Intuition The big pictureYou have a linear system x ˙ = A x + B u \dot{x} = Ax + Bu x ˙ = A x + B u eigenvalues of A A A Pole placement says: feed the state back, u = − K x u = -Kx u = − K x closed-loop matrix A − B K A-BK A − B K exactly where you want them . Ackermann's formula is a one-shot recipe for K K K
The solution of x ˙ = A x \dot x = Ax x ˙ = A x e λ i t e^{\lambda_i t} e λ i t λ i \lambda_i λ i A A A
Re ( λ i ) > 0 \text{Re}(\lambda_i) > 0 Re ( λ i ) > 0 Re ( λ i ) < 0 \text{Re}(\lambda_i) < 0 Re ( λ i ) < 0 Im ( λ i ) ≠ 0 \text{Im}(\lambda_i) \neq 0 Im ( λ i ) = 0
So the poles are the dynamics. If we can choose them, we choose settling time, damping, everything.
Definition State feedbackInstead of an external command, drive the input from the measured state:
u = − K x , K = [ k 1 k 2 … k n ] . u = -Kx, \qquad K = [k_1\ k_2\ \dots\ k_n]. u = − K x , K = [ k 1 k 2 … k n ] . x ˙ = A x + B u \dot x = Ax + Bu x ˙ = A x + B u x ˙ = A x − B K x = ( A − B K ) x . \dot x = Ax - BKx = (A-BK)\,x. x ˙ = A x − B K x = ( A − B K ) x . closed-loop matrix is A c l = A − B K A_{cl} = A - BK A c l = A − B K
We are given desired poles μ 1 , … , μ n \mu_1,\dots,\mu_n μ 1 , … , μ n desired characteristic polynomial
α ( s ) = ( s − μ 1 ) ( s − μ 2 ) ⋯ ( s − μ n ) = s n + a n − 1 s n − 1 + ⋯ + a 1 s + a 0 . \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. α ( s ) = ( s − μ 1 ) ( s − μ 2 ) ⋯ ( s − μ n ) = s n + a n − 1 s n − 1 + ⋯ + a 1 s + a 0 .
Goal: find K K K det ( s I − ( A − B K ) ) = α ( s ) \det(sI-(A-BK)) = \alpha(s) det ( s I − ( A − B K )) = α ( s )
The condition for this to work: C \mathcal{C} C invertible — i.e. the system is controllable . If you cannot steer some direction, you cannot place its pole.
The trick is the controllable canonical form (CCF) , where pole placement is trivial, then transform back.
Suppose A , B A,B A , B A c = [ 0 1 0 ⋯ 0 0 1 ⋯ ⋮ − a 0 ′ − a 1 ′ ⋯ − a n − 1 ′ ] , B c = [ 0 ⋮ 0 1 ] . A_c=\begin{bmatrix}0&1&0&\cdots\\0&0&1&\cdots\\ \vdots&&&\\ -a_0'&-a_1'&\cdots&-a_{n-1}'\end{bmatrix},\quad B_c=\begin{bmatrix}0\\ \vdots\\0\\1\end{bmatrix}. A c = 0 0 ⋮ − a 0 ′ 1 0 − a 1 ′ 0 1 ⋯ ⋯ ⋯ − a n − 1 ′ , B c = 0 ⋮ 0 1 . det ( s I − A c ) = s n + a n − 1 ′ s n − 1 + ⋯ + a 0 ′ \det(sI-A_c)=s^n+a_{n-1}'s^{n-1}+\cdots+a_0' det ( s I − A c ) = s n + a n − 1 ′ s n − 1 + ⋯ + a 0 ′ bottom row . Since B c B_c B c K c K_c K c directly overwrites that row :
A c − B c K c has bottom row [ − ( a 0 ′ + k 1 ) … ] . A_c - B_cK_c \text{ has bottom row } [-(a_0'+k_1)\ \dots]. A c − B c K c has bottom row [ − ( a 0 ′ + k 1 ) … ] . α ( s ) \alpha(s) α ( s ) k i = a i − 1 − a i − 1 ′ k_i = a_{i-1} - a_{i-1}' k i = a i − 1 − a i − 1 ′
Every matrix satisfies its own characteristic polynomial. So α ( A c ) \alpha(A_c) α ( A c ) desired polynomial — is not zero; it equals α ( A c ) − χ ( A c ) ⏟ = 0 \alpha(A_c)-\underbrace{\chi(A_c)}_{=0} α ( A c ) − = 0 χ ( A c ) χ \chi χ actual char. poly. Because α \alpha α χ \chi χ α ( A c ) \alpha(A_c) α ( A c ) last row equals exactly the required feedback gains K c K_c K c , while everything above cancels. Concretely one can show
[ 0 ⋯ 0 1 ] α ( A c ) = K c . \begin{bmatrix}0&\cdots&0&1\end{bmatrix}\alpha(A_c)=K_c. [ 0 ⋯ 0 1 ] α ( A c ) = K c .
