Stability

Stability

Stability of an LTI system is governed by the eigenvalues of $A$.

Internal (zero-input) stability concerns the homogeneous response $x(t)=e^{At}x(0)$. The system is:

• Asymptotically stable iff every eigenvalue satisfies $\mathrm{Re}({\lambda }_i)<0$ (continuous time) or $|{\lambda }_i|<1$ (discrete time). Every initial condition decays to $0$.
• Marginally stable iff every eigenvalue satisfies $\mathrm{Re}({\lambda }_i)\le 0$ (CT) or $|{\lambda }_i|\le 1$ (DT), with simple (non-repeated) roots on the boundary. Trajectories stay bounded but do not decay.
• Unstable iff any eigenvalue violates the above — or a boundary eigenvalue is repeated (defective), producing $t^k{e}^{\lambda t}$ or $i^k{\lambda }^i$ growth.

Bounded-input bounded-output (BIBO) stability concerns the forced response $y(t)$: the system is BIBO stable iff every bounded input $u$ produces a bounded output $y$. For an LTI system with transfer function $G(s)$, BIBO stability is equivalent to all poles of $G(s)$ lying in the open left half-plane (CT) or the open unit disk (DT).

Asymptotic stability $\Rightarrow$ BIBO stability. The converse fails when unstable internal modes are canceled by zeros in $G(s)$ (pole-zero cancellation) and therefore do not appear in the input-output map — but they are still present internally and can blow up via disturbances or initial conditions.

The stable region is the open left half-plane for continuous time and the open unit disk for discrete time. This is the fundamental mapping: $\lambda \mapsto e^{\lambda T}$ sends the left half of the $\lambda$-plane to the interior of the unit disk, connecting the two criteria.

Marginal Stability Is Fragile $|\lambda |=1$) is on the boundary between stable and unstable. Any tiny perturbation — numerical roundoff, modeling error, unmodeled dynamics — can push it to either side. In practice, controllers require strict asymptotic stability with a safety margin ($\mathrm{Re}(\lambda )<-ϵ$ for some $ϵ>0$), not just $\mathrm{Re}(\lambda )\le 0$.

A marginally stable system (e.g., purely imaginary eigenvalues,

Intuition

A stable LTI system is one whose natural response decays on its own — “flick it, and it settles back down.” The eigenvalues of $A$ are the growth rates of its modes: $e^{{\lambda }_{i}t}{v}_i$. If every ${\lambda }_i$ has negative real part, every mode decays, and therefore every linear combination of modes decays too. One positive-real-part eigenvalue is enough to blow up the whole system, no matter how many other modes are stable.

The left half-plane / unit disk criterion reflects the two ways to grow. In continuous time, growth is exponential: $|e^{\lambda t}|=e^{\mathrm{Re}(\lambda )t}$ decays iff $\mathrm{Re}(\lambda )<0$. In discrete time, growth is power: $|{\lambda }^i|$ decays iff $|\lambda |<1$. Sampling a CT system at rate $T$ turns eigenvalues $\lambda$ into $e^{\lambda T}$ — and $\mathrm{Re}(\lambda )<0$ is exactly $|e^{\lambda T}|<1$. The two pictures are the same, viewed through different time representations.

BIBO stability asks a different question: can a bounded input drive the output unbounded? Even an unstable system ($A$ with a positive eigenvalue) might be BIBO stable if that bad mode isn’t reachable from $u$ or visible from $y$ — but that relies on exact cancellation of a pole by a zero, which is never exact in practice. The safe engineering standard is asymptotic (internal) stability, which implies BIBO and is robust.

Properties

Eigenvalue Criterion

(CT asymptotic stability) Every eigenvalue in the open left half-plane: $\mathrm{Re}({\lambda }_i)<0$ for all $i$.

This is both necessary and sufficient. Each mode contributes $e^{{\lambda }_{i}t}$; the full response is a finite sum of these. The sum decays to $0$ iff every summand does. Strict inequality is needed because $\lambda =0$ gives a constant mode $e^{0\cdot t}=1$ that never decays.

(DT asymptotic stability) Every eigenvalue strictly inside the unit disk: $|{\lambda }_i|<1$ for all $i$.

The discrete-time analog. Modes evolve as ${\lambda }^i$; decay requires $|\lambda |<1$. Eigenvalues on the unit circle give persistent oscillation (${\lambda }^i$ circles around the unit circle without shrinking); eigenvalues outside give exponential growth.

(CT-DT mapping) Sampling sends $\lambda \mapsto e^{\lambda T}$: LHP maps to the open unit disk.

If a CT system with eigenvalues ${\lambda }_i$ is sampled at interval $T$, the discrete-time dynamics has eigenvalues $e^{{\lambda }_{i}T}$. The mapping is: LHP $\to$ open unit disk, imaginary axis $\to$ unit circle, RHP $\to$ outside the disk. Stability classifications transfer: CT-stable systems remain DT-stable under sampling (though convergence rates change).

(Marginal stability) Boundary eigenvalues give bounded but non-decaying response, only if simple.

Pure imaginary eigenvalues $\pm i\omega$ (CT) or unit-circle eigenvalues $e^{\pm i\theta }$ (DT) give oscillations that never grow or shrink. Zero eigenvalue (CT) or $\lambda =1$ (DT) give constant modes. But if a boundary eigenvalue is repeated and the $A$ matrix is defective (Jordan block), the mode grows as $t^k$ (CT) or $i^k$ (DT) — that is unstable, not marginal.

