Pole placement — Ackermann's formula

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

WHY — poles ko move karna kyun chahte hain?

$\dot{x}=Ax$ ka solution $e^{{\lambda }_{i}t}$ jaise terms se banta hai jahan ${\lambda }_i$, $A$ ke eigenvalues hain.

• $\text{Re}({\lambda }_i)>0$ ⟹ blow up ho jaata hai (unstable).
• $\text{Re}({\lambda }_i)<0$ ⟹ decay karta hai (stable); zyada negative = zyada fast.
• $\text{Im}({\lambda }_i)\ne 0$ ⟹ oscillation hoti hai.

Toh poles hi dynamics hain. Agar hum unhe choose kar sakein, toh settling time, damping, sab kuch choose kar sakte hain.

WHAT — Ackermann kaunsi problem solve karta hai?

Humein desired poles ${\mu }_1,\dots ,{\mu }_n$ diye gaye hain. Desired characteristic polynomial banao: $\alpha (s)=(s-{\mu }_1)(s-{\mu }_2)\cdots (s-{\mu }_n)=s^n+a_{n-1}{s}^{n-1}+\cdots +a_{1}s+a_0.$

Goal: $K$ nikalo taki $\det (sI-(A-BK))=\alpha (s)$ ho.

Is kaam karne ki shart: $\mathcal{C}$ invertible hona chahiye — matlab system controllable ho. Agar aap kisi direction ko steer nahi kar sakte, toh uska pole place nahi kar sakte.

HOW — Ackermann ko scratch se derive karna

Trick hai controllable canonical form (CCF), jahan pole placement trivial hai, phir wapas transform karo.

Step 1 — CCF mein poles place karna trivial hai

Maano $A,B$ pehle se CCF mein hain: $A_c=\left[\begin{array}{cccc}0& 1& 0& \cdots \\ 0& 0& 1& \cdots \\ ⋮& & & \\ -a_0^{\prime }& -a_1^{\prime }& \cdots & -a_{n-1}^{\prime }\end{array}\right],B_c=\left[\begin{array}{c}0\\ ⋮\\ 0\\ 1\end{array}\right].$ Yahan $\det (sI-A_c)=s^n+a_{n-1}^{\prime }{s}^{n-1}+\cdots +a_0^{\prime }$ — coefficients bottom row par rehte hain. Kyunki $B_c$ sirf last row ko touch karta hai, feedback $K_c$ directly us row ko overwrite karta hai: $A_c-B_c{K}_c\text{ ki bottom row }[-(a_0^{\prime }+k_1)\dots ]\text{ hoti hai}.$ Desired $\alpha (s)$ paane ke liye bas $k_i=a_{i-1}-a_{i-1}^{\prime }$ set karo. Aasaan — lekin sirf CCF mein.

Step 2 — Cayley–Hamilton engine hai

Har matrix apni khud ki characteristic polynomial satisfy karta hai. Toh $\alpha (A_c)$ — desired polynomial use karke — zero nahi hota; yeh equals hota hai $\alpha (A_c)-\underset{=0}{\underbrace{\chi (A_c)}}$ jahan $\chi$ actual char. poly hai. Kyunki $\alpha$ aur $\chi$ sirf coefficients mein alag hain, $\alpha (A_c)$ is tarah collapse karta hai ki uski last row exactly required feedback gains $K_c$ ke barabar hoti hai, jabki upar sab cancel ho jaata hai. Concretely dikhaya ja sakta hai ki: $\left[\begin{array}{cccc}0& \cdots & 0& 1\end{array}\right]\alpha (A_c)=K_c.$

Step 3 — Coordinate change undo karo

Ek general controllable $(A,B)$, CCF mein map hota hai $z=Tx$ ke zariye jahan $T=\tilde{\mathcal{C}}{\mathcal{C}}^{-1}$ (canonical aur actual controllability matrices ka product) hai. CCF result ko $T$ ke through wapas push karke aur simplify karke, transformation matrices merge ho jaati hain aur general formula nikal aata hai: $K=\left[\begin{array}{cccc}0& \cdots & 0& 1\end{array}\right]{\mathcal{C}}^{-1}\alpha (A).$ ${\mathcal{C}}^{-1}$ literally CCF se aapke coordinates mein "return ticket" hai.

