The Concept of Stability

Chapter 7 read a time constant off a decaying exponential. Chapter 8 read overshoot and settling time off a decaying oscillation. Chapter 9 applied the final-value theorem and attached a warning to it: the result is meaningful only if the closed loop is stable. Every one of those results rests on the same silent assumption — that the part of the response generated by the system itself eventually disappears, leaving only the part forced by the input. This chapter examines that assumption and finds the exact condition under which it holds.

The everyday picture is a ball resting on a surface. Nudge it, and one of three things happens. In a valley it rolls back and settles at the bottom: the disturbance dies away, and the equilibrium is stable. On a hilltop the smallest nudge sends it away and it never returns: the equilibrium is unstable. On level ground it moves to a new spot and stays there, neither returning nor running away: the equilibrium is neutral, or in our vocabulary, marginally stable.

The analogy is worth keeping because it isolates the right idea: stability is about what the system does on its own after being disturbed, not about how large the disturbance was. A stable valley returns the ball whether the nudge was gentle or firm; an unstable hilltop loses it either way. For a linear time-invariant system the same is true — stability is a property of the system's poles alone, entirely independent of the input applied to it.

The engineering definition of stability asks a practical question: if we promise never to drive the system beyond some finite limit, can the system promise never to produce an output beyond some finite limit? A system that keeps that promise is called BIBO stable — bounded-input, bounded-output.

To turn this into a condition on the system, recall from Chapter 3 that the output of a linear time-invariant system is the convolution of the input with the impulse response \(h(t)\):

Suppose the input is bounded, so \(|r(t)| \le M\) for all \(t\) and some finite \(M\). Take the magnitude of both sides and move it inside the integral, where it can only grow:

So if the area under \(|h(t)|\) is finite, the output is bounded by \(M\) times that area, and the system is BIBO stable. The condition is also necessary: if the area is infinite, one can construct a bounded input that drives the output arbitrarily large — the input \(r(t-\tau)=\operatorname{sgn}\{h(\tau)\}\), which never exceeds unity, makes every contribution to the integral positive and reproduces \(\int|h|\) exactly. Absolute integrability of \(h(t)\) is therefore not merely sufficient but equivalent to stability.

The integral condition is exact but awkward to apply. It becomes a simple geometric statement once we write \(h(t)\) out. Let the transfer function be \(T(s)=N(s)/D(s)\) with distinct poles \(p_1,\dots,p_n\), the roots of \(D(s)\). Partial fractions give

Each pole contributes one mode \(e^{p_i t}\), and the whole natural response is a sum of them. Writing a pole as \(p_i=\sigma_i + j\omega_i\) separates that mode into an envelope and an oscillation:

The imaginary part \(\omega_i\) only decides how fast the mode oscillates; it has no effect on size. Everything rests on the real part \(\sigma_i\). If \(\sigma_i < 0\) the envelope \(e^{\sigma_i t}\) decays to zero and that mode contributes a finite area. If \(\sigma_i > 0\) the envelope grows without bound. If \(\sigma_i = 0\) the envelope is constant, and the mode neither dies nor grows — it persists as a constant term (for \(\omega_i=0\)) or an undamped sinusoid (for \(\omega_i\neq 0\)), and the area under it is infinite.

Repeated poles need one extra remark. A pole of multiplicity \(m\) contributes terms \(t^{k}e^{p_i t}\) for \(k=0,\dots,m-1\). In the left half-plane the exponential defeats any power of \(t\), so repetition changes nothing — the mode still decays. On the imaginary axis there is no decaying exponential to do that work, so a repeated pole at \(s=\pm j\omega\) produces \(t\sin\omega t\), an oscillation whose amplitude grows linearly. Repeated poles on the imaginary axis are unstable, even though single ones are not.

The pole map partitions systems into three classes, and control engineering uses a distinct name for each.

A system is absolutely stable — usually just stable — when all poles are in the open left half-plane. Its natural response decays to zero, the final-value theorem of Chapter 9 applies, and every bounded input gives a bounded output.

A system is unstable when at least one pole lies in the right half-plane, or when a repeated pole sits on the imaginary axis. Some component of the natural response grows without limit. In hardware the growth stops only when something saturates, bends, or burns.

A system is marginally stable when it has simple poles on the imaginary axis and no poles at all in the right half-plane. This is the borderline case, and it deserves care. The natural response neither decays nor grows: an undamped oscillation at \(\omega\) persists for ever at whatever amplitude the initial disturbance set. Such a system is not BIBO stable, because \(\int|\sin\omega t|\,dt\) diverges; feed it a sinusoid at exactly \(\omega\) and the output grows linearly in time — plain resonance.

