Eigenvalues of A — system modes, stability

3.5.30 · Physics › Guidance, Navigation & Control (GNC)

Hum eigenvalues ki parwah kyun karte hain?

KYA hamare paas hai: ek state vector $x (t) \in R^{n}$ (jaise spacecraft attitude error aur uski rate) jo $\dot{x} = A x$ follow karta hai.

KYUN mushkil hai: equations coupled hain — $\overset{x}{˙}_{1}$ depend karta hai $x_{2}$ par, $\overset{x}{˙}_{2}$ $x_{1}$ par, wagera. Inhe ek ek karke solve nahi kar sakte.

KAISE eigenvalues bachate hain: woh directions dhundho jahan $A$ sirf pure scaling ki tarah kaam kare. Un directions ke along poora tangle $n$ alag scalar ODEs mein collapse ho jaata hai, har ek trivially solvable.

Scratch se derivation: solution guess karo

Step 1 — Ansatz. Ek aisi solution try karo jo ek fixed shape $v$ rakhe aur sirf time mein size badlaaye: $x (t) = e^{λ t} v$ . Ye step kyun? Agar koi bhi solution ek hi direction mein pointed rehti hai, toh vector ODE scalar ban jaata hai — sabse simple cheez jo kaam kar sakti hai, toh hum ise test karte hain.

Step 2 — Substitute karo. Differentiate karo: $\dot{x} = λ e^{λ t} v$ . $\dot{x} = A x$ mein daalo: $λ e^{λ t} v = A e^{λ t} v .$ Ye step kyun? Hum ansatz ko actually physics follow karate hain.

Step 3 — $e^{λ t}$ cancel karo (kabhi zero nahi): $A v = λ v \Rightarrow (A - λ I) v = 0 .$ Ye step kyun? Hum discover karte hain ki ansatz tabhi kaam karta hai jab $(λ, v)$ ek eigenpair ho. Exponentials $A$ mein bake in hain.

Step 4 — Non-trivial solution condition. $(A - λ I) v = 0$ ka nonzero $v$ tab hoga jab matrix singular ho: $det (A - λ I) = 0$ $λ$ mein ye polynomial characteristic equation hai; iske roots eigenvalues hain. Ye step kyun? Ek invertible matrix $v = 0$ force kar deta, jo hum ne ban kiya tha.

Har eigenvalue kya karta hai — complex plane padhna

Ek complex eigenvalue $λ = σ + j ω$ likhte hain. Phir, Euler use karke, $e^{λ t} = e^{σ t} (cos ω t + j sin ω t) .$

$σ = Re (λ)$ → envelope $e^{σ t}$ ki growth/decay rate.
$ω = Im (λ)$ → us mode ki oscillation frequency.

Figure — Eigenvalues of A — system modes, stability

KYUN sirf real part matter karta hai: $∣ e^{λ t} ∣ = e^{σ t}$ . Oscillation $∣ cos + j sin ∣ = 1$ kabhi grow nahi karta; sirf real-part exponential decide karta hai ki magnitude badhega ya ghategaa.

Ek physical 2nd-order mode $λ = - ζ ω_{n} \pm j ω_{n} 1 - ζ^{2}$ ke liye: $σ = - ζ ω_{n}$ damping deta hai, $ω_{d} = ω_{n} 1 - ζ^{2}$ damped frequency.

Worked Example 1 — ek stable oscillator (spring–damper as state space)

$m \overset{q}{¨} + c \overset{q}{˙} + k q = 0$ . Maan lo $x_{1} = q, x_{2} = \overset{q}{˙}$ . Tab $A = (0 - k / m 1 - c / m) .$ $k / m = 4, c / m = 2$ lo.

Step 1 — characteristic eqn: $det (A - λ I) = λ^{2} + \frac{c}{m} λ + \frac{k}{m} = λ^{2} + 2 λ + 4 = 0.$ Kyun? Ek scalar 2nd-order ODE ke companion form ke liye, $det (A - λ I)$ ODE ka khud ka characteristic polynomial reproduce karta hai — ek achha sanity check.

Step 2 — roots: $λ = \frac{- 2 \pm 4 - 16}{2} = - 1 \pm j 3 .$ Kyun? Quadratic formula.

Step 3 — padho: $σ = - 1 < 0$ → stable; $ω_{d} = 3$ → decay karte hue oscillate karta hai. $ω_{n} = k / m = 2$ , $ζ = σ / (- ω_{n}) = 0.5$ (underdamped). Response: $e^{- t} cos (3 t + ϕ)$ .

Worked Example 2 — eigenvectors se decouple karna

$A = (- 3 0 1 - 2) .$ Step 1 — triangular ⇒ eigenvalues diagonal par: $λ_{1} = - 3, λ_{2} = - 2$ . Kyun? Triangular matrices ke liye $det (A - λ I) = (- 3 - λ) (- 2 - λ)$ .

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

Step 3 — solution $x (t) = c_{1} e^{- 3 t} (1, 0)^{T} + c_{2} e^{- 2 t} (1, 1)^{T}$ . $- 3$ wala mode jaldi marta hai; long-term behaviour sabse slow (least-negative) eigenvalue $- 2$ se dominated hai. Ye kyun matter karta hai: imaginary axis ke sabse kareeb eigenvalue settling time set karta hai.

