From Disturbance to Dynamic Modes

Reference motion and perturbation equilibrium

An equilibrium in perturbation coordinates means that the departure is zero along the declared reference motion. It does not require every position coordinate of the full aircraft to be constant. Straight reference flight can have constant body velocities and attitude while the geographic position changes continuously.

The decomposition u = $u_0$ + $\Delta u$ applies at every time. The suffix 0 is a reference label; $\Delta u$(0) is the forward-speed departure at the chosen initial instant. The post-disturbance initial state may contain several nonzero entries. A short input establishes that initial state; after release its continuing perturbation value is zero.

A release experiment

In flight-test language, a brief pulse followed by a return to the reference input can expose a free transient. A sustained input must remain in the forced-response model. Choosing a different reference condition also changes the local matrix, so a stability statement always identifies the operating point and the perturbations under consideration.

The course state velocities are ground-relative body components. Aerodynamic angles use air-relative velocity. In wind, subtract the wind expressed in body axes before forming aerodynamic angles; constant navigation-frame wind still changes body components as attitude changes. Zero wind is the explicit simplification in this lesson. Body pitch rate q and Euler pitch-angle rate coincide only under the declared symmetric longitudinal linearization.

A derivative is one ingredient of a dynamic model

A static derivative evaluates a local force or moment change with specified quantities held fixed. The characteristic roots instead combine the state couplings, inertia, kinematics and derivatives in the complete local model. For example, a canonical scalar oscillator has the same restoring term for every damping value in the visual, while its rate term changes the subsequent envelope.

The phrase “for every sufficiently small initial perturbation” matters. A stable observed component is compatible with an unexcited unstable internal component. Bounded neutral response is also different from convergence: an undamped sinusoid can remain near the reference without tending to it.

After Section 6, revisit the imaginary axis. A repeated eigenvalue on that axis gives bounded response only if its associated Jordan blocks have size one. The defective zero-root example in Student Study Section 6 produces linear drift despite having no positive-real-part root. This matrix caveat does not change the bounded undamped scalar example shown here.

For an actual nonlinear aircraft, strictly negative real parts support local asymptotic stability under the usual smoothness assumptions. Axis roots leave the linearization test inconclusive about nonlinear local stability. These are local statements, not guarantees for large disturbances.

Solve the free equation

For a nonzero initial displacement, divide the homogeneous equation by x and separate the variables. Integrating between the release time and t fixes the constant of integration. The zero initial state is handled separately and remains identically zero.

Setting the amplitude ratio to 1/e, 1/2 or 2 yields the time metrics. The stable fixture a=2 $s^{-1}$ has τ=0.5 s and half-time ln(2)/2=0.3466 s. The unstable λ=0.5 $s^{-1}$ fixture doubles in 1.3863 s. At a=0, x(t)=x(0), so no finite decay time exists.

A forced first-order example

For a constant forcing, subtract the equilibrium b/a when a is nonzero. The error then obeys the homogeneous equation. The equilibrium is approached only for a positive coefficient a.

Here τ=2/3 s, x(0)=0 and the final value is 2. Differentiating gives $\dot{x}$=3$e^{\frac{-3t}{2}}$, so 2$\dot{x}$+3x=6 exactly. At τ the completed change is 2(1−$e^{-1}$); the remaining error is 2$e^{-1}$. If a=0, the constant forcing integrates into x(0)+bt rather than approaching b/a.

A mechanical analogy, after the aircraft example

A mass $m_s$>0, spring $k_s$>0 and viscous damper $c_s$ obey the following equation when displacement ξ is measured from static equilibrium. Dividing by mass identifies the canonical parameters. Negative $c_s$ is a manufactured negative-damping case, not a passive damper.

For free motion, insert x=C$e^{\lambda t}$. The resulting quadratic gives the roots in EQ-L02-10. Distinct exponentials are independent solutions; a repeated root requires the second independent solution t$e^{\lambda t}$. Complex conjugates combine into a real cosine/sine form.

