Worked examples — Controllability matrix — rank test

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

Yahan sab kuch parent note mein pehle se build kiye gaye ideas se hai: State-space representation $\dot{x}=Ax+Bu$, matrix multiplication, determinant/rank, aur Cayley–Hamilton theorem — jisse pata chalta hai ki hum $A^{n-1}B$ pe kyun rukate hain.

Scenario matrix

Koi bhi numbers aane se pehle, aao possible cases ka space map karte hain. Har controllability problem in cells mein se kisi ek mein hoti hai. Neeche ke 8 examples mein se har ek pe (C2) jaisa tag hai jo batata hai wo kis cell ko fill karta hai.

C1 — Full rank, single input

1. $AB$ compute karo (hume $A^{n-1}B=A^{1}B$ tak chahiye kyunki $n=2$). $AB=\left[\begin{array}{cc}0& 1\\ -2& -3\end{array}\right]\left[\begin{array}{c}0\\ 1\end{array}\right]=\left[\begin{array}{c}1\\ -3\end{array}\right].$ Ye step kyun? Cayley–Hamilton theorem guarantee karta hai ki $A^{2}B$ kuch naya nahi add karta, isliye $\{B,AB\}$ complete list hai.

2. Stack karke $\mathcal{C}$ banao. $\mathcal{C}=[BAB]=\left[\begin{array}{cc}0& 1\\ 1& -3\end{array}\right].$ Ye step kyun? Reachable subspace, $\mathcal{C}$ ka column span hai; side by side stack karna hi tarika hai saari reachable directions collect karne ka.

3. Determinant lo (square matrix ke liye fastest rank test). $\det \mathcal{C}=0\cdot (-3)-1\cdot 1=-1\ne 0.$ Ye step kyun? Non-zero determinant $\Rightarrow$ full rank $=2=n$.

✅ Controllable.

C2 — Decoupled modes, actuator ek ko miss karta hai

1. $AB$ compute karo. $AB=\left[\begin{array}{cc}-1& 0\\ 0& -4\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]=\left[\begin{array}{c}-1\\ 0\end{array}\right].$ Ye step kyun? Diagonal $A$ har coordinate ko scale karta hai; $B$ ki row 2 ka zero, zero hi rehta hai.

2. Stack karo. $\mathcal{C}=\left[\begin{array}{cc}1& -1\\ 0& 0\end{array}\right].$ Ye step kyun? Row 2 poori tarah zero hai — ek bada hint.

3. Rank. Doosri row sab zeros $\Rightarrow \mathrm{rank}=1<2$. Ye step kyun? Ek independent row ka matlab ek reachable dimension. $x_2$ dead direction hai.

❌ Controllable nahi. Mode ${\dot{x}}_2=-4x_2$ bina kisi input ke chalta rehta hai.

C3 — Same $A$, coupling in $B$ isko bacha leta hai

1. $AB$ compute karo. $AB=\left[\begin{array}{c}-1\\ -4\end{array}\right].$ Ye step kyun? Har mode apni eigenvalue se scale hota hai, toh $B$ aur $AB$ ab proportion mein differ karte hain, sirf magnitude mein nahi.

2. Stack karo. $\mathcal{C}=\left[\begin{array}{cc}1& -1\\ 1& -4\end{array}\right].$

3. Determinant. $\det =1\cdot (-4)-(-1)\cdot 1=-4+1=-3\ne 0.$ Ye step kyun? Non-zero $\Rightarrow$ rank $2$.

✅ Controllable.

C4 — Degenerate input $B=0$

1. $AB$ compute karo. $AB=A\cdot 0=\left[\begin{array}{c}0\\ 0\end{array}\right].$ Ye step kyun? $A$ times zero vector zero hi hota hai — $A$ ki har power $0$ pe apply karo, $0$ hi milega.

2. Stack karo. $\mathcal{C}=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right].$

3. Rank. Sab zeros $\Rightarrow \mathrm{rank}=0<2$.

❌ Uncontrollable — reachable subspace sirf origin hai. Ye extreme limiting case hai: zero rank, zero reachable dimensions.

C5 — Multi-input: do weak inputs, ek strong system

1. $AB$ compute karo. Kyunki $A=0$, $AB=0$ aur $A^{2}B=0$. Ye step kyun? $A=0$ ke saath, higher power blocks gayab ho jaate hain — sirf $B$ khud reachable directions carry karta hai.

2. Stack karo $\mathcal{C}=[BAB{A}^{2}B]$, size $3\times 6$: $\mathcal{C}=\left[\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\end{array}\right].$

3. Rank. Sirf do non-zero rows $\Rightarrow \mathrm{rank}=2<3$.

❌ Uncontrollable. Direction 3 unreachable hai — koi thruster nahi, koi coupling nahi.

