Pole placement — Ackermann's formula: Exercises — Hinglish deep dive

3.5.34 · D4 · Physics › Guidance, Navigation & Control (GNC) › Pole placement — Ackermann's formula

Throughout, system hai $\overset{x}{˙} = A x + B u$ , feedback hai $u = - K x$ , closed-loop matrix hai $A - B K$ , controllability matrix hai $C = [B A B \dots A^{n - 1} B]$ , aur $α (s)$ woh desired characteristic polynomial hai jo tumhare target poles se bani hai.

Level 1 — Recognition

Kya tum machine ko padh sakte ho aur uske parts ko naam de sakte ho?

L1.1

$n = 2$ size ke ek system ke liye jisme $A = [0 - 3 1 - 4]$ , $B = [01]$ hai, controllability matrix $C$ likhо aur decide karo ki Ackermann's formula apply ho sakta hai ya nahi.

Recall Solution

Hum kya karte hain: $B$ ke saath $A B$ ko stack karo. $A B = [0 - 3 1 - 4] [01] = [1 - 4], C = [B A B] = [01 1 - 4] .$ Ye kyun matter karta hai: Ackermann ko $C$ invertible chahiye. Yahan $det C = (0) (- 4) - (1) (1) = - 1 \neq = 0$ , toh system controllable hai aur Ackermann apply hota hai.

L1.2

Tumhe desired poles $μ_{1} = - 3, μ_{2} = - 5$ diye gaye hain. Desired characteristic polynomial $α (s)$ likho aur $a_{1}$ aur $a_{0}$ ko $α (s) = s^{2} + a_{1} s + a_{0}$ mein identify karo.

Recall Solution

Hum kya karte hain: root factors ko multiply karo. $α (s) = (s + 3) (s + 5) = s^{2} + 8 s + 15.$ Read off karo: $a_{1} = 8, a_{0} = 15$ . Ye coefficients wahi target dynamics hain jo tum chahte ho ki $A - B K$ ke paas ho.

L1.3

True ya false, har ek ke saath ek-line reason: (a) Ackermann $α (A)$ ke andar open-loop characteristic polynomial use karta hai. (b) $u = - K x$ ke under closed-loop matrix $A - B K$ hai. (c) Agar $det C = 0$ ho toh bhi tum poles ko jahan chahо wahan place kar sakte ho.

Recall Solution

(a) False. Tum $A$ ko desired polynomial $α$ mein substitute karte ho. Actual char poly use karne se $α (A) = 0$ milta hai (Cayley–Hamilton theorem se) aur isliye $K = 0$ . (b) True. $u = - K x$ ko $\overset{x}{˙} = A x + B u$ mein substitute karo: $\overset{x}{˙} = (A - B K) x$ . (c) False. $det C = 0$ ka matlab hai koi ek direction uncontrollable hai; woh poles stuck hain. Dekho Eigenvalues and system stability kyun stuck unstable poles fatal hote hain.

Level 2 — Application

Poora recipe end to end chalao.

L2.1

$A = [0010]$ , $B = [01]$ . Poles $μ = - 3, - 3$ par place karo. $K$ dhundho.

Recall Solution

Step 1 — desired polynomial. $α (s) = (s + 3)^{2} = s^{2} + 6 s + 9$ , toh $a_{1} = 6, a_{0} = 9$ . Step 2 — controllability. $A B = [10]$ , $C = [0110]$ , $det = - 1$ , $C^{- 1} = [0110]$ . Step 3 — $α (A)$ . $A^{2} = 0$ , toh $α (A) = 6 A + 9 I = [9069]$ . Step 4 — assemble karo. $K = [01] [0110] [9069] = [01] [0996] = [96] .$ Check: $A - B K = [0 - 9 1 - 6]$ , char poly $s^{2} + 6 s + 9$ ✓, poles $- 3, - 3$ .

L2.2

$A = [0 - 2 1 - 3]$ , $B = [01]$ . Poles $μ = - 4, - 5$ par place karo.

Recall Solution

Step 1. $α (s) = (s + 4) (s + 5) = s^{2} + 9 s + 20$ . Step 2. $A B = [1 - 3]$ , $C = [01 1 - 3]$ , $det = - 1$ , $C^{- 1} = [3110]$ . Step 3. $A^{2} = [- 2 6 - 3 7]$ . $α (A) = A^{2} + 9 A + 20 I = [- 2 6 - 3 7] + [0 - 18 9 - 27] + [200020] = [18 - 12 60] .$ Step 4. $K = [01] [3110] [18 - 12 60] = [01] [4218186] = [186] .$ Check: $A - B K = [0 - 20 1 - 9]$ , char poly $s^{2} + 9 s + 20$ ✓, poles $- 4, - 5$ .