Worked Example 3 — Forecast-then-Verify

$A = (02 1 - 1) .$ Forecast: off-diagonal mein $+ 2$ positive feedback jaisa lagta hai → shayad unstable hai. Verify: $det (A - λ I) = λ^{2} + λ - 2 = (λ + 2) (λ - 1)$ ⇒ $λ = - 2, + 1$ . $+ 1 > 0$ hone se → unstable. Forecast confirm hua; growing mode $e^{+ t} v$ hai.

Common mistake Common errors ko steel-man karo

1. "Negative determinant/trace matlab stable." Kyun sahi lagta hai: 2×2 ke liye, $det A = λ_{1} λ_{2}$ aur $tr A = λ_{1} + λ_{2}$ , toh log inhe stability se confuse karte hain. Fix: stability ko $tr A < 0$ AUR $det A > 0$ chahiye (2×2 ke liye). Real test = eigenvalues ke real parts, ya higher order ke liye Routh–Hurwitz use karo.

2. "Complex eigenvalues ⇒ unstable." Kyun sahi lagta hai: complex scary/oscillatory lagta hai. Fix: oscillation ≠ instability. Stability sirf real part se hai; $- 1 \pm 100 j$ beautifully stable hai, bas bahut oscillatory hai.

3. Axis par repeated eigenvalues bhool jaana. Double $λ = 0$ ek term $t e^{0 t} = t$ de sakta hai, jo grow karta hai! Marginal stability ke liye axis eigenvalues simple (semisimple) hone chahiye.

4. $A$ ki entries ko seedha rates samajhna. Diagonal entries modes NAHI hain jab tak $A$ already diagonal/triangular na ho. $det (A - λ I) = 0$ solve karna zaroori hai.

Recall Feynman: 12-saal ke bachche ko samjhao

Socho ek trampoline par kaafi bachche bounce kar rahe hain aur ek doosre ko dhakka de rahe hain — messy aur coupled. Lekin kuch khaas bouncing tarike (patterns) hote hain jahan sablog ek saath rahe aur pattern sirf bina shape badlaaye smoothly badhta ya ghata hai. Ye khaas patterns eigenvectors hain, aur har pattern kitni tezi se badhta ya ghatta hai woh uski eigenvalue hai. Agar har pattern time ke saath ghatta hai, toh trampoline settle ho jaata hai (stable). Agar ek bhi pattern badhta hai, toh poori cheez eventually wild ho jaati hai (unstable). Hume bas har pattern ka shrink/grow number check karna hai.

Flashcards

\dot{x} = A x

ko asymptotically stable banane ke liye eigenvalues par kya condition chahiye?

Har eigenvalue ka strictly negative real part ho,

Re (λ_{i}) < 0

(sab open left half-plane mein).

A

ke eigenvalues kaunsi equation deti hai?

Characteristic equation

det (A - λ I) = 0

Ansatz

x = e^{λ t} v

ek coupled system ko scalar mein kyun badal deta hai?

Ek eigenvector ke along

A

sirf scale karta hai, toh

A v = λ v

vector ODE ko

\overset{s}{˙} = λ s

tak reduce kar deta hai.

λ = σ + j ω

mein,

σ

kya control karta hai?

ω

kya control karta hai?

σ

= envelope

e^{σ t}

ki growth/decay rate;

ω

= oscillation frequency.

Sirf

λ

ka real part stability kyun decide karta hai?

Kyunki

∣ e^{λ t} ∣ = e^{σ t}

; oscillatory factor ki magnitude 1 hai, toh sirf

σ

decide karta hai ki magnitude badhe ya ghate.

2×2

A

ke liye stability guarantee karne wali do conditions kya hain?

tr (A) < 0

aur

det (A) > 0

Long-term response mein kaunsa eigenvalue dominate karta hai?

Woh jo real part ke saath imaginary axis ke sabse kareeb ho (slowest-decaying / least-negative mode).

True/False: complex eigenvalues matlab hamesha instability.

False — oscillation ≠ instability; real part check karo.

2nd-order mode

λ = - ζ ω_{n} \pm j ω_{n} 1 - ζ^{2}

ke liye damped frequency kya hai?

ω_{d} = ω_{n} 1 - ζ^{2}

Imaginary axis par ek repeated eigenvalue unstable kyun ho sakta hai?

Ye ek

t e^{λ t}

term produce kar sakta hai jo without bound badhta hai (defective/non-simple case).

Connections

State-Space Representation — jahan se $\dot{x} = A x + B u$ aata hai.
Characteristic Polynomial & Determinants — $det (A - λ I)$ kaise compute hota hai.
Diagonalization and Modal Decomposition — $A = V Λ V^{- 1}$ modes ko explicit banata hai.
Routh–Hurwitz Criterion — $n > 2$ ke liye roots nikale bina stability.
Pole Placement & LQR — GNC control design eigenvalues ko left half-plane mein move karta hai.
Damping Ratio and Natural Frequency — complex eigenvalues ki physical reading.
Transfer Functions and Poles — poles = $A$ ke eigenvalues (minimal realization ke).