C6 — $n=3$ jahan $A^{2}B$ genuinely matter karta hai

1. $AB$ compute karo. $AB=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right].$ Ye step kyun? $A$ "1" ko ek row upar shift karta hai — ye cascade action mein hai.

2. $A^{2}B=A(AB)$ compute karo. $A^{2}B=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right].$ Ye step kyun? Ab "1" top state tak pahunch gaya — ye final power hi hai jo last dimension fill karta hai. Isse skip karna galat taur pe rank-2 dikhata.

3. Stack karo aur check karo. $\mathcal{C}=\left[\begin{array}{ccc}0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right].$ Ye anti-diagonal identity hai: $\det =+1\cdot (0\cdot 0-1\cdot 1)\cdots$ — ye note karna zyada clean hai ki ye ek permutation matrix hai, isliye $\det =\pm 1\ne 0$, rank $3$. Ye step kyun? Full rank $=3=n$.

✅ Controllable. Isliye parent kehta hai "$A^{n-1}B$ pe ruko" aur pehle nahi — yahan $A^{2}B$ essential tha.

C7 — Repeated-eigenvalue trap

1. Case (a): $A_{a}B$ compute karo. $A_{a}B=\left[\begin{array}{cc}2& 1\\ 0& 2\end{array}\right]\left[\begin{array}{c}0\\ 1\end{array}\right]=\left[\begin{array}{c}1\\ 2\end{array}\right].$ Stack: ${\mathcal{C}}_a=\left[\begin{array}{cc}0& 1\\ 1& 2\end{array}\right]$, $\det =0\cdot 2-1\cdot 1=-1\ne 0$. Controllable ✅ Ye step kyun? Off-diagonal $1$ do states ko mix karta hai, toh ek single input dono tak pahunchta hai.

2. Case (b): $A_{b}B$ compute karo. $A_{b}B=\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]\left[\begin{array}{c}1\\ 1\end{array}\right]=\left[\begin{array}{c}2\\ 2\end{array}\right].$ Stack: ${\mathcal{C}}_b=\left[\begin{array}{cc}1& 2\\ 1& 2\end{array}\right]$, $\det =1\cdot 2-2\cdot 1=0$. Uncontrollable ❌ Ye step kyun? Ek diagonalizable repeated eigenvalue ke saath, do modes kisi bhi single input ke liye indistinguishable hote hain — $B$ aur $A_{b}B$ parallel hain.

C8 — Word problem + exam twist

1. Physics se $(A,B)$ likho. $A=\left[\begin{array}{cc}0& 1\\ -k& 0\end{array}\right],B=\left[\begin{array}{c}0\\ b\end{array}\right].$ Ye step kyun? Do scalar equations se $\dot{x}=Ax+Bu$ padhna — ye State-space representation step hai.

2. $AB$ compute karo. $AB=\left[\begin{array}{cc}0& 1\\ -k& 0\end{array}\right]\left[\begin{array}{c}0\\ b\end{array}\right]=\left[\begin{array}{c}b\\ 0\end{array}\right].$ Ye step kyun? Sirf $A^{1}B$ chahiye kyunki $n=2$.

3. Stack karo aur symbolically determinant lo. $\mathcal{C}=\left[\begin{array}{cc}0& b\\ b& 0\end{array}\right],\det \mathcal{C}=0\cdot 0-b\cdot b=-b^2.$ Ye step kyun? Symbolic determinant se hum directly failure condition solve kar sakte hain.

4. $\det =0$ solve karo. $-b^2=0⟺b=0$. Ye step kyun? Exactly tab uncontrollable hoga jab determinant zero ho.

Jawab: har $b\ne 0$ ke liye controllable ($k$ ki kisi bhi value pe!), aur sirf $b=0$ pe uncontrollable — physically obvious "dead thruster" case. Stiffness $k$ controllability ko kabhi affect nahi karta.

Recap: kis cell ne kya sikhaya

Practice reflexes:

Connections

• Controllability matrix — rank test (parent — wo machine jinhe ye examples exercise karte hain)
• Observability matrix — rank test (dual: same drill $A^T,C^T$ pe chalao)
• State-space representation (C8 ka translation step)
• Cayley–Hamilton theorem (har example mein $A^{n-1}B$ pe kyun rukate hain)
• Pole placement & state feedback (C1/C3/C6 controllability se saare poles place ho sakte hain)
• Kalman decomposition (C2/C4/C5 uncontrollable modes yahan split hote hain)
• Matrix exponential $e^{At}$ (reachability ke peeche propagator)
• Stabilizability (C7b ka uncontrollable mode unstable hai → stabilizable bhi nahi)

Concept Map