Eigenvalues of A — system modes, stability: Exercises

3.5.30 · D4 · Physics › Guidance, Navigation & Control (GNC) › Eigenvalues of A — system modes, stability

Definition Imaginary axis, aur teen stability words

Complex plane ke do axes hote hain. Horizontal wala real axis hai ( $σ$ ki values); vertical wala imaginary axis hai ( $ω$ ki values) — yeh un points ka set hai jahan $σ = 0$ hai, yaani purely up-down numbers jaise $2 j$ . Jab is page pe "the axis" bina kisi aur word ke aaye, toh hamesha iska matlab yahi imaginary axis hai — Figure s01 mein vertical dividing line. Teen words jo hum constantly use karenge:

Asymptotically stable — har eigenvalue ka $σ < 0$ hai (axis ke strictly left mein); har mode decay karta hai aur $x \to 0$ .
Marginally stable — koi bhi eigenvalue $σ > 0$ wala nahi hai, aur koi bhi eigenvalue jo axis pe baitha ho ( $σ = 0$ ) woh simple hai (repeated nahi). Response na badhti hai na marti hai — bounded rehti hai.
Unstable — koi eigenvalue $σ > 0$ wala hai, ya axis pe ek repeated eigenvalue baitha hai. Kam se kam ek mode unbounded grow karta hai.

Har eigenvalue jo hum dhundhte hain woh $λ = σ + j ω$ likhte hain, jahan $σ$ (real part) growth/decay rate hai aur $ω$ (imaginary part) oscillation frequency hai. Complex plane ki picture apne dimaag mein poore waqt rakho:

Figure — Eigenvalues of A — system modes, stability

Figure s01 — stability map (visual key). Isse ek baar dhyan se padho, kyunki neeche ke har problem mein bas yahi puchha jaata hai "mera pole kis region mein hai?":

Horizontal white line real axis hai; red dashed vertical line imaginary axis hai — settle hone aur blow up hone ke beech ki border.

Red line ke left mein sab kuch (cyan shaded, $σ < 0$ ) stable hai; right mein sab kuch ( $σ > 0$ ) unstable hai.

Har × ek eigenvalue hai (ek "pole"). Cyan ×'s stable hain: ek real wala $- 2$ pe aur ek complex pair $- 1 \pm 3 j$ pe. Amber × $+ 0.2$ pe unstable hai. White ×'s $\pm 2 j$ pe exactly imaginary axis pe baithe hain ( $σ = 0$ ) — marginal edge case.

L1.2 amber aur white points ko directly use karta hai, aur L5.4 poori tarah aise white pure-imaginary pair pe based hai.

Level 1 — Recognition

L1.1 — Plane se seedha stability padho

Ek system $\dot{x} = A x$ ke eigenvalues ${- 2, - 5}$ hain. Kya yeh asymptotically stable hai, marginally stable hai, ya unstable hai?

Recall Solution

Dono eigenvalues real aur negative hain, isliye dono strictly imaginary axis ke left mein baithe hain (Figure s01 mein $- 2$ pe cyan × dekho). Har mode $e^{- 2 t}$ aur $e^{- 5 t}$ zero pe decay karta hai. $\Rightarrow ==asymptotically stable== .$ Picture: negative real axis pe do dots, dono left half-plane ke andar.

L1.2 — Unstable wala dhundho

In eigenvalue sets mein se kaun sa unstable system ka hai? $(a) {- 1 \pm 3 j} (b) {- 4, 0} (c) {- 1, + 0.2}$

Recall Solution

Har ek ka real part check karo — Figure s01 mein horizontal position:

(a) $σ = - 1 < 0$ dono ke liye → stable (oscillatory but decaying); ye woh cyan pair hai $- 1 \pm 3 j$ pe.
(b) real parts $- 4$ aur $0$ → $0$ imaginary axis pe hai aur simple hai, isliye marginally stable (bounded, upar di gayi definition ke anusaar), unstable nahi.
(c) $+ 0.2 > 0$ → ek mode $e^{0.2 t}$ ki tarah grow karta hai → ==(c) unstable hai==; yeh amber × hai $+ 0.2$ pe. Sirf strictly positive real part ka matlab unstable hota hai.