Evaluate both initial-condition constants

At ζ=1, λ=−${\omega }_n$. At ζ=−1, λ=+${\omega }_n$. For the common displacement-one, rate-zero fixture, the stable critical response is (1+2t)$e^{-2t}$ and the negative-critical response is (1−2t)$e^{2t}$. Direct substitution and evaluation at t=0 verify both conditions. A polynomial times a decaying exponential still tends to zero.

Keep forced response distinct

The constant particular solution 1 becomes the final value only when the homogeneous transient decays. With negative damping it remains a particular solution without being an attained steady state. Yechout Examples7.3–7.6 extend this prerequisite algebra; the fixtures here are independently declared mathematics, not numerical reconstructions of those aircraft examples.

Geometry and time scales

Circles centered at the origin have constant natural frequency. In the stable half-plane, rays from the origin have constant damping ratio because the angle χ is fixed. Along one ray, moving outward speeds both the exponential and sinusoidal time scales; along a vertical line, only damped frequency changes.

For λ=−0.4±j1.2 $s^{-1}$, the displayed values are ${\omega }_n$=1.2649 rad/s, ζ=0.3162, T=5.2360 s, envelope time constant 2.5 s, and half-time 1.7329 s. Internal calculations retain the exact square root and logarithms. The oscillatory transient is not strictly periodic when its envelope changes, although T is the period of its sinusoidal factor.

A real-root example

The slower −2 $s^{-1}$ term governs late decay. This quadratic has a coefficient-defined damping ratio 1.25; neither individual real root is assigned the oscillatory pair formula by itself. Arbitrary coefficients can make a sum of real exponentials turn once without becoming sinusoidal.

Optional extension: for an isolated stable pair, the logarithmic decrement between peaks one sinusoidal period apart is −σT. It measures the envelope ratio. A root moving closer to the origin can slow one or both time scales, depending on the path; a larger damping ratio alone does not uniquely specify recovery speed.

Derive the eigenproblem from the aircraft model

Use the sign-neutral control order [$\Delta {\delta }_e$, $\Delta {\delta }_T$]ᵀ and retain the reference values. For free motion only the control perturbation vector vanishes. For the exponential trial state, differentiation gives λv$e^{\lambda t}$, while the right side is Av$e^{\lambda t}$. Cancel the nonzero exponential and require a nonzero vector.

Changing determinant order multiplies the determinant by (−1)ⁿ, leaving its zeros unchanged. The determinant is a polynomial in λ, rather than a test of det(A) alone. For mixed-unit states, symbolic A entries have row/column-dependent dimensions; blanket inverse-second units apply to the dimensionless mathematical fixture, not every aircraft entry.

The general solution operator

Differentiate the power series term by term to recover $\dot{x}$=Ax and set t=0 to recover the initial state. When V is nonsingular, A=VΛ${\mathbf{V}}^{-1}$ and $e^{\mathbf{A}t}$=V$e^{\mathbf{\Lambda }t}$${\mathbf{V}}^{-1}$. Repeated stable roots may introduce polynomial factors that still decay; a complete eigenvector basis is not assumed in the matrix-exponential statement.

Why axis-root multiplicity needs care

Both mathematical states are dimensionless. With x(0)=[0,1]ᵀ the response is [t/(1 s),1]ᵀ: the first state grows linearly. The double zero eigenvalue is defective. By contrast, A=0 has semisimple zero eigenvalues and every state stays constant. Thus imaginary-axis location alone is insufficient for bounded matrix response. These Category C examples are course-derived analysis, not aircraft data.

Calculate the exact Category C eigenpair

The first row gives $v_2$=λ(1 s)$v_1$. With $v_1$=1, substitution into the second row gives the same quadratic. The residual Av−λv is zero up to floating-point roundoff. The factor one second makes the two eigenvector components consistent with dimensionless states; no aircraft variable is attached to either coordinate.

For x(0)=[1,0]ᵀ, c=1/2−j/(2√15). Both conjugate contributions are required for a real state. This two-state fixture contains one real oscillatory mode; its conjugate roots are not two separately excitable real aircraft modes. Different initial states alter amplitude and phase, not its eigenvalues.