L2.3

$A = [1023]$ , $B = [11]$ . Poles $μ = - 1, - 2$ par place karo.

Recall Solution

Step 1. $α (s) = (s + 1) (s + 2) = s^{2} + 3 s + 2$ . Step 2. $A B = [1023] [11] = [33]$ , $C = [1133]$ . $det C = (1) (3) - (3) (1) = 0$ — invertible nahi! System uncontrollable hai; Ackermann yahan apply nahi hota. Kyun: $B$ aur $A B$ parallel hain, toh input sirf ek direction mein push kar sakta hai; ek pole frozen hai. Yeh deliberately "degenerate input" case hai — hamesha $det C$ pehle check karo algebra grind karne se.

Level 3 — Analysis

Signs, cases, aur numbers ka matlab samjho.

L3.1

$A = [0510]$ , $B = [01]$ . Open-loop poles $\pm 5$ hain (ek unstable). Poles $- 2 \pm 2 j$ par place karo. $K$ dhundho aur confirm karo ki closed loop stable hai.

Recall Solution

Step 1. $α (s) = (s + 2)^{2} + 4 = s^{2} + 4 s + 8$ . Step 2. $A B = [10]$ , $C = [0110]$ , $C^{- 1} = [0110]$ . Step 3. $A^{2} = [5005] = 5 I$ . $α (A) = 5 I + 4 A + 8 I = 13 I + 4 A = [1320413] .$ Step 4. $K = [01] [0110] [1320413] = [01] [2013134] = [134] .$ Check: $A - B K = [0 5 - 13 1 0 - 4] = [0 - 8 1 - 4]$ , char poly $s^{2} + 4 s + 8$ ✓, roots $- 2 \pm 2 j$ . Dono ka $Re = - 2 < 0$ hai ⟹ stable aur oscillatory (neeche pole plot dekho).

Figure — Pole placement — Ackermann's formula

L3.2

Parent ke Worked Example 1 ko use karke ( $K = [44]$ double integrator ke liye poles $- 2, - 2$ ke saath), maano ek student input $u = + K x$ likhta hai lekin $K = [44]$ rakhta hai. Unhe actual closed-loop poles kya milenge, aur kya system stable hai?

Recall Solution

Kya hota hai: $u = + K x$ ke saath closed loop $A + B K$ hai, $A - B K$ nahi. $A + B K = [0010] + [01] [44] = [0414] .$ Char poly: $s^{2} - 4 s - 4 = 0 \Rightarrow s = 2 \pm 22 \approx 4.83, - 0.83$ . Ek pole positive hai ⟹ unstable. Sign flip ne sirf label nahi badla — usne poles ko right half-plane mein move kar diya. Feedback sign cosmetic nahi hai.

Level 4 — Synthesis

Ideas combine karo, topics cross karo, $n = 3$ par ya design specs ke under kaam karo.

L4.1

$A = 000100010$ (triple integrator), $B = 001$ . Teeno poles $- 1$ par place karo. $K$ dhundho.

Recall Solution

Yeh $A, B$ pehle se Controllable canonical form mein hai, toh answer CCF shortcut ka preview karta hai. Step 1. $α (s) = (s + 1)^{3} = s^{3} + 3 s^{2} + 3 s + 1$ , toh $a_{2} = 3, a_{1} = 3, a_{0} = 1$ . Step 2 — controllability. $A B = 010$ , $A^{2} B = 100$ . $C = [B A B A^{2} B] = 001010100, C^{- 1} = C (yeh apna khud ka inverse hai) .$ Step 3 — $α (A)$ . Is nilpotent $A$ ki powers: $A^{2} = 000000100$ , $A^{3} = 0$ . $α (A) = A^{3} + 3 A^{2} + 3 A + I = 100310331 .$ Step 4. Selector row $[001]$ times $C^{- 1}$ top row of $α (A)$ pick karta hai (kyunki $C^{- 1}$ yahan order reverse karta hai): $[001] C^{- 1} = [100], K = [100] α (A) = [133] .$ Check: CCF ke liye gains $k_{i} = a_{i - 1} - a_{i - 1}^{'}$ hain jisme open-loop coefficients sab zero hain, toh $K = [a_{0} a_{1} a_{2}] = [133]$ ✓. Closed-loop char poly $s^{3} + 3 s^{2} + 3 s + 1 = (s + 1)^{3}$ ✓.

L4.2

Design spec: double integrator $A = [0010]$ , $B = [01]$ ke liye, tum critically damped response chahte ho jisme natural frequency $ω_{n} = 3$ ho. Iska matlab desired poles $μ = - ω_{n}, - ω_{n} = - 3, - 3$ hain. $K$ dhundho aur resulting $A - B K$ state karo. Conceptually compare karo ki LQR kya deta.

