Worked examples — Eigenvalues of A — system modes, stability

3.5.30 · D3 · Physics › Guidance, Navigation & Control (GNC) › Eigenvalues of A — system modes, stability

Kuch bhi shuru karne se pehle, teen choti vocabulary ki baatein jo parent ne use ki thi lekin ek newcomer ko pakki honi chahiye.

Ab woh map jis par har answer rehta hai:

• Horizontal axis hai $\sigma =\text{Re}(\lambda )$ — real part, growth/decay knob. Vertical line ke left = decays = accha.
• Vertical axis hai $\omega =\text{Im}(\lambda )$ — imaginary part, wobble knob. Jitna upar = utni tez oscillation.
• Vertical line khud ($\sigma =0$) imaginary axis hai, stable aur unstable ke beech ka knife-edge.

Neeche sab kuch bas yeh hai: "$\det (A-\lambda I)=0$ solve karo, phir is map par dot dhundho."

---

The scenario matrix

Neeche ke $2\times 2$ examples har distinct eigenvalue picture ko cover karte hain jo ek real $2\times 2$ produce kar sakta hai; final $3\times 3$ dikhata hai ki wahi reading higher order mein kaise extend hoti hai (yeh ek representative $3\times 3$ — ek stable companion form — treat karta hai, na ki har $3\times 3$ sub-case).

Cell	Eigenvalue signature	Map par kahan	Behaviour	Example
A	Do real, dono $<0$, distinct	Dono dots left axis par	Stable, koi oscillation nahi (overdamped)	Ex 1
B	Complex pair, $\sigma <0$	Mirror pair, left side	Stable, oscillates (underdamped)	Ex 2
C	Complex pair, $\sigma >0$	Mirror pair, right side	Unstable, growing oscillation	Ex 3
D	Do real, opposite signs	Ek left, ek right	Unstable (saddle)	Ex 4
E	Pure imaginary, $\sigma =0$, simple	Pair on vertical line	Marginally stable (constant wobble)	Ex 5
F	Repeated $\lambda =0$ (defective)	Double dot at origin	Unstable via $t$-growth	Ex 6
G	Repeated real $<0$ ($\zeta =1$, critically damped)	Double dot on left axis	Stable, fastest non-oscillating	Ex 7a
H	Limiting sweep: $\zeta$ from $0\to 1\to \infty$	Dots slide toward axis / merge / split	Damping continuum	Ex 7b
I	$3\times 3$ real-world: attitude + integrator	Teen dots	Design/stability read	Ex 8

Ab hum har cell cover karte hain.

---

Cell A — do real negative eigenvalues (overdamped)

Forecast: bada damping ($4$) versus modest stiffness ($3$) — guess: stable, aur oscillate karne ke liye zyada hi damped. Do real negative roots.

Step 1 — characteristic polynomial. $\det (A-\lambda I)={\lambda }^2+\frac{c}{m}\lambda +\frac{k}{m}={\lambda }^2+4\lambda +3=0.$ Yeh step kyun? Yeh ek companion matrix hai, toh (upar ki definition se) determinant sirf ODE ka polynomial hai jo coefficients se padha — fastest route.

Step 2 — factor. ${\lambda }^2+4\lambda +3=(\lambda +1)(\lambda +3)=0\Rightarrow {\lambda }_1=-1,{\lambda }_2=-3.$ Yeh step kyun? Real, distinct, dono negative → Cell A, map ke far-left par dono dots, koi imaginary part nahi, toh koi oscillation nahi.

Step 3 — read damping. ${\omega }_n=\sqrt{k/m}=\sqrt{3}$, aur $2\zeta {\omega }_n=c/m=4$, toh $\zeta =\frac{4}{2\sqrt{3}}=\frac{2}{\sqrt{3}}\approx 1.155>1.$ Yeh step kyun? $\zeta >1$ confirm karta hai overdamped — do real roots ka mathematical signature.

Verify: discriminant $=4^2-4\cdot 3=4>0$ → do real roots (matches). Slowest root $-1$ settling time set karta hai; response hai $c_1{e}^{-t}+c_2{e}^{-3t}$, purely decaying. ✓

---

Cell B — complex pair, left half-plane (underdamped stable)

Forecast: chhota damping ($2$) versus stiffness ($4$) — guess: stable lekin ringing (complex pair, left side).

Step 1 — polynomial. $\det (A-\lambda I)={\lambda }^2+2\lambda +4=0.$ Yeh step kyun? Same companion-form shortcut.