Scale coordinates and normalize once

With z=${\mathbf{S}}^{-1}$x and S=diag($s_1$,$s_2$), compare |${\mathbf{v}}_z$,i|/max|${\mathbf{v}}_z$,j|. Relative phase is measured from a declared nonzero component. For S=I, |$v_2$|/|$v_1$|=2 and the upper-root relative phase is arg(−0.5+j√15/2)=1.8235 rad. If S=diag(1,2), both scaled amplitudes are equal. This is a coordinate change, not altered dynamics. Multiplying every component by the same nonzero complex number preserves these relative quantities.

Laplace prerequisite recap

Use causal transforms over t≥0 and a region of convergence to the right of the signal growth bound. The uppercase X(s) denotes the transform of the scalar or vector perturbation; it does not redefine the course reference suffix. With no impulsive state jump, pre- and post-release initial values coincide for the derivative rule.

Partial fractions reduce a rational response to real exponential, real conjugate-pair, or repeated-factor terms. For a declared scalar equation a double integrator/first-order denominator b/[s²(s+a)] inverts to b(at−1+$e^{-at}$)/a², for nonzero a. Its zero initial displacement/rate and late-time ramp slope b/a verify the coefficient; no unresolved source aircraft/control-sign example is imported.

Retain the initial state before forming a transfer function

Transforming the state equation gives (sI−A)ΔX=$\Delta \mathbf{x}$(0)+B$\Delta {\mathbf{U}}_{\text{c}}$. Solving for the transformed state yields EQ-L02-28. Add the output equation Δy=C$\Delta \mathbf{x}$+D$\Delta {\mathbf{u}}_{\text{c}}$. Only the zero-initial-condition part defines the input/output transfer matrix:

The transfer function is a ratio of transforms under zero initial conditions. The Laplace transform does not distribute over a time-domain quotient. Its inverse g(t) is the impulse-response kernel; the zero-state output is the convolution integral of g(t−τ)u(τ) from 0 to t, plus any direct feedthrough term. Zeros shape forced response and may cancel modal poles in a particular channel.

The initial-value theorem requires suitable regularity and no impulse at the origin. The final-value theorem requires all poles of sY(s), after valid simplification, to be in the open left half-plane. A growing or persistently oscillating response fails that condition.

A stable visible channel can hide instability

All states and input are dimensionless. The zero-state visible transfer function is G(s)=1/(s+1), while the internal eigenvalues are −1 and +1 $s^{-1}$. The second state is neither driven by this input nor observed by this output; a nonzero second initial state grows as $e^t$. Choosing output C=[0,1] reveals that free growth. This explicit nonminimal realization explains why input/output poles cannot be read as an unconditional list of internal modes.

Coordinates before mode labels

The primary longitudinal state remains [$\Delta u$,$\Delta w$,$\Delta q$,$\Delta \theta$]ᵀ. The preview lateral state is [$\Delta v$,$\Delta p$,$\Delta r$,$\Delta \phi$]ᵀ. Alpha and beta are aerodynamic angles formed from air-relative velocity; replace those components with the course ground-relative states only under the explicit zero-wind assumption.

At a trim-aligned reference $w_{\text{A0}}$=0, $\Delta \alpha$≈$\Delta w_{\text{A}}$/$V_{\text{A0}}$. At zero reference sideslip, $\Delta \beta$≈$\Delta v_{\text{A}}$/$V_{\text{A0}}$. These are conditional first-order relationships; they do not turn body angular rates p,q,r into Euler-angle rates. The lateral linearized bank kinematics are φ̇≈p+tan(${\theta }_0$)r, reducing to p only when the reference pitch makes that approximation appropriate.

The longitudinal factorization has two quadratics; the lateral pattern has one quadratic and two linear factors. Added position or heading integrators belong to an augmented state declaration and are listed separately. They are not silently renamed spiral. Detailed mode physics and reduction derivations remain in the next two lectures.

Source-linked numerical aircraft examples require their own complete state, unit, trim and model-version records. The present bridge uses no aircraft matrix. Its qualitative root symbols describe a common pattern rather than universal immutable identities.