The poles that matter are those of the closed-loop transfer function. For the standard loop of Chapter 4, with forward path \(G(s)\) and feedback path \(H(s)\),

Its poles are the values of \(s\) that make the denominator vanish. Writing \(G(s)H(s)=N(s)/D(s)\) as a ratio of polynomials and clearing the fraction gives the polynomial whose roots we must locate.

Two consequences deserve emphasis, because both contradict a natural first guess.

First, the closed-loop poles are not the open-loop poles. Feedback moves them. Take \(G(s)=K/(s+1)\) with unity feedback: the open-loop pole sits at \(s=-1\), but the characteristic equation \(s+1+K=0\) puts the closed-loop pole at \(s=-(1+K)\), which moves further left as gain rises. Knowing where a plant's poles are tells you nothing directly about whether the loop around it is stable.

Second, feedback can create instability where none existed, and can remove instability that did. The plant \(G(s)=K/\big(s(s+1)(s+5)\big)\) is not itself unstable — its poles are at \(0,-1,-5\), all in the closed left half-plane. Yet closing the loop gives \(s^3+6s^2+5s+K=0\), and Example 3 below shows that its roots cross into the right half-plane once \(K\) exceeds 30. Conversely, the unstable plant \(G(s)=K/(s-2)\) closes to \(s-2+K=0\), a stable loop for any \(K>2\). Stabilisation by feedback is precisely this effect used deliberately.

Suppose the characteristic polynomial is

Root-finding for \(n\ge 3\) is unpleasant by hand and impossible symbolically for \(n\ge 5\). Fortunately a great deal can be read straight from the coefficients. Suppose all the roots are indeed in the left half-plane. Then \(D(s)\) factors into real first-order terms \((s+\sigma)\) with \(\sigma>0\), and real quadratic terms \((s^2+2\zeta\omega_n s+\omega_n^2)\) with \(\zeta>0\) and \(\omega_n>0\) coming from complex pairs. Every one of those factors has strictly positive coefficients, and multiplying polynomials with positive coefficients can only produce positive coefficients — no cancellation is possible, because there is nothing negative to cancel with.

The condition is necessary but not sufficient, and it is important to see a case where it fails. Consider

Every coefficient of \(D(s)\) is positive, so the test raises no objection. But the quadratic factor has roots at \(s = 0.5 \pm j1.936\), squarely in the right half-plane. The positive coefficient of \(s^2\) in \(D(s)\) came from the real root at \(-2\) outweighing the unstable pair, and the sum hides what the factors reveal. Passing the coefficient test buys us nothing more than permission to keep looking.

There is one more thing we can extract by hand: the exact boundary between stable and unstable. A root sits on the imaginary axis when \(s=j\omega\) satisfies \(D(s)=0\). Substituting and separating real from imaginary parts gives two real equations, and where they can be satisfied simultaneously we have both the critical parameter value and the frequency at which the loop would oscillate.

At \(K=30\) the loop carries a pole pair exactly at \(s=\pm j\sqrt{5}\) and oscillates at \(\sqrt5 \approx 2.24\) rad/s for ever. Below that gain the pair sits in the left half-plane; above it, in the right. So the loop is stable for \(0 < K < 30\).

This substitution works, but only because we were told what to vary and because there was a single unknown. For a fixed polynomial of degree five with no parameter, it tells us nothing; for two parameters it becomes a mess. What we want is a test that takes any characteristic polynomial and reports how many of its roots lie in the right half-plane, without root-finding and without guesswork. That test is the subject of Chapter 11.

Knowing that all poles are in the left half-plane settles a yes-or-no question. Design needs more. A pole at \(s=-0.01\) is technically stable, but its mode takes four hundred seconds to fade; a loop built on it is stable and useless. The useful quantity is how far into the left half-plane the poles have got, and this is called relative stability.

The natural measure is the real part of the pole closest to the imaginary axis. If every pole satisfies \(\operatorname{Re}\{p_i\} \le -\sigma_1\), then every mode decays at least as fast as \(e^{-\sigma_1 t}\), and the whole natural response is trapped inside that envelope. Chapter 8's settling-time estimate follows immediately.

The test for a required \(\sigma_1\) is a change of variable. Substituting \(s = z - \sigma_1\) slides the whole \(s\)-plane so that the vertical line \(\operatorname{Re}\{s\}=-\sigma_1\) becomes the imaginary axis of the \(z\)-plane. A pole lies to the left of that line in \(s\) exactly when its image lies in the left half of the \(z\)-plane. So the question "are all poles at least \(\sigma_1\) deep?" becomes the ordinary stability question applied to \(D(z-\sigma_1)\), and any stability test can be reused unchanged. Example 6 carries this out.

Classify each system before computing anything, and state explicitly which class it falls into. Assume unity feedback where a loop is described by \(G(s)\) alone. Difficulty rises down the list.