Level 2 — Application

L2.1 — 2×2 ke eigenvalues dhundho

$A = (- 1 0 2 - 3) .$ Eigenvalues compute karo aur stability classify karo.

Recall Solution

$A$ upper-triangular hai (diagonal ke neeche wali entry $0$ hai), isliye iske eigenvalues bas diagonal entries hain — Characteristic Polynomial & Determinants dekho: $det (A - λ I) = (- 1 - λ) (- 3 - λ) = 0 \Rightarrow λ_{1} = - 1, λ_{2} = - 3.$ Dono $σ < 0$ → asymptotically stable. Slow mode $- 1$ settling time dominate karta hai.

L2.2 — 2nd-order ODE ki companion form

Ek rotor $\ddot{θ} + 3 \dot{θ} + 2 θ = 0$ follow karta hai. State-space $A$ likho jahan $x_{1} = θ, x_{2} = \dot{θ}$ ho, phir eigenvalues dhundho.

Recall Solution

$x_{1} = θ$ , $x_{2} = \dot{θ}$ ke saath: $\overset{x}{˙}_{1} = x_{2}$ aur $\overset{x}{˙}_{2} = \ddot{θ} = - 2 θ - 3 \dot{θ}$ : $A = (0 - 2 1 - 3) .$ Characteristic equation (is companion form ke liye yeh ODE ko exactly reproduce karta hai): $det (A - λ I) = λ^{2} + 3 λ + 2 = (λ + 1) (λ + 2) = 0 \Rightarrow λ = - 1, - 2.$ Dono real, dono negative → stable, overdamped (do distinct real roots, koi oscillation nahi).

Level 3 — Analysis

L3.1 — Damping ratio aur natural frequency

Ek mode ke eigenvalues $λ = - 1 \pm j 3$ hain. Natural frequency $ω_{n}$ , damped frequency $ω_{d}$ , aur damping ratio $ζ$ dhundho. Yeh under-, over-, ya critically damped hai?

Recall Solution

Standard form kahan se aata hai. Ek damped 2nd-order system likha jaata hai $\overset{y}{¨} + 2 ζ ω_{n} \overset{y}{˙} + ω_{n}^{2} y = 0$ ke roop mein. Yeh coefficients kyun use kiye jaate hain? Iska characteristic polynomial (plug $y = e^{λ t}$ ) hai $λ^{2} + 2 ζ ω_{n} λ + ω_{n}^{2} = 0.$ Quadratic formula se solve karne par:

= -\zeta\omega_n \pm \omega_n\sqrt{\zeta^2-1}.$$ $0<\zeta<1$ ke liye root ke andar wala term negative hai, isliye $j$ nikalo: $\lambda=-\zeta\omega_n\pm j\,\omega_n\sqrt{1-\zeta^2}$. Toh polynomial ka **constant term** $\omega_n^2$ hai aur **$\lambda$-coefficient** $2\zeta\omega_n$ hai — isi wajah se kisi polynomial ko is template se match karke $\omega_n$ aur $\zeta$ padha jaata hai (dekho [[Damping Ratio and Natural Frequency]]). **Apne eigenvalues match karo** $\lambda=-1\pm j\sqrt3$: - $\omega_n=|\lambda|=\sqrt{(-1)^2+(\sqrt3)^2}=\sqrt{1+3}=2.$ - $\omega_d=\operatorname{Im}(\lambda)=\sqrt3\approx 1.732.$ - $\sigma=-\zeta\omega_n=-1 \Rightarrow \zeta=\dfrac{1}{\omega_n}=\dfrac{1}{2}=0.5.$ Kyunki $0<\zeta<1$ → ==underdamped== (yeh decay karte waqt oscillate karta hai). **$\zeta$ ka geometric matlab — Figure s02 dekho.** Complex plane mein pole $\lambda=-1+j\sqrt3$ plot karo. Origin se pole tak straight line (length $\omega_n$) kheencho. ==$\phi$== woh angle ho jो *negative real axis se, pole ki taraf khulte hue* measure kiya gaya ho. Pole axis ke left mein jitna baitha hai woh horizontal distance $\omega_n\cos\phi$ hai, aur humne woh distance $|\sigma|=\zeta\omega_n$ naam diya tha. Dono ko barabar karo: $$\zeta\omega_n=\omega_n\cos\phi \;\Rightarrow\; \boxed{\zeta=\cos\phi}.$$ Toh negative real axis ke *paas* wala pole ($\phi$ chhota) heavily damped hai ($\zeta\to 1$), aur imaginary axis ke *paas* wala pole ($\phi\to 90^\circ$) barely damped hai ($\zeta\to 0$). Hamare liye $\cos\phi=0.5\Rightarrow\phi=60^\circ$.