Internal vs BIBO

(Internal $\Rightarrow$ BIBO) Asymptotic internal stability implies BIBO stability.

If every internal mode decays, the state response to any bounded input stays bounded, and so does the output $y=Cx+Du$. The impulse response $h(t)=C{e}^{At}B$ decays exponentially, so ${\int }_0^{\infty }|h(\tau )|d\tau <\infty$, which is the standard BIBO condition.

(BIBO $\nRightarrow$ internal) BIBO stability does not imply internal stability.

If an unstable internal mode is uncontrollable (not reachable from $u$) or unobservable (not seen in $y$), it does not appear in the transfer function $G(s)=C(sI-A{)}^{-1}B+D$ — it is canceled by a zero. The system looks stable from input-output measurements but still has a hidden divergent mode. Disturbances or nonzero initial conditions can excite it.

(Transfer function poles) BIBO stability iff all poles of $G(s)$ are in the open LHP (CT) / open unit disk (DT).

The poles of $G(s)$ are a subset of the eigenvalues of $A$ — specifically, the eigenvalues of the controllable-observable part. Checking pole locations is the standard frequency-domain stability test. Tools: Routh–Hurwitz criterion (algebraic test on polynomial coefficients), Nyquist criterion (contour integral on the frequency response), root locus (pole locations as a parameter varies).

Robustness

(Stability margin) Distance of the nearest eigenvalue from the stability boundary quantifies robustness.

${\min }_i|\mathrm{Re}({\lambda }_i)|$ (CT) or $1-{\max }_i|{\lambda }_i|$ (DT) measures how much perturbation the system can tolerate before becoming unstable. Control design often aims for a target margin, not just for stability.

(Parameter sensitivity) Eigenvalues move continuously in the matrix entries; stability can be lost under perturbation.

Small changes to $A$ cause small changes to eigenvalues. A system with eigenvalues near the boundary can flip from stable to unstable under unmodeled dynamics or parameter drift. This motivates robust control techniques ($H_{\infty }$, $\mu$-synthesis) which design for a family of plants rather than a single model.

(Nonlinear implications) Linearization stability $\Rightarrow$ local nonlinear stability (Lyapunov’s first method).

If the linearization of $\dot{x}=f(x)$ at an equilibrium $x_0$ has all eigenvalues in the open LHP, then $x_0$ is a locally asymptotically stable equilibrium of the nonlinear system. If even one eigenvalue is in the open RHP, $x_0$ is unstable. Eigenvalues on the imaginary axis are inconclusive — higher-order terms decide. See 7.2. Linearization.

Examples

Stability from Eigenvalues

Is $A=\left[\begin{array}{cc}-2& 0\\ 3& -1\end{array}\right]$ asymptotically stable?

Solution $\lambda =-2,-1$. Both in the open LHP.

Lower-triangular, so eigenvalues are the diagonal entries:

Asymptotically stable. Stability margin $|\mathrm{Re}({\lambda }_{\text{closest}})|=1$.

Marginal Stability: Undamped Oscillator

Is $A=\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]$ stable?

Solution ${\lambda }^2+1=0$, eigenvalues $\pm i$. Pure imaginary, simple, on the boundary.

Characteristic polynomial

Marginally stable. Trajectories are closed orbits (see center); bounded but not decaying. Any dissipation (however small) moves eigenvalues into the LHP and makes the system asymptotically stable; any energy injection moves them into the RHP.

Discrete-Time Stability

Is the discrete-time system with $A=\left[\begin{array}{cc}0.5& 0.3\\ 0.4& -0.2\end{array}\right]$ stable?

Solution $\det (\lambda I-A)={\lambda }^2-0.3\lambda -(0.1+0.12)={\lambda }^2-0.3\lambda -0.22=0$.

$\lambda =\frac{0.3\pm \sqrt{0.09+0.88}}{2}=\frac{0.3\pm 0.985}{2}$, so ${\lambda }_1\approx 0.643$, ${\lambda }_2\approx -0.343$.

Both $|{\lambda }_i|<1$. Asymptotically stable. Stability margin $1-\max |{\lambda }_i|\approx 0.357$.

Defective Repeated Eigenvalue

Is $A=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$ stable?

Solution $0$, both on the boundary. But $A$ has only one eigenvector $(1,0{)}^T$ — it is a defective Jordan block.

Both eigenvalues are

Solution: $e^{At}=I+At=\left[\begin{array}{cc}1& t\\ 0& 1\end{array}\right]$. Response to $x(0)=(0,1{)}^T$ is $(t,1{)}^T$ — unbounded.

Unstable. The repeated boundary eigenvalue with defective $A$ produces polynomial growth even though $|\lambda |=0$.

See Also

• 7.3. Solving LTI Systems — the closed-form solution whose decay/growth stability characterizes
• 7.4. Phase Portraits — geometric visualization of stability in 2D
• 7.2. Linearization — stability criterion applied to the linearized $A$ of a nonlinear system
• 7.6. State Feedback — stabilization by relocating eigenvalues via feedback
• eigenvalues and eigenvectors — the underlying linear-algebra primitive