, pick any $B$ and a $t$ with $\int_{0}^{t} ∣ g ∣ d τ > B$ . The input $u (t - τ) = sgn g (τ)$ ( $\pm 1$ as $g$ is positive or negative, $0$ where $g = 0$ ) on $0 \leq τ \leq t$ has $∣ u ∣ \leq 1$ and gives $y (t) = \int_{0}^{t} ∣ g ∣ d τ > B$ . So BIBO stable means $\int_{0}^{\infty} ∣ g ∣ d τ < \infty$ .

IMPULSE RESPONSE OF A RATIONAL TRANSFER FUNCTION

g (t) = k \sum r_{k} e^{p_{k} t}, \int_{0}^{\infty} e^{p_{k} t} d t = \int_{0}^{\infty} e^{σ_{k} t} d t = - \frac{1}{σ _{k}} (σ_{k} < 0)

Strictly proper rational $G$ in lowest terms; distinct poles $p_{k} = σ_{k} + j ω_{k}$ (1/s), residues $r_{k}$ . As $∣ e^{j ω_{k} t} ∣ = 1$ , only $σ_{k}$ matters: if every $σ_{k} < 0$ , $\int_{0}^{\infty} ∣ g ∣ d t \leq \sum_{k} ∣ r_{k} ∣/∣ σ_{k} ∣$ (a repeated pole's $t^{m} e^{p_{k} t}$ is integrable too). Conversely, if that integral is finite, $∣ G (s) ∣ \leq \int_{0}^{\infty} ∣ g ∣ d t$ wherever $σ \geq 0$ , since $∣ e^{- s t} ∣ \leq 1$ , so no pole has $σ \geq 0$ . A constant $d$ in a proper $G$ adds only $d u$ to $y$ . So stable means every pole in the LHP.

WATCH OUT FOR

Simple poles on the $j ω$ axis, the rest in the LHP, make a system marginally stable: $g$ stays bounded but does not decay. That is not BIBO stable. Test: $g = sin ω_{0} t$ driven by $u = sin ω_{0} t$ gives $y = \frac{s i n ω _{0} t}{2 ω _{0}} - \frac{t}{2} cos ω_{0} t$ , which grows without limit.

CHARACTERISTIC EQUATION

T (s) = \frac{G ( s )}{1 + G ( s ) H ( s )}, 1 + G (s) H (s) = 0

Negative feedback, $T (s) = C (s) / R (s)$ with $R$ the reference and $C$ the output. With $G = N_{G} / D_{G}$ and $H = N_{H} / D_{H}$ (polynomials) the equation becomes $D_{G} D_{H} + N_{G} N_{H} = 0$ , a polynomial $a_{n} s^{n} + \dots + a_{0} = 0$ with real $a_{i}$ and $a_{n} \neq = 0$ . Its roots are the closed-loop poles, provided no factor cancels in $T$ . $T$ is then BIBO stable exactly when all of them lie in the LHP.

Signs, and the crossing of the axis

You rarely want the roots themselves, only whether any lie outside the LHP. Call a polynomial stable when all its roots are in the LHP. First, a stable polynomial is a product of factors with positive coefficients, which fixes the signs of its coefficients. Second, roots move continuously as the coefficients change, while $a_{n}$ stays nonzero. The $j ω$ axis separates the LHP from the RHP, so a root can enter the RHP only by passing through it.

FACTORS WITH POSITIVE COEFFICIENTS

(s + α), (s + β)^{2} + γ^{2} = s^{2} + 2 β s + β^{2} + γ^{2}

A real root $s = - α$ and a complex pair $s = - β \pm j γ$ , with $α, β > 0$ (1/s) and $γ$ in rad/s. Every coefficient is positive. A stable polynomial is $a_{n}$ times a product of such factors, and multiplying them only adds positive products. So every coefficient is nonzero, with the sign of $a_{n}$ . The reverse fails from degree 3 up.