Step 2 — roots. $\lambda =\frac{-2\pm \sqrt{4-16}}{2}=\frac{-2\pm \sqrt{-12}}{2}=-1\pm j\sqrt{3}.$ Yeh step kyun? Negative discriminant $=-12<0$ → $\sqrt{-12}=2\sqrt{3}j$ mein $j$ aata hai, jo ek complex pair mark karta hai; $\sigma =-1<0$ → Cell B, map ke left half mein mirror dots.

Step 3 — physical numbers. ${\omega }_n=\sqrt{4}=2$; $\zeta {\omega }_n=1\Rightarrow \zeta =0.5$; damped frequency ${\omega }_d={\omega }_n\sqrt{1-{\zeta }^2}=2\sqrt{1-0.25}=\sqrt{3}\approx \mathrm{1.732.}$ Yeh step kyun? ${\omega }_d$ exactly imaginary part hai — woh actual wobble rate jo tum measure karoge.

Verify: $\sigma =-\zeta {\omega }_n=-0.5\cdot 2=-1$ ✓; $|\lambda |=\sqrt{1+3}=2={\omega }_n$ ✓. Stable, underdamped, $\sqrt{3}$ rad/s par ringing karta hai. Parent ke phase-plane picture mein inward spiral dekho.

---

Cell C — complex pair, right half-plane (growing oscillation)

Forecast: positive "damping" energy inject karta hai — guess: phir bhi oscillate karta hai (stiffness unchanged) lekin ab grows. Complex pair right par.

Step 1 — polynomial. $\det (A-\lambda I)={\lambda }^2-2\lambda +4=0.$ Yeh step kyun? $+2$ ki wajah se $\lambda$ coefficient negative ho jaata hai — pehla warning sign.

Step 2 — roots. $\lambda =\frac{2\pm \sqrt{4-16}}{2}=+1\pm j\sqrt{3}.$ Yeh step kyun? Discriminant abhi bhi $-12<0$ → complex pair (phir $j$), lekin ab $\sigma =+1>0$ → Cell C, map ke right par mirror dots.

Step 3 — verdict. Envelope $e^{+t}$ oscillation ko multiply karta hai → magnitude ringing karte hue blow up hoti hai. Unstable. Yeh step kyun? Sirf $\sigma$ stability decide karta hai; wobble blow-up ke liye irrelevant hai.

Verify: Routh–Hurwitz Criterion se compare karo: ${\lambda }^2+a_1\lambda +a_0$ ke liye stability chahiye $a_1>0$ aur $a_0>0$. Yahan $a_1=-2<0$ → fail → unstable ✓.

---

Cell D — real, opposite signs (saddle)

Forecast: $+2$ positive feedback hai → guess ek growing, ek decaying mode. Ek saddle: Cell D.

Step 1 — polynomial. $\det (A-\lambda I)={\lambda }^2+\lambda -2=0.$ Yeh step kyun? Negative constant term $-2$ note karo — $2\times 2$ ke liye, constant term $=\det A={\lambda }_1{\lambda }_2<0$ turant opposite signs force karta hai.

Step 2 — factor. $(\lambda +2)(\lambda -1)=0\Rightarrow \lambda =-2,+1.$ Yeh step kyun? Real, ek negative ek positive → confirms Cell D.

Step 3 — verdict. $e^{+t}$ mode apne eigenvector ke along grow karta hai → unstable saddle.

Verify: $\det A=(0)(-1)-(1)(2)=-2={\lambda }_1{\lambda }_2=(-2)(1)=-2$ ✓. $\text{tr}A=-1={\lambda }_1+{\lambda }_2=-2+1=-1$ ✓.

---

Cell E — pure imaginary, simple (marginal, constant wobble)

Forecast: ek frictionless spring — energy conserved, toh guess: forever oscillate karta hai, kabhi grow nahi karta, kabhi die nahi karta. Dots imaginary axis par. Cell E.

Step 1 — polynomial. $\det (A-\lambda I)={\lambda }^2+9=0.$ Yeh step kyun? Koi first-order term nahi kyunki damping zero hai.

Step 2 — roots. ${\lambda }^2=-9\Rightarrow \lambda =\pm 3j.$ Yeh step kyun? $\sigma =0$ exactly, $\omega =3$. Dots map ki vertical line par baithte hain.

Step 3 — verdict. Roots axis par simple hain (distinct) → marginally stable: $\mathbf{x}(t)$ phase space mein ek bounded circle par rehta hai, hamesha $3$ rad/s par oscillate karta hai. Yeh step kyun? Marginal stability sirf isliye allowed hai kyunki axis roots non-repeated hain (Cell F se contrast karo).

Verify: ${\omega }_n=\sqrt{9}=3$, $\zeta =0$ → ${\omega }_d={\omega }_n\sqrt{1-0}=3$ matches $\text{Im}(\lambda )$ ✓. $|e^{3jt}|=1$ sabhi $t$ ke liye → bounded, decay nahi karta ✓.