[ 0 … 0 1 ] [0\dots0\,1] [ 0 … 0 1 ] last coordinate is "actuated." That selector row picks out the row of α ( A c ) \alpha(A_c) α ( A c )
A general controllable ( A , B ) (A,B) ( A , B ) z = T x z = Tx z = T x T = C ~ C − 1 T = \tilde{\mathcal C}\,\mathcal C^{-1} T = C ~ C − 1 T T T K = [ 0 ⋯ 0 1 ] C − 1 α ( A ) . K = \begin{bmatrix}0&\cdots&0&1\end{bmatrix}\mathcal C^{-1}\,\alpha(A). K = [ 0 ⋯ 0 1 ] C − 1 α ( A ) . C − 1 \mathcal C^{-1} C − 1
System A = [ 0 1 0 0 ] A=\begin{bmatrix}0&1\\0&0\end{bmatrix} A = [ 0 0 1 0 ] B = [ 0 1 ] B=\begin{bmatrix}0\\1\end{bmatrix} B = [ 0 1 ] μ = − 2 , − 2 \mu=-2,-2 μ = − 2 , − 2
Step 1 — desired polynomial. α ( s ) = ( s + 2 ) 2 = s 2 + 4 s + 4 \alpha(s)=(s+2)^2=s^2+4s+4 α ( s ) = ( s + 2 ) 2 = s 2 + 4 s + 4 a 1 = 4 , a 0 = 4 a_1=4,a_0=4 a 1 = 4 , a 0 = 4 Why: these coefficients ARE the target dynamics.
Step 2 — controllability matrix.
C = [ B A B ] = [ 0 1 1 0 ] , C − 1 = [ 0 1 1 0 ] . \mathcal C=[B\ AB]=\begin{bmatrix}0&1\\1&0\end{bmatrix},\quad \mathcal C^{-1}=\begin{bmatrix}0&1\\1&0\end{bmatrix}. C = [ B A B ] = [ 0 1 1 0 ] , C − 1 = [ 0 1 1 0 ] . Why: need det C = − 1 ≠ 0 \det\mathcal C=-1\neq0 det C = − 1 = 0
Step 3 — α ( A ) \alpha(A) α ( A ) A 2 = [ 0 0 0 0 ] A^2=\begin{bmatrix}0&0\\0&0\end{bmatrix} A 2 = [ 0 0 0 0 ] α ( A ) = A 2 + 4 A + 4 I = [ 4 4 0 4 ] . \alpha(A)=A^2+4A+4I=\begin{bmatrix}4&4\\0&4\end{bmatrix}. α ( A ) = A 2 + 4 A + 4 I = [ 4 0 4 4 ] . Why: substitute the matrix into the desired polynomial (matrix powers, identity for constant).
Step 4 — assemble.
K = [ 0 1 ] [ 0 1 1 0 ] [ 4 4 0 4 ] = [ 0 1 ] [ 0 4 4 4 ] = [ 4 4 ] . K=[0\ 1]\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}4&4\\0&4\end{bmatrix}=[0\ 1]\begin{bmatrix}0&4\\4&4\end{bmatrix}=[4\ 4]. K = [ 0 1 ] [ 0 1 1 0 ] [ 4 0 4 4 ] = [ 0 1 ] [ 0 4 4 4 ] = [ 4 4 ] .
