Worked examples — Observability matrix — rank test

3.5.33 · D3 · Physics › Guidance, Navigation & Control (GNC) › Observability matrix — rank test

Hum sirf parent ke core objects ko hi touch karte hain: observability matrix $\mathcal{O}$, uska rank aur null space, aur — last example ke liye — PBH test. Kuch naya assume nahi kiya gaya.

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Scenario matrix

Har observability problem in case classes mein se kisi ek mein aata hai. Is page ka goal hai har row ko kam se kam ek baar hit karna.

#	Case class	Isme tricky kya hai	Kahan aata hai
A	Full-rank, distinct eigenvalues, single output	"Textbook yes" wala case	Ex 1
B	Hidden decoupled state	$\mathcal{O}$ ka poora ek column zero ho jaata hai	Ex 2
C	Repeated eigenvalue, 2 eigenvectors	Scalar output unhe alag nahi kar sakta	Ex 3
D	Repeated eigenvalue, 1 eigenvector (Jordan block)	Coupling observability bacha leti hai	Ex 4
E	Zero / degenerate input: $C=0$	Blind sensor — limiting case	Ex 5
F	$3\times 3$ system, kya 2nd sensor add karne se fayda hoga?	$p$ ke saath rank badhta hai	Ex 6
G	Real-world: spacecraft attitude + gyro bias	Word problem, $\ker \mathcal{O}$ ka physical meaning	Ex 7
H	Exam twist: rank poora lagta hai but PBH disagree karta hai?	Do tests cross-check karo — agree hote hain	Ex 8

Recall Recipe (ek baar yaad karo, neeche reuse karo)

1. $n$ padhho ($A$ ki size), $p$ padhho ($C$ ki rows).
2. $C,CA,C{A}^2,\dots ,C{A}^{n-1}$ compute karo.
3. Unhe $\mathcal{O}$ mein stack karo (size $pn\times n$).
4. $\mathrm{rank}(\mathcal{O})=n\Rightarrow$ observable; kam $\Rightarrow$ $\ker \mathcal{O}$ mein unobservable direction dhoondo.

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Example 1 — Case A: textbook "yes"

Step 1 — doosri row $CA$ banao. $CA=\left[\begin{array}{cc}1& 1\end{array}\right]\left[\begin{array}{cc}-1& 0\\ 0& -3\end{array}\right]=\left[\begin{array}{cc}-1& -3\end{array}\right].$ Yeh step kyun? $n=2$ ke saath hume exactly $C$ aur $CA$ chahiye — ek power upar, aur nahi (recipe step 2).

Step 2 — stack karo. $\mathcal{O}=\left[\begin{array}{cc}1& 1\\ -1& -3\end{array}\right].$ Yeh step kyun? $\mathcal{O}$ ki rows woh directions hain jo sensor $x(0)$ aur uski derivative ke through "dekh" sakta hai.

Step 3 — rank. $\det \mathcal{O}=(1)(-3)-(1)(-1)=-3+1=-2\ne 0$, toh $\mathrm{rank}=2=n$ ⇒ observable. ✅ Yeh step kyun? Nonzero determinant matlab $x(0)$ ki koi bhi direction zero output nahi deti.

Verify: dono modes ke distinct eigenvalues $-1,-3$ hain, aur $C$ ka har eigenvector par nonzero entry hai ($C=[11]$). Distinct eigenvalues + $C$ har eigenvector ko touch karna — bilkul wahi condition hai jo PBH test chahta hai — dono answers agree karte hain.

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Example 2 — Case B: ek hidden decoupled state

Step 1 — $CA$. $CA=\left[\begin{array}{cc}1& 0\end{array}\right]\left[\begin{array}{cc}-1& 0\\ 0& -3\end{array}\right]=\left[\begin{array}{cc}-1& 0\end{array}\right].$ Yeh step kyun? Recipe — derivative information lo.

Step 2 — stack karo aur columns dekho. $\mathcal{O}=\left[\begin{array}{cc}1& 0\\ -1& 0\end{array}\right].$ Poora doosra column zero hai. Yeh kyun matter karta hai: zero column matlab direction $x_0=[0,1{]}^{\top }$ multiply hokar kuch nahi deti.