---

Cell F — repeated zero, defective (woh sneaky unstable wala)

Forecast: eigenvalues axis par hain, toh tempting hai marginal call karna. Lekin repeated root dhyaan se dekho.

Step 1 — polynomial. $\det (A-\lambda I)={\lambda }^2=0\Rightarrow \lambda =0,0.$ Yeh step kyun? Origin par repeated eigenvalue — danger flag.

Step 2 — count eigenvectors. $A\mathbf{v}=0$ solve karo: $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)\mathbf{v}=0$ sirf $\mathbf{v}=(1,0{)}^T$ deta hai — double root ke liye ek eigenvector. Matrix defective hai. Yeh step kyun? Repeated-on-axis sirf tab marginal hota hai jab uske paas eigenvectors ka full set ho (semisimple). Yahan nahi hai.

Step 3 — actual solution. Ek missing eigenvector ke saath solution mein $t$ term aata hai: $\mathbf{x}(t)=\left(\begin{array}{c}{x}_1(0)+t{x}_2(0)\\ x_2(0)\end{array}\right).$ $t{x}_2(0)$ term without bound grow karta hai → unstable. Yeh step kyun? Yeh parent ka mistake #3 hai, concrete roop mein: $t{e}^{0t}=t\to \infty$.

Verify: $\ddot{q}=0$ directly integrate karo: $q(t)=q(0)+\dot{q}(0)t$ — ek straight line, unbounded jab bhi $\dot{q}(0)\ne 0$ ✓. Repeated axis root + missing eigenvector = unstable ✓.

---

Cell G — repeated real negative eigenvalue (critically damped, $\zeta =1$)

Forecast: $\zeta =\frac{c/m}{2{\omega }_n}=\frac{4}{4}=1$ exactly — Cell A (do real, oscillation-free lekin sluggish) aur Cell B (complex, ringing) ke beech ka tipping point. Guess: ek single repeated real negative root, koi oscillation nahi, bina ringing ke fastest possible decay.

Step 1 — polynomial. $\det (A-\lambda I)={\lambda }^2+4\lambda +4=0.$ Yeh step kyun? Phir companion-form shortcut.

Step 2 — roots. ${\lambda }^2+4\lambda +4=(\lambda +2{)}^2=0\Rightarrow \lambda =-2,-2.$ Yeh step kyun? Discriminant $=16-16=0$ exactly → ek repeated real root. Yeh knife-edge $\zeta =1$ hai: Cell G, left axis par $-2$ par double dot.

Step 3 — verdict. Dono roots mein $\sigma =-2<0$ hai → stable, aur koi imaginary part nahi → koi oscillation nahi. Cell F ke unlike, yahan axis root strictly negative hai, toh $t{e}^{-2t}$ term bhi phir bhi decay karta hai (exponential linear growth ko beat karta hai). Yeh critically damped hai: bina ringing ke fastest decay jo tum pa sakte ho. Yeh step kyun? Cell F se carefully contrast karo — ek repeated root sirf tab dangerous hai jab woh imaginary axis par baitha ho; left half-plane mein ek repeated root bilkul stable hai.

Verify: discriminant $=4^2-4\cdot 4=0$ ✓; $\zeta =\frac{4}{2\cdot 2}=1$ ✓; $t{e}^{-2t}\to 0$ as $t\to \infty$ (exponential dominates) → stable ✓.

---

Cell H — limiting sweep: dots ko move karte dekho jaise $\zeta$ $0\to 1\to \infty$ jaata hai

Forecast: jaise hum damping add karte hain oscillation fade honi chahiye; guess: dono dots imaginary axis se slide karke inward aayenge, $\zeta =1$ par real axis par collide karenge, phir real axis ke along split ho jayenge.

Step 1 — general roots. ${\lambda }^2+4\zeta \lambda +4=0\Rightarrow \lambda =-2\zeta \pm 2\sqrt{{\zeta }^2-1}.$ Yeh step kyun? Ek formula poora sweep cover karta hai.

Step 2 — regimes, har ek apne cell se tagged.

• $\zeta =0$: $\lambda =\pm 2j$ → Cell E, axis par pure wobble.
• $\zeta =0.5$: $\lambda =-1\pm j\sqrt{3}$ → Cell B, spiralling in (yeh exactly Ex 2 hai).
• $\zeta =1$: $\lambda =-2,-2$ → Cell G, critically damped, dots real axis par collide karte hain (yeh exactly Ex 7a hai).
• $\zeta =2$: $\lambda =-4\pm 2\sqrt{3}=-0.54,-7.46$ → Cell A, do real, overdamped.