Worked Example 1 — ek double integrator

System $A=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$, $B=\left[\begin{array}{c}0\\ 1\end{array}\right]$. Poles $\mu =-2,-2$ par place karo.

Step 1 — desired polynomial. $\alpha (s)=(s+2{)}^2=s^2+4s+4$, toh $a_1=4,a_0=4$. Kyun: ye coefficients target dynamics HAIN.

Step 2 — controllability matrix. $\mathcal{C}=[BAB]=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right],{\mathcal{C}}^{-1}=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right].$ Kyun: $\det \mathcal{C}=-1\ne 0$ chahiye ⟹ controllable hai, toh Ackermann apply hoga.

Step 3 — $\alpha (A)$. $A^2=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$, toh $\alpha (A)=A^2+4A+4I=\left[\begin{array}{cc}4& 4\\ 0& 4\end{array}\right].$ Kyun: matrix ko desired polynomial mein substitute karo (matrix powers, constant ke liye identity).

Step 4 — assemble karo. $K=[01]\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}4& 4\\ 0& 4\end{array}\right]=[01]\left[\begin{array}{cc}0& 4\\ 4& 4\end{array}\right]=[44].$

Check: $A-BK=\left[\begin{array}{cc}0& 1\\ -4& -4\end{array}\right]$, char. poly $s^2+4s+4$. ✓ Poles $-2,-2$ par hain.

Worked Example 2 — ek unstable pole ko stabilise karna

$A=\left[\begin{array}{cc}0& 1\\ 2& -1\end{array}\right]$, $B=\left[\begin{array}{c}0\\ 1\end{array}\right]$. Open-loop poles $s^2+s-2=0\Rightarrow s=1,-2$ se solve hote hain ($+1$ unstable hai). $-1\pm j$ par place karo.

Step 1. $\alpha (s)=(s+1{)}^2+1=s^2+2s+2$, toh $a_1=2,a_0=2$.

Step 2. $\mathcal{C}=[BAB]=\left[\begin{array}{cc}0& 1\\ 1& -1\end{array}\right]$, $\det =-1$, ${\mathcal{C}}^{-1}=\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]$.

Step 3. $A^2=\left[\begin{array}{cc}2& -1\\ -2& 3\end{array}\right]$. $\alpha (A)=A^2+2A+2I=\left[\begin{array}{cc}2& -1\\ -2& 3\end{array}\right]+\left[\begin{array}{cc}0& 2\\ 4& -2\end{array}\right]+\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]=\left[\begin{array}{cc}4& 1\\ 2& 3\end{array}\right].$

Step 4. $K=[01]\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}4& 1\\ 2& 3\end{array}\right]=[01]\left[\begin{array}{cc}6& 4\\ 4& 1\end{array}\right]=[41].$

Check: $A-BK=\left[\begin{array}{cc}0& 1\\ -2& -2\end{array}\right]$, char poly $s^2+2s+2$ ⟹ roots $-1\pm j$. ✓

Recap

Flashcards

Connections

• Controllability and the controllability matrix
• Cayley–Hamilton theorem
• Eigenvalues and system stability
• Controllable canonical form
• State feedback control
• LQR — optimal pole placement alternative
• Observer design and duality (Ackermann for observers)
• Characteristic polynomial

Concept Map

Deep Dive

• D1 · Foundations — har symbol zero se
• D2 · Visual walkthrough — derivation pictures mein
• D3 · Worked examples — har scenario
• D4 · Exercises — graded, full solutions ke saath
• D5 · Question bank — concept traps