Check: A − B K = [ 0 1 − 4 − 4 ] A-BK=\begin{bmatrix}0&1\\-4&-4\end{bmatrix} A − B K = [ 0 − 4 1 − 4 ] s 2 + 4 s + 4 s^2+4s+4 s 2 + 4 s + 4 − 2 , − 2 -2,-2 − 2 , − 2
A = [ 0 1 2 − 1 ] A=\begin{bmatrix}0&1\\2&-1\end{bmatrix} A = [ 0 2 1 − 1 ] B = [ 0 1 ] B=\begin{bmatrix}0\\1\end{bmatrix} B = [ 0 1 ] s 2 + s − 2 = 0 ⇒ s = 1 , − 2 s^2+s-2=0 \Rightarrow s=1,-2 s 2 + s − 2 = 0 ⇒ s = 1 , − 2 + 1 +1 + 1 − 1 ± j -1\pm j − 1 ± j
Step 1. α ( s ) = ( s + 1 ) 2 + 1 = s 2 + 2 s + 2 \alpha(s)=(s+1)^2+1=s^2+2s+2 α ( s ) = ( s + 1 ) 2 + 1 = s 2 + 2 s + 2 a 1 = 2 , a 0 = 2 a_1=2,a_0=2 a 1 = 2 , a 0 = 2
Step 2. C = [ B A B ] = [ 0 1 1 − 1 ] \mathcal C=[B\ AB]=\begin{bmatrix}0&1\\1&-1\end{bmatrix} C = [ B A B ] = [ 0 1 1 − 1 ] det = − 1 \det=-1 det = − 1 C − 1 = [ 1 1 1 0 ] \mathcal C^{-1}=\begin{bmatrix}1&1\\1&0\end{bmatrix} C − 1 = [ 1 1 1 0 ]
Step 3. A 2 = [ 2 − 1 − 2 3 ] A^2=\begin{bmatrix}2&-1\\-2&3\end{bmatrix} A 2 = [ 2 − 2 − 1 3 ] α ( A ) = A 2 + 2 A + 2 I = [ 2 − 1 − 2 3 ] + [ 0 2 4 − 2 ] + [ 2 0 0 2 ] = [ 4 1 2 3 ] . \alpha(A)=A^2+2A+2I=\begin{bmatrix}2&-1\\-2&3\end{bmatrix}+\begin{bmatrix}0&2\\4&-2\end{bmatrix}+\begin{bmatrix}2&0\\0&2\end{bmatrix}=\begin{bmatrix}4&1\\2&3\end{bmatrix}. α ( A ) = A 2 + 2 A + 2 I = [ 2 − 2 − 1 3 ] + [ 0 4 2 − 2 ] + [ 2 0 0 2 ] = [ 4 2 1 3 ] .
Step 4. K = [ 0 1 ] [ 1 1 1 0 ] [ 4 1 2 3 ] = [ 0 1 ] [ 6 4 4 1 ] = [ 4 1 ] . K=[0\ 1]\begin{bmatrix}1&1\\1&0\end{bmatrix}\begin{bmatrix}4&1\\2&3\end{bmatrix}=[0\ 1]\begin{bmatrix}6&4\\4&1\end{bmatrix}=[4\ 1]. K = [ 0 1 ] [ 1 1 1 0 ] [ 4 2 1 3 ] = [ 0 1 ] [ 6 4 4 1 ] = [ 4 1 ] .
Check: A − B K = [ 0 1 − 2 − 2 ] A-BK=\begin{bmatrix}0&1\\-2&-2\end{bmatrix} A − B K = [ 0 − 2 1 − 2 ] s 2 + 2 s + 2 s^2+2s+2 s 2 + 2 s + 2 − 1 ± j -1\pm j − 1 ± j
Common mistake Steel-manned common errors1. "Just set A − B K A-BK A − B K μ i \mu_i μ i
Why it feels right: for tiny 2 × 2 2\times2 2 × 2 The fix: it's ad hoc and fails for n ≥ 3 n\ge3 n ≥ 3
2. "Use the OPEN-loop characteristic polynomial in α ( A ) \alpha(A) α ( A )
Why it feels right: Cayley–Hamilton says the actual char poly of A A A some polynomial into A A A The fix: you must use the DESIRED polynomial α \alpha α α ( A ) = 0 ⇒ K = 0 \alpha(A)=0 \Rightarrow K=0 α ( A ) = 0 ⇒ K = 0
3. "It always works."