Yeh step kyun? Yeh continuous geometry dikhata hai: $\zeta$ badhane se $\sigma =-2\zeta$ leftward drag hota hai jabki imaginary part $2\sqrt{1-{\zeta }^2}$ $\zeta =1$ par zero ho jaata hai — aur har stop ek aise cell mein land karta hai jo hum pehle se work kar chuke hain.

Step 3 — picture padho. $\zeta <1$ ke liye dots radius ${\omega }_n=2$ ke circle par ride karte hain (kyunki $|\lambda |={\omega }_n$); $\zeta =1$ par woh $-2$ par milte hain; $\zeta >1$ ke liye woh real axis par walk karte hain, ek $0$ ki taraf, ek $-\infty$ ki taraf. Yeh step kyun? Circle key limiting-behaviour insight hai: underdamped eigenvalues hamesha origin se ${\omega }_n$ distance par baithe hote hain.

Verify: $\zeta =0.5$ par, $|\lambda |=\sqrt{1+3}=2={\omega }_n$ ✓. $\zeta =1$ par discriminant $=4({\zeta }^2-1)=0$ → repeated $-2$ ✓. $\zeta =2$ par, roots ka product $=(-0.536)(-7.464)=4=k/m$ ✓.

---

Cell I — ek $3\times 3$ real-world design read

Forecast: teen states, toh teen eigenvalues. Ek well-designed loop unhe sabko left half-plane mein dalna chahiye. Guess: stable.

Step 1 — characteristic polynomial. Kyunki ek $3\times 3$ companion matrix ka bottom row $-a_0,-a_1,-a_2$ list karta hai (yahan $-6,-11,-6$), hum read off karte hain $\det (A-\lambda I)={\lambda }^3+6{\lambda }^2+11\lambda +6=0.$ Yeh step kyun? Same companion-matrix property jaise $2\times 2$ cases mein thi, ek dimension upar — general determinant expand karne ki zarurat nahi.

Step 2 — factor. ${\lambda }^3+6{\lambda }^2+11\lambda +6=(\lambda +1)(\lambda +2)(\lambda +3)=0\Rightarrow \lambda =-1,-2,-3.$ Yeh step kyun? Chhote integer roots try karo ($\lambda =-1$ kaam karta hai), phir factor down karo. Teeno real aur negative → sabhi dots left par. Stable.

Step 3 — Routh–Hurwitz se cross-check (taaki agli baar factor na karna pade). ${\lambda }^3+a_2{\lambda }^2+a_1\lambda +a_0$ ke liye jisme $a_2=6,a_1=11,a_0=6$ ho, Routh–Hurwitz Criterion chahiye ki saare coefficients positive hoon aur $a_2{a}_1>a_0$: $6\cdot 11=66>6.✓$ Yeh step kyun? Higher order ke liye, factoring aksar impossible hoti hai; Routh–Hurwitz akele coefficients se same stable/unstable verdict deta hai. Yeh Characteristic Polynomial & Determinants aur Pole Placement & LQR se connect hota hai (jahan tum yeh coefficients choose karte ho).

Verify: slowest (least-negative) eigenvalue $-1$ hai → settling time $\approx 4/1=4$ s set karta hai. Sum of roots $=-6=-a_2$ ✓; product $=(-1)(-2)(-3)=-6$, aur ek cubic ${\lambda }^3+\cdots +a_0$ ke liye roots ka product $=-a_0=-6$ ✓.

---

Recall Kaun sa cell kaun sa hai — quick self-test

$\sigma <0$ ke saath complex roots kaun se cell mein rehte hain? ::: Cell B — stable, underdamped (oscillating decay). $\det A<0$ wala ek $2\times 2$ hamesha kaun sa cell hota hai? ::: Cell D — opposite sign ke real roots (saddle), hamesha unstable. Sirf ek eigenvector ke saath repeated $\lambda =0$ stable hai ya unstable? ::: Unstable (Cell F) — missing eigenvector ek $t$-term produce karta hai jo grow karta hai. $-2$ par repeated real root (dono copies) — stable ya unstable? ::: Stable (Cell G) — left half-plane mein ek repeated root phir bhi decay karta hai, kyunki $t{e}^{-2t}\to 0$. ${\omega }_n$ fixed ke saath $\zeta \to 0$ hone par eigenvalues kahan jaate hain? ::: Imaginary axis par $\pm j{\omega }_n$ par (Cell E), marginally stable. Underdamped roots origin se kitni door hote hain? ::: Exactly ${\omega }_n$ distance par, kyunki $|\lambda |={\omega }_n$.

Related: State-Space Representation · Diagonalization and Modal Decomposition · Damping Ratio and Natural Frequency · Transfer Functions and Poles.