Figure s02 — damping angle $ϕ$ . Cyan line jis ki length

ω_{n}

hai, origin se pole

- 1 + j 3

tak jaati hai; amber angle

ϕ

negative real axis se measure kiya gaya hai; iska horizontal leg

ζ ω_{n}

hai aur vertical leg

ω_{d}

hai. Isi wajah se

ζ = cos ϕ

L3.2 — Sabse slow mode settling time set karta hai

$A = (- 10 1 0 - 0.5) .$ Kaun sa eigenvalue long-term settling govern karta hai, aur roughly response settle hone mein kitna waqt lagega?

Recall Solution

Lower-triangular → eigenvalues diagonal hain: $λ_{1} = - 10, λ_{2} = - 0.5$ . Mode $e^{- 10 t}$ almost instantly khatam ho jaata hai; $e^{- 0.5 t}$ sabse zyada der tak rehta hai. Imaginary axis ke sabse paas wala eigenvalue ( $- 0.5$ ) dominate karta hai.

$4/∣ σ ∣$ rule kahan se aata hai. Ek decaying mode ka envelope $e^{- ∣ σ ∣ t}$ hota hai. Hum response ko "settled" tab kehte hain jab woh envelope ek chhoti si band tak shrink ho jaaye (usual engineering convention $2%$ of its start value hai). Woh time dhundho jab envelope us band tak pahunche: $e^{- ∣ σ ∣ t_{s}} = 0.02 \Rightarrow ∣ σ ∣ t_{s} = ln \frac{1}{0.02} = ln 50 \approx 3.91.$ $ln 50 \approx 4$ round karne se handy rule milta hai $t_{s} \approx 4/∣ σ ∣$ — yeh bas "kitne time-constants mein $2%$ tak fade ho jaao" hai (dekho Damping Ratio and Natural Frequency). Ise slowest pole pe apply karo: $t_{s} \approx \frac{4}{∣ σ _{slow} ∣} = \frac{4}{0.5} = 8 s .$ Long-term behaviour least-negative eigenvalue se set hoti hai, fast wale se nahi.

Level 4 — Synthesis

L4.1 — Quadratic solve kiye bina stability

$\dot{x} = A x$ ke liye jahan $A = (0 - 6 1 - 5),$ 2×2 rule (trace aur determinant) use karke stability decide karo, phir eigenvalues se confirm karo.

Recall Solution

Trace–determinant test kyun kaam karta hai. Kisi bhi 2×2 matrix ke liye characteristic polynomial sirf uske trace aur determinant se likha ja sakta hai: $det (A - λ I) = λ^{2} - (tr A) λ + det A = 0.$ Iske do roots satisfy karte hain $λ_{1} + λ_{2} = tr A$ aur $λ_{1} λ_{2} = det A$ . Ab poochho: dono roots left half-plane mein kab hote hain ( $Re < 0$ )?

Agar roots real hain, "dono negative" ka matlab hai unka sum negative ho ( $tr A < 0$ ) aur unka product positive ho ( $det A > 0$ ; do negatives multiply hoke positive dete hain).
Agar roots ek complex pair $σ \pm j ω$ hain, toh sum $= 2 σ = tr A$ , isliye $σ < 0 ⟺ tr A < 0$ ; aur product $= σ^{2} + ω^{2} = det A > 0$ automatically, jo bas $det A > 0$ confirm karta hai.