WATCH OUT FOR

Coefficients all of one sign do not guarantee stability. Test: $s^{3} + s^{2} + 2 s + 8 = (s + 2) (s^{2} - s + 4)$ has all coefficients positive, and the pair $\frac{1}{2} \pm j \frac{15}{2}$ is in the RHP.

IMAGINARY-AXIS CROSSING OF A CUBIC

a_{0} - a_{2} ω^{2} ω (a_{1} - a_{3} ω^{2}) = 0 = 0

Cubic $a_{3} s^{3} + a_{2} s^{2} + a_{1} s + a_{0} = 0$ with real coefficients and $a_{3} \neq = 0$ . Put $s = j ω$ ( $ω$ in rad/s). Since $j^{2} = - 1$ and $j^{3} = - j$ : $- j a_{3} ω^{3} - a_{2} ω^{2} + j a_{1} ω + a_{0} = 0$ . The real part and the imaginary part must each vanish. A root at $ω = 0$ needs $a_{0} = 0$ ; otherwise take $ω \neq = 0$ in the second line.

CONDITION FOR A ROOT ON THE IMAGINARY AXIS

a_{2} a_{1} = a_{0} a_{3}, ω^{2} = \frac{a _{0}}{a _{2}} = \frac{a _{1}}{a _{3}}

Needs $ω \neq = 0$ and $a_{2}, a_{3} \neq = 0$ : equate the two values of $ω^{2}$ . Conversely, with all four coefficients positive and equality, $ω = a_{1} / a_{3}$ satisfies both lines, so $\pm j ω$ are roots.

Which side is stable? Hold $a_{3}, a_{2}, a_{1} > 0$ fixed and vary $a_{0} > 0$ . While the inequality keeps one direction, no root is on the axis, so the number of RHP roots stays fixed. If $a_{2} a_{1} > a_{0} a_{3}$ , lower $a_{0}$ toward $0$ (the inequality still holds). The roots approach $0$ and those of $a_{3} s^{2} + a_{2} s + a_{1}$ , which have negative real parts by the quadratic formula, as its coefficients are positive. Complex roots come in conjugate pairs, so the single root near $0$ is real; it is negative because the cubic is positive for real $s \geq 0$ . So the cubic is stable.

If $a_{2} a_{1} < a_{0} a_{3}$ , raise $a_{0}$ without limit; the inequality still holds. Write each root as $s = (a_{0} / a_{3})^{1/3} z$ . Dividing the cubic by $a_{0}$ leaves $z^{3} + 1$ plus terms that shrink to $0$ as $a_{0} \to \infty$ , so each $z$ approaches a cube root of $- 1$ : $- 1$ or $\frac{1}{2} \pm j \frac{3}{2}$ . For large $a_{0}$ , two roots therefore have positive real parts. So the cubic has two RHP roots and is unstable.

WORKED EXAMPLE

A unity-feedback loop ( $H = 1$ , negative feedback) has $G (s) = \frac{K}{s ( s + 1 ) ( s + 2 )}$ with $K > 0$ . Find the range of $K$ for BIBO stability, and the frequency of oscillation at the limit.

Solution

The closed-loop poles are the roots of $1 + G = 0$ . Multiply by the denominator to get a polynomial: $s (s + 1) (s + 2) + K = 0$ .
Expand: $s (s^{2} + 3 s + 2) = s^{3} + 3 s^{2} + 2 s$ . So $s^{3} + 3 s^{2} + 2 s + K = 0$ , with $a_{3} = 1$ , $a_{2} = 3$ , $a_{1} = 2$ , $a_{0} = K$ .
With $K > 0$ all four coefficients are positive, so the sign test passes and the cubic condition decides.
The cubic is stable exactly when $a_{2} a_{1} > a_{0} a_{3}$ : $3 \cdot 2 = 6 > K$ .
At $K = 6$ the axis roots have $ω^{2} = a_{1} / a_{3} = 2$ , matching $a_{0} / a_{2} = 6/3 = 2$ . So $ω = 2 \approx 1.41$ rad/s.
Confirm by factoring: $(s^{2} + 2) (s + 3) = s^{3} + 3 s^{2} + 2 s + 6$ , so the roots are $\pm j 2$ and $- 3$ .