Why it feels right: the formula looks universal. The fix: it needs C \mathcal C C invertible (controllable) and single input . Multi-input needs different methods (or extra freedom).
4. Sign of K K K With u = − K x u=-Kx u = − K x A − B K A-BK A − B K u = + K x u=+Kx u = + K x A + B K A+BK A + B K
Recall Feynman: explain to a 12-year-oldImagine a wobbly cart on a track. How wobbly, how fast it settles — that's baked into some secret numbers called poles. Left alone the cart might tip over. But you can watch where it is and gently push it based on what you see. Ackermann's formula is a cheat sheet that tells you exactly how hard to push for each thing you measure so the cart settles the way you want — smooth, quick, no tipping. It only works if your pushes can actually reach every wobble (that's "controllable").
Mnemonic Remember the assembly order"Zero-One, Inverse-C, Alpha-of-A" — read right to left it builds the gain:
feed A A A desired polynomial → un-transform with C − 1 \mathcal C^{-1} C − 1 select the bottom row with [ 0 … 0 1 ] [0\dots0\,1] [ 0 … 0 1 ]
What must be nonzero for Ackermann to apply? ⟹ det C \det\mathcal C det C
Which polynomial goes into α ( A ) \alpha(A) α ( A ) desired one.
What is the closed-loop matrix? ⟹ A − B K A-BK A − B K
What is the closed-loop system matrix under u = − K x u=-Kx u = − K x State Ackermann's formula for a single-input system. K = [ 0 ⋯ 0 1 ] C − 1 α ( A ) K=[0\,\cdots\,0\,1]\,\mathcal C^{-1}\,\alpha(A) K = [ 0 ⋯ 0 1 ] C − 1 α ( A ) What is C \mathcal C C [ B A B A 2 B ⋯ A n − 1 B ] [B\ AB\ A^2B\ \cdots\ A^{n-1}B] [ B A B A 2 B ⋯ A n − 1 B ] Which polynomial is α \alpha α α ( A ) \alpha(A) α ( A ) The DESIRED characteristic polynomial from target poles
μ i \mu_i μ i .
Necessary condition for arbitrary pole placement? The pair
( A , B ) (A,B) ( A , B ) is controllable (
C \mathcal C C invertible).
Why does the selector row [ 0 ⋯ 0 1 ] [0\cdots0\,1] [ 0 ⋯ 0 1 ] In controllable canonical form only the last coordinate is actuated, so gains sit in the bottom row.
What theorem underlies the derivation? Cayley–Hamilton (a matrix satisfies its characteristic polynomial).
If you mistakenly use the actual char. poly of A A A K K K K = 0 K=0 K = 0 , since
χ ( A ) = 0 \chi(A)=0 χ ( A ) = 0 .
How do poles relate to system dynamics? Eigenvalues
λ i \lambda_i λ i give modes
e λ i t e^{\lambda_i t} e λ i t : sign of real part sets stability, imaginary part sets oscillation.
Unstable slow oscillatory
Ackermann K = e_n C-inv alpha A
Intuition Hinglish mein samjhoDekho, kisi bhi linear system x ˙ = A x + B u \dot x = Ax + Bu x ˙ = A x + B u poles (matrix A A A state feedback lagate hain: u = − K x u = -Kx u = − K x A − B K A - BK A − B K K K K
Ackermann's formula ek ready-made recipe hai us K K K K = [ 0 ⋯ 0 1 ] C − 1 α ( A ) K = [0\,\cdots\,0\,1]\,\mathcal C^{-1}\,\alpha(A) K = [ 0 ⋯ 0 1 ] C − 1 α ( A ) C = [ B A B ⋯ A n − 1 B ] \mathcal C = [B\ AB\ \cdots\ A^{n-1}B] C = [ B A B ⋯ A n − 1 B ] α ( A ) \alpha(A) α ( A ) desired characteristic polynomial mein matrix A A A K = 0 K=0 K = 0
Ek zaroori condition: system controllable hona chahiye, yaani C \mathcal C C