Step 3 — rank. $\mathrm{rank}=1<2$ ⇒ not observable. ❌

Step 4 — ghost ko naam do. $\ker \mathcal{O}=\mathrm{span}\{[0,1{]}^{\top }\}$. Check karo: $C{e}^{At}[0,1{]}^{\top }=\left[\begin{array}{cc}1& 0\end{array}\right]\left[\begin{array}{c}0\\ e^{-3t}\end{array}\right]=0$ har $t$ ke liye.

Verify: unobservable state $y(t)\equiv 0$ deta hai — bilkul invisible, exactly jaisa parent note ki boxed definition of unobservable state ne promise kiya tha.

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Example 3 — Case C: repeated eigenvalue, do eigenvectors

Step 1 — $CA$. $CA=\left[\begin{array}{cc}1& 1\end{array}\right]\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]=\left[\begin{array}{cc}2& 2\end{array}\right].$

Step 2 — stack karo. $\mathcal{O}=\left[\begin{array}{cc}1& 1\\ 2& 2\end{array}\right].$ Row 2 $=2\times$ Row 1. Yeh kyun matter karta hai: rows parallel hain, toh sirf 1-D space span karte hain.

Step 3 — rank. $\det =1\cdot 2-1\cdot 2=0$ ⇒ $\mathrm{rank}=1<2$ ⇒ not observable. ❌

Step 4 — ghost. $\ker \mathcal{O}=\mathrm{span}\{[1,-1{]}^{\top }\}$: dono states ka difference invisible hai.

Verify: eigenvalue $2$ ki geometric multiplicity $2$ hai (do independent eigenvectors). Ek scalar output kabhi yeh resolve nahi kar sakta — tumhe $C$ mein $p\ge 2$ rows chahiye hongi. Parent ki "modal degeneracy" warning se match karta hai.

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Example 4 — Case D: repeated eigenvalue, EK eigenvector (coupling bacha leta hai)

Step 1 — $CA$. $CA=\left[\begin{array}{cc}1& 0\end{array}\right]\left[\begin{array}{cc}2& 1\\ 0& 2\end{array}\right]=\left[\begin{array}{cc}2& 1\end{array}\right].$ Yeh step kyun? Coupling term $1$, state 2 ko measured state 1 ki derivative mein leak karta hai.

Step 2 — stack karo. $\mathcal{O}=\left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right].$

Step 3 — rank. $\det =1\cdot 1-0\cdot 2=1\ne 0$ ⇒ $\mathrm{rank}=2=n$ ⇒ observable! ✅

Verify: Ex 3 se fark geometric multiplicity hai. Ek repeated eigenvalue jiska single eigenvector ho (ek genuine Jordan block) ki geometric multiplicity $1$ hoti hai; PBH test ko phir sirf yeh chahiye ki $C$ us ek eigenvector ko touch kare — jo $C=[10]$ karta hai. Toh repeated eigenvalues automatically fatal nahi hote: faisla eigenvalue count nahi, eigenvector count karta hai.

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Example 5 — Case E: blind sensor ($C=0$)

Step 1 — $CA$. $CA=\left[\begin{array}{cc}0& 0\end{array}\right]A=\left[\begin{array}{cc}0& 0\end{array}\right].$ Yeh step kyun? $A$ ki koi bhi power ko zero row se left-multiply karo, result zero hi rahega.

Step 2 — stack karo. $\mathcal{O}=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right].$

Step 3 — rank. $\mathrm{rank}=0<2$ ⇒ not observable — maximum degree tak. ❌

Step 4 — ghost sab kuch hai. $\ker \mathcal{O}={\mathbb{R}}^2$: har initial state $y\equiv 0$ deta hai.

Verify: rank $0$ rank test ka floor hai; "observable directions ki sankhya" $\mathrm{rank}(\mathcal{O})=0$ ke barabar hai, toh $n=2$ states mein se zero recoverable hain — ek dead sensor ke liye yeh bilkul sahi limiting answer hai.

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Example 6 — Case F: kya second sensor help karta hai? ($3\times 3$)

Step 1 — $A$ ki powers times $C_1$. $C_1=\left[\begin{array}{ccc}1& 0& 0\end{array}\right],C_{1}A=\left[\begin{array}{ccc}0& 1& 0\end{array}\right],C_1{A}^2=\left[\begin{array}{ccc}0& 0& 1\end{array}\right].$ Yeh step kyun? $n=3$ hai toh $A^2$ tak jaana padega.