Answer: Stable for $0 < K < 6$ . At $K = 6$ the poles are $\pm j 2$ and $- 3$ , so the loop oscillates at $2 \approx 1.41$ rad/s.

The Routh array and the auxiliary equation

For a cubic you now have the whole test in one line. The Routh array arranges the same test so that it works for any degree; it is stated here, not proved. For the cubic, its $s^{1}$ entry is zero exactly at the crossing, $a_{2} a_{1} = a_{0} a_{3}$ . For a higher degree, a zero in the first column shows only that the polynomial is not stable, not that a root is on the $j ω$ axis.

NEXT ROW OF THE ROUTH ARRAY

c_{j} = \frac{q _{1} p _{j + 1} - p _{1} q _{j + 1}}{q _{1}}

Rows run from $s^{n}$ down to $s^{0}$ ; the first two hold $a_{n}, a_{n - 2}, \dots$ and $a_{n - 1}, a_{n - 3}, \dots$ . Each new row $c_{1}, c_{2}, \dots$ uses the two rows above it, upper $p_{1}, p_{2}, \dots$ and lower $q_{1}, q_{2}, \dots$ , with $q_{1} \neq = 0$ ; an entry past the end of a row counts as $0$ . With $a_{n} \neq = 0$ , all roots are in the LHP exactly when the array completes with every first-column entry nonzero and of one sign. With no zero there, the sign changes down the first column count the RHP roots. Cubic rows: $a_{3}, a_{1}$ ; $a_{2}, a_{0}$ ; $(a_{2} a_{1} - a_{3} a_{0}) / a_{2}$ ; $a_{0}$ .

AUXILIARY EQUATION AS THE CROSSING

a_{3} s^{3} + a_{2} s^{2} + a_{1} s + a_{0} = (a_{2} s^{2} + a_{0}) (\frac{a _{3}}{a _{2}} s + 1) when a_{2} a_{1} = a_{0} a_{3}

All four coefficients positive. Expanding, the $s$ term is $a_{0} a_{3} / a_{2} = a_{1}$ . The roots are $\pm j a_{0} / a_{2}$ , the same $ω$ as before, and $- a_{2} / a_{3}$ in the LHP. In the array the $s^{1}$ row is all zeros, so no $s^{0}$ row follows ( $q_{1} = 0$ ); the row above, read as $a_{2} s^{2} + a_{0}$ , is the auxiliary polynomial. For any degree, a zero row gives an auxiliary polynomial that divides the characteristic one, with roots symmetric about the origin: on the axis, or with one in the RHP. Either way the polynomial is not stable.

WORKED EXAMPLE

A unity-feedback loop has $G (s) = \frac{K}{( s + 1 ) ( s + 3 ) ( s + 5 )}$ , $H = 1$ , $K > 0$ . Use the Routh array to find the range of $K$ for stability, and the auxiliary equation to find the oscillation frequency at the limit.

Solution

Expand the denominator: $(s + 1) (s + 3) (s + 5) = s^{3} + 9 s^{2} + 23 s + 15$ . So the characteristic equation $1 + G = 0$ becomes $s^{3} + 9 s^{2} + 23 s + (15 + K) = 0$ .
Rows $s^{3}$ and $s^{2}$ are $1, 23$ and $9, 15 + K$ . The $s^{1}$ row, with $p = (1, 23)$ and $q = (9, 15 + K)$ , is $c_{1} = (9 \cdot 23 - 1 \cdot (15 + K)) /9 = (192 - K) /9$ .
The $s^{0}$ row, from the $s^{2}$ and $s^{1}$ rows with missing entries taken as $0$ , is $15 + K$ .
The first column is $1, 9, (192 - K) /9, 15 + K$ . For $K > 0$ the first, second and last entries are positive, so there is no sign change exactly when $192 - K > 0$ .
At $K = 192$ the $s^{1}$ row is zero. The auxiliary equation from the $s^{2}$ row is $9 s^{2} + (15 + 192) = 9 s^{2} + 207 = 0$ , so $s^{2} = - 23$ .
So $ω = 23 \approx 4.80$ rad/s. The third root is $- a_{2} / a_{3} = - 9$ . Check: $(9 s^{2} + 207) (s /9 + 1) = s^{3} + 9 s^{2} + 23 s + 207$ , and $207 = 15 + 192$ .