Recall Solution

Spec translate karo: critically damped $\Rightarrow$ repeated real poles; $ω_{n} = 3 \Rightarrow$ dono $- 3$ par. Toh $α (s) = (s + 3)^{2} = s^{2} + 6 s + 9$ — L2.1 se identical hai. Isliye $K = [96]$ , aur $A - B K = [0 - 9 1 - 6]$ . Conceptual comparison: state feedback Ackermann ke zariye poles exactly wahan place karta hai jahan tum dictate karo; LQR instead ek cost $\int (x^{T} Q x + u^{T} R u) d t$ minimise karta hai aur optimal poles khud aate hain — tum pole locations directly tune karne ki bajay $Q, R$ tune karte ho. Ackermann = "main poles choose karta hoon"; LQR = "main trade-off choose karta hoon, poles emerge hote hain."

Level 5 — Mastery

Prove karo, generalise karo, ya duality exploit karo.

L5.1

Dikhao ki kisi bhi single-input system ke liye jo pehle se controllable canonical form mein hai (saare open-loop coefficients $a_{i}^{'}$ ke saath), Ackermann direct rule $k_{i} = a_{i - 1} - a_{i - 1}^{'}$ mein reduce ho jaata hai. Yeh $n = 2$ case ke liye karo: $A_{c} = [0 - a_{0}^{'} 1 - a_{1}^{'}]$ , $B_{c} = [01]$ aur desired $α (s) = s^{2} + a_{1} s + a_{0}$ .

Recall Solution

Step 1 — CCF ki controllability. $A_{c} B_{c} = [1 - a_{1}^{'}]$ , $C = [01 1 - a_{1}^{'}]$ , $det = - 1$ , $C^{- 1} = [a_{1}^{'} 1 10]$ . Step 2 — selector row shortcut. $[01] C^{- 1} = [01] [a_{1}^{'} 1 10] = [10]$ . Toh $K = [10] α (A_{c})$ = $α (A_{c})$ ki top row. Step 3 — $α (A_{c})$ ki top row. $A_{c}^{2} = [- a_{0}^{'} a_{0}^{'} a_{1}^{'} - a_{1}^{'} a_{1}^{'2} - a_{0}^{'}]$ ke saath, $α (A_{c}) = A_{c}^{2} + a_{1} A_{c} + a_{0} I .$ Top row: $A_{c}^{2}$ se: $(- a_{0}^{'}, - a_{1}^{'})$ ; $a_{1} A_{c}$ se: $(0, a_{1})$ ; $a_{0} I$ se: $(a_{0}, 0)$ . Sum: $top row = (a_{0} - a_{0}^{'}, a_{1} - a_{1}^{'}) .$ Conclusion: $K = [a_{0} - a_{0}^{'}, a_{1} - a_{1}^{'}]$ , yaani $k_{1} = a_{0} - a_{0}^{'}, k_{2} = a_{1} - a_{1}^{'}$ — exactly wahi "bottom row overwrite" rule jo parent ke Step 1 se hai. Ackermann aur CCF shortcut agree karte hain. $■$

L5.2 (Duality)

Usi double integrator ke liye, ek observer ko gain $L$ chahiye jo $A - L C$ ke poles $C = [10]$ ke saath place kare. Duality se, $(A, C)$ ke liye $L$ dual pair $(A^{T}, C^{T})$ ke controller gain $K$ ka transpose hai. Observer poles $- 4, - 4$ par place karo aur $L$ dhundho.

Recall Solution

Dual system. $A^{T} = [0100]$ , $B_{dual} = C^{T} = [10]$ . Step 1. $α (s) = (s + 4)^{2} = s^{2} + 8 s + 16$ . Step 2. $A^{T} C^{T} = [01]$ , $C = [C^{T} A^{T} C^{T}] = [1001] = I$ , toh $C^{- 1} = I$ . Step 3. $(A^{T})^{2} = 0$ , toh $α (A^{T}) = 8 A^{T} + 16 I = [168016]$ . Step 4. $K_{dual} = [01] I α (A^{T}) = [01] [168016] = [816]$ . Transpose back karo: $L = K_{dual}^{T} = [816]$ . Check: $A - L C = [0010] - [816] [10] = [- 8 - 16 10]$ , char poly $s^{2} + 8 s + 16 = (s + 4)^{2}$ ✓.

Recall One-line answer key

L2.1 $K = [96]$ ::: L2.2 $K = [186]$ ::: L2.3 uncontrollable ::: L3.1 $K = [134]$ ::: L3.2 unstable, poles $2 \pm 22$ ::: L4.1 $K = [133]$ ::: L4.2 $K = [96]$ ::: L5.1 $k_{i} = a_{i - 1} - a_{i - 1}^{'}$ ::: L5.2 $L = [816]^{T}$