Step 2 — single sensor ke liye stack karo. ${\mathcal{O}}_1=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=I_3,\mathrm{rank}=3=n.$ Pehle se hi Observable! Kyun: position → velocity → acceleration; har derivative agla state reveal kar deti hai.

Step 3 — two-sensor case (kya isse nuksan hoga?). $C_{2}A=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 0\end{array}\right],C_2{A}^2=\left[\begin{array}{ccc}0& 0& 1\\ 0& 0& 0\end{array}\right].$ ${\mathcal{O}}_2=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\\ 0& 0& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right],\mathrm{rank}=3=n.$ Yeh step kyun? Extra sensor rows rank ko sirf badha ya maintain kar sakte hain — kabhi ghata nahi sakte — lekin rank $n=3$ par cap hai.

Verify: dono $\mathrm{rank}=3$ dete hain. Lesson: integrators ki chain apni pehli state se akele observable hoti hai; extra sensor yahan redundant hai (rank pehle se saturated hai). Rows add karna observability kabhi reduce nahi karta — bas $n$ se zyada nahi ho sakti.

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Example 7 — Case G: real-world spacecraft attitude + gyro bias

Step 1 — $CA$. $CA=\left[\begin{array}{cc}1& 0\end{array}\right]\left[\begin{array}{cc}0& -1\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}0& -1\end{array}\right].$ Yeh step kyun? Bias, $\dot{\theta }$ mein couple hota hai, toh measured angle ki derivative mein $b$ ki information hoti hai.

Step 2 — stack karo. $\mathcal{O}=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right],\det =-1\ne 0,\mathrm{rank}=2=n.$

Step 3 — verdict. Observable ✅ — star tracker attitude aur bias ko alag kar sakta hai.

Verify (physical + units): $\dot{\theta }=-b$ (with $u=0$), toh nonzero constant bias measured angle ko linearly drift karaata hai: $\theta (t)={\theta }_0-bt$. Drift ka slope $b$ reveal karta hai (units: rad/s) aur intercept ${\theta }_0$ reveal karta hai (rad) — do observations, do unknowns, exactly wahi jo rank $2$ guarantee karta hai. Isi liye ek Kalman filter (Kalman Filter) ek single angle measurement se gyro bias estimate aur remove kar sakta hai.

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Example 8 — Case H: exam twist, kya rank test aur PBH agree karte hain?

Step 1 — rank test. $CA=\left[\begin{array}{cc}-1& 0\end{array}\right],\mathcal{O}=\left[\begin{array}{cc}1& 0\\ -1& 0\end{array}\right],\mathrm{rank}=1<2.$ ⇒ not observable. ❌

Step 2 — PBH cross-check. Yeh step kyun? PBH test har eigenvalue $\lambda$ par check karta hai ki $\left[\begin{array}{c}A-\lambda I\\ C\end{array}\right]$ ka full column rank $n$ hai ya nahi. $\lambda =-1$ par: $\left[\begin{array}{c}A-(-1)I\\ C\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\\ 1& 0\end{array}\right],\mathrm{rank}=1<2.$

Step 3 — reconcile karo. Dono tests kehte hain not observable. Student ki galti: $-1$ ek repeated eigenvalue hai jiske do eigenvectors hain (yeh $-I$ hai, geometric multiplicity $2$), aur ek scalar $C$ do identical modes resolve nahi kar sakta — yeh bilkul Case C wali failure hai disguise mein.

Verify: $\ker \mathcal{O}=\mathrm{span}\{[0,1{]}^{\top }\}$, aur $C{e}^{At}[0,1{]}^{\top }=\left[\begin{array}{cc}1& 0\end{array}\right]\left[\begin{array}{c}0\\ e^{-t}\end{array}\right]=0$. Rank test aur PBH zaroor agree karte hain — yeh same geometry ke do alag windows hain, jaisa parent ke connections mein bataya gaya hai.

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Connections

• Observability matrix — rank test (parent)
• PBH test (Ex 8 mein cross-check ke liye use kiya)
• Kalman Filter (Ex 7 bias estimation)
• Kalman decomposition (Ex 2, 3, 5 ke ghost states ko alag karta hai)
• Rank and null space of a matrix (har step yahan rehta hai)
• Cayley–Hamilton theorem (isliye hum $A^{n-1}$ par rukते hain)
• Controllability matrix — rank test (dual)
• State-space representation