Har case mein conditions ki pair exactly yahi hai: $stable ⟺ tr A < 0 AND det A > 0.$ Yeh Routh–Hurwitz Criterion ka 2×2 special case hai. Yahan $tr A = 0 + (- 5) = - 5 < 0$ ✓ aur $det A = (0) (- 5) - (1) (- 6) = 6 > 0$ ✓ → stable. Confirm: $det (A - λ I) = λ^{2} + 5 λ + 6 = (λ + 2) (λ + 3) = 0 \Rightarrow λ = - 2, - 3.$ ✓

L4.2 — Target design: parameter choose karke poles place karo

Ek tunable system hai $A = (0 - 4 1 - a),$ jahan gain $a > 0$ hum choose kar sakte hain. $a$ choose karo taaki closed loop critically damped ho ( $ζ = 1$ ), aur resulting (repeated) eigenvalue do.

Recall Solution

Characteristic equation: $λ^{2} + aλ + 4 = 0$ . L3.1 mein derive ki gayi standard form $λ^{2} + 2 ζ ω_{n} λ + ω_{n}^{2}$ se compare karo (constant term $= ω_{n}^{2}$ , $λ$ -coefficient $= 2 ζ ω_{n}$ ):

$ω_{n}^{2} = 4 \Rightarrow ω_{n} = 2$ .
$2 ζ ω_{n} = a$ . Critical damping ke liye $ζ = 1$ : $a = 2 (1) (2) = 4$ .

$a = 4$ ke saath: $λ^{2} + 4 λ + 4 = (λ + 2)^{2} = 0 \Rightarrow λ = - 2$ (repeated). Yeh no overshoot ke saath fastest response hai — woh sweet spot jis par control engineers aim karte hain (dekho Pole Placement & LQR).

Level 5 — Mastery

L5.1 — Axis pe repeated eigenvalue: hidden growth

$A = (0010) .$ Dono eigenvalues $0$ hain. Kya system marginally stable hai ya unstable? Justify karne ke liye $x (0) = (x_{10}, x_{20})^{T}$ se $x (t)$ solve karo.

Recall Solution

$det (A - λ I) = λ^{2} = 0 \Rightarrow λ = 0$ (repeated). Dono real parts $0$ hain, isliye naive check kehta hai "axis pe → shayad marginal". Lekin dono zeros sirf ek eigenvector share karte hain (ek defective/repeated eigenvalue), isliye ek $t$ -term appear hota hai. Seedha solve karo: $\overset{x}{˙}_{1} = x_{2}, \overset{x}{˙}_{2} = 0 \Rightarrow x_{2} (t) = x_{20}, x_{1} (t) = x_{10} + x_{20} t .$ $x_{20} t$ term unbounded grow karta hai jab bhi $x_{20} \neq = 0$ → ==unstable==. Marginal stability ke liye axis eigenvalues simple (non-repeated) hone chahiye. Yeh wala us test mein fail karta hai.

$A = (- 3 0 1 - 2), x (0) = (21) .$ Eigenvalues, eigenvectors, constants $c_{i}$ , full solution dhundho, aur batao kaun sa mode $t \to \infty$ ke saath dominate karta hai.

Recall Solution

Eigenvalues (triangular): $λ_{1} = - 3, λ_{2} = - 2$ . Eigenvectors (dekho Diagonalization and Modal Decomposition):

$λ_{1} = - 3$ : $(A + 3 I) v = (0011) v = 0 \Rightarrow v_{1} = (1, 0)^{T}$ .
$λ_{2} = - 2$ : $(A + 2 I) v = (- 1 0 10) v = 0 \Rightarrow v_{2} = (1, 1)^{T}$ .

Constants $x (0) = c_{1} v_{1} + c_{2} v_{2}$ se: $c_{1} + c_{2} = 2, c_{2} = 1 \Rightarrow c_{2} = 1, c_{1} = 1.$ Solution:

=\begin{pmatrix} e^{-3t}+e^{-2t} \\ e^{-2t}\end{pmatrix}.$$ $t\to\infty$ hone par $e^{-3t}$ mode pehle mar jaata hai; **slower** $e^{-2t}$ mode (axis ke sabse paas) tail dominate karta hai. Dono decay karte hain → asymptotically stable.

L5.3 — Routh–Hurwitz se third-order stability

Ek system ke eigenvalues uske characteristic polynomial ke roots hain. Woh polynomial eigenvalue variable $λ$ mein likhne par, maan lo $det (A - λ I) = λ^{3} + 2 λ^{2} + 2 λ + K = 0.$ Gain $K$ ki woh range dhundho jiske liye system asymptotically stable hai.