Answer: Stable for $0 < K < 192$ . At $K = 192$ the poles are $\pm j 23 \approx \pm j 4.80$ and $- 9$ , so the loop oscillates at $4.80$ rad/s.

Check yourself

CHECK YOURSELF 1

Which impulse response, zero for $t < 0$ , belongs to a BIBO stable system?

(A)
$g (t) = 1$ for $t \geq 0$ , the unit step
(B)
$g (t) = t e^{- t}$ for $t \geq 0$
(C)
$g (t) = e^{t}$ for $t \geq 0$
(D)
$g (t) = sin 3 t$ for $t \geq 0$

Answer and solution

Answer: (B)

BIBO stability needs $\int_{0}^{\infty} ∣ g ∣ d t < \infty$ .
A has $∣ g ∣ = 1$ and C grows, so both integrals diverge. D has $∣ sin 3 t ∣$ with no decay, so it diverges too.
B: $\int_{0}^{\infty} t e^{- t} d t = 1$ , finite. This is a double pole at $- 1$ , in the LHP.

CHECK YOURSELF 2

A unity negative-feedback loop has $G (s) = \frac{K}{s ( s + 3 ) ( s + 4 )}$ with $K > 0$ . The closed loop is stable for $0 < K < K_{m a x}$ . Find $K_{m a x}$ .

Answer and solution

Answer: $84$ (accepted range $\pm 0.5$ )

Characteristic equation: $s (s + 3) (s + 4) + K = s^{3} + 7 s^{2} + 12 s + K = 0$ .
Here $a_{3} = 1$ , $a_{2} = 7$ , $a_{1} = 12$ , $a_{0} = K$ , all positive.
Stable when $a_{2} a_{1} > a_{0} a_{3}$ : $7 \cdot 12 = 84 > K$ . So $K_{m a x} = 84$ .

CHECK YOURSELF 3

Use the Routh array to find how many roots of $s^{3} + 2 s^{2} + s + 6$ lie in the RHP.

Answer and solution

Answer: $2$

Rows $s^{3}$ and $s^{2}$ : $1, 1$ and $2, 6$ .
Row $s^{1}$ : $(2 \cdot 1 - 1 \cdot 6) /2 = - 2$ . Row $s^{0}$ : $6$ .
The first column $1, 2, - 2, 6$ has no zero and changes sign twice, so 2 roots lie in the RHP.

CHECK YOURSELF 4One or more options are correct

Which of these characteristic polynomials have at least one root outside the LHP (on the $j ω$ axis or in the RHP)?

(A)
$s^{3} + 2 s^{2} - s + 5$
(B)
$s^{3} + 4 s^{2} + 5$
(C)
$s^{3} + 2 s^{2} + 3 s + 10$
(D)
$s^{3} + 4 s^{2} + 5 s + 2$

Answer and solution

Answer: (A), (B), (C)

A has a negative coefficient among positive ones. A stable polynomial has all coefficients of one sign, so A is not stable.
B has no $s$ term. A stable polynomial has every coefficient nonzero, so B is not stable.
C has positive coefficients but $a_{2} a_{1} = 6 < 10 = a_{0} a_{3}$ , so it is unstable.
D has $a_{2} a_{1} = 20 > 2 = a_{0} a_{3}$ with positive coefficients, so it is stable. Indeed it is $(s + 1)^{2} (s + 2)$ .

Stability, and where it is lost

What you will be able to do

Before you start

Stable means an integrable impulse response

Signs, and the crossing of the axis

The Routh array and the auxiliary equation

Check yourself

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