Recall Solution

Is cubic ka har root $λ$ ek eigenvalue hai, isliye "asymptotically stable" ka matlab hai har root ka $Re (λ) < 0$ ho. Routh–Hurwitz Criterion humein roots dhundhe bina yeh test karne deta hai. Ek cubic $λ^{3} + a_{2} λ^{2} + a_{1} λ + a_{0}$ ke liye saari chaar conditions simultaneously hold karni chahiye: $a_{2} > 0, a_{1} > 0, a_{0} > 0, aur a_{2} a_{1} > a_{0} .$ (Pehli teen demand karti hain ki har coefficient positive ho; aakhri woh extra "inner" determinant condition hai jo ek cubic ko chahiye.) Yahan $a_{2} = 2, a_{1} = 2, a_{0} = K$ :

$a_{2} = 2 > 0$ ✓ hamesha.
$a_{1} = 2 > 0$ ✓ hamesha.
$a_{0} = K > 0 \Rightarrow K > 0$ .
$a_{2} a_{1} = 2 \cdot 2 = 4 > K \Rightarrow K < 4$ .

Combine karne par: $0 < K < 4$ . $K = 4$ pe eigenvalues ki ek pair exactly imaginary axis pe baithti hai (marginal); $K = 0$ pe ek eigenvalue origin pe hota hai.

L5.4 — Pure-imaginary poles: undamped edge case

Ek frictionless flywheel deta hai $A = (0 - 9 10) .$ Eigenvalues dhundho, stability classify karo, aur motion $x (t)$ describe karo.

Recall Solution

$det (A - λ I) = λ^{2} + 9 = 0 \Rightarrow λ^{2} = - 9 \Rightarrow λ = \pm 3 j$ ( $j = - 1$ use karke). Yeh pure imaginary hain: $σ = 0, ω = \pm 3$ — yeh Figure s01 mein exactly imaginary axis pe baithe hain (jaise white × marks, lekin $\pm 3 j$ pe).

Classification. Dono eigenvalues ka real part $σ = 0$ hai, isliye koi mode exponentially grow nahi karta — system not unstable hai. Aur dono roots distinct hain ( $+ 3 j \neq = - 3 j$ ), yaani axis eigenvalues simple hain, isliye koi dangerous $t$ -term appear nahi hota (contrast L5.1 se, jahan repeated $0$ ne marginal stability tod di thi). Definition ki dono requirements meet hain, isliye flywheel ==marginally stable== hai.

Motion. Envelope $e^{σ t} = e^{0 \cdot t} = 1$ hai, ek constant, isliye amplitude na badhti hai na ghatti hai. Euler's formula $e^{\pm 3 j t} = cos 3 t \pm j sin 3 t$ use karke, real solution ek pure sustained oscillation hai frequency $ω = 3 rad/s$ par: $x (t) = (c_{1} cos 3 t + c_{2} sin 3 t - 3 c_{1} sin 3 t + 3 c_{2} cos 3 t),$ jahan $c_{1} = x_{1} (0)$ aur $c_{2} = x_{2} (0) /3$ initial state se aate hain. State $(x_{1}, x_{2})$ plane mein hamesha ke liye ek closed loop (ek ellipse) trace karta hai — constant amplitude par spin karta, na settle hota na blow up karta. Damping language mein yeh $ζ = 0$ (no damping) hai $ω_{n} = 9 = 3$ ke saath, isliye $ω_{d} = ω_{n} 1 - ζ^{2} = 3 = ω_{n}$ . Yeh knife-edge hai: kisi pole ko imaginary axis se thoda left nudge karo → decaying spiral (stable); thoda right → growing spiral (unstable).

Recall Ek-line ladder recap

L1 plane padho ::: real part ka sign stability decide karta hai. L2 eigenvalues compute karo ::: $det (A - λ I) = 0$ solve karo, diagonal se nahi. L3 interpret karo ::: pole location se $ζ, ω_{n}, ω_{d}$ ; slowest pole $t_{s}$ set karta hai. L4 design karo ::: trace/det rule + pole placement behaviour choose karta hai. L5 master karo ::: repeated axis roots grow karte hain, simple imaginary pairs sirf oscillate karte hain; higher order ke liye Routh–Hurwitz.