Pole placement — Ackermann's formula: Exercises

Throughout, the system is $\overset{x}{˙} = A x + B u$ , feedback is $u = - K x$ , the closed-loop matrix is $A - B K$ , the controllability matrix is $C = [B A B \dots A^{n - 1} B]$ , and $α (s)$ is the desired characteristic polynomial built from your target poles.

Level 1 — Recognition

Can you read the machine and name its parts?

L1.1

For a system of size $n = 2$ with $A = [0 - 3 1 - 4]$ , $B = [01]$ , write down the controllability matrix $C$ and decide whether Ackermann's formula can be applied.

Recall Solution

What we do: stack $B$ next to $A B$ . $A B = [0 - 3 1 - 4] [01] = [1 - 4], C = [B A B] = [01 1 - 4] .$ Why it matters: Ackermann needs $C$ invertible. Here $det C = (0) (- 4) - (1) (1) = - 1 \neq = 0$ , so the system is controllable and Ackermann applies.

L1.2

You are given desired poles $μ_{1} = - 3, μ_{2} = - 5$ . Write the desired characteristic polynomial $α (s)$ and identify $a_{1}$ and $a_{0}$ in $α (s) = s^{2} + a_{1} s + a_{0}$ .

Recall Solution

What we do: multiply the root factors. $α (s) = (s + 3) (s + 5) = s^{2} + 8 s + 15.$ Read off: $a_{1} = 8, a_{0} = 15$ . These coefficients are the target dynamics you want $A - B K$ to have.

L1.3

True or false, with a one-line reason each: (a) Ackermann uses the open-loop characteristic polynomial inside $α (A)$ . (b) The closed-loop matrix under $u = - K x$ is $A - B K$ . (c) If $det C = 0$ you can still place poles anywhere you like.

Recall Solution

(a) False. You substitute $A$ into the desired polynomial $α$ . Using the actual char poly gives $α (A) = 0$ (by Cayley–Hamilton theorem) and hence $K = 0$ . (b) True. Substitute $u = - K x$ into $\overset{x}{˙} = A x + B u$ : $\overset{x}{˙} = (A - B K) x$ . (c) False. $det C = 0$ means some direction is uncontrollable; those poles are stuck. See Eigenvalues and system stability for why stuck unstable poles are fatal.

Level 2 — Application

Run the full recipe end to end.

L2.1

$A = [0010]$ , $B = [01]$ . Place poles at $μ = - 3, - 3$ . Find $K$ .

Recall Solution

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

L2.2

$A = [0 - 2 1 - 3]$ , $B = [01]$ . Place poles at $μ = - 4, - 5$ .

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

$A = [1023]$ , $B = [11]$ . Place poles at $μ = - 1, - 2$ .

Recall Solution

Step 1. $α (s) = (s + 1) (s + 2) = s^{2} + 3 s + 2$ . Step 2. $A B = [1023] [11] = [33]$ , $C = [1133]$ . $det C = (1) (3) - (3) (1) = 0$ — not invertible! The system is uncontrollable; Ackermann does not apply here. Why: $B$ and $A B$ are parallel, so the input can only push along one direction; one pole is frozen. This is a deliberate "degenerate input" case — always check $det C$ before grinding the algebra.

Level 3 — Analysis

Reason about signs, cases, and what the numbers mean.

L3.1

$A = [0510]$ , $B = [01]$ . The open-loop poles are $\pm 5$ (one unstable). Place poles at $- 2 \pm 2 j$ . Find $K$ and confirm the closed loop is stable.

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 $- 2 \pm 2 j$ . Both have $Re = - 2 < 0$ ⟹ stable and oscillatory (see the pole plot below).

Figure — Pole placement — Ackermann's formula

L3.2

Using Worked Example 1 of the parent ( $K = [44]$ for the double integrator with poles $- 2, - 2$ ), suppose a student instead writes the input as $u = + K x$ but keeps $K = [44]$ . What are the actual closed-loop poles they get, and is the system stable?

Recall Solution

What happens: with $u = + K x$ the closed loop is $A + B K$ , not $A - B K$ . $A + B K = [0010] + [01] [44] = [0414] .$ Char poly: $s^{2} - 4 s - 4 = 0 \Rightarrow s = 2 \pm 22 \approx 4.83, - 0.83$ . One pole is positive ⟹ unstable. The sign flip did not merely mislabel — it moved the poles into the right half-plane. Feedback sign is not cosmetic.

Level 4 — Synthesis

Combine ideas, cross topics, work at $n = 3$ or under design specs.

L4.1

$A = 000100010$ (triple integrator), $B = 001$ . Place all three poles at $- 1$ . Find $K$ .

Recall Solution

This $A, B$ is already in Controllable canonical form, so the answer previews the CCF shortcut. Step 1. $α (s) = (s + 1)^{3} = s^{3} + 3 s^{2} + 3 s + 1$ , so $a_{2} = 3, a_{1} = 3, a_{0} = 1$ . Step 2 — controllability. $A B = 010$ , $A^{2} B = 100$ . $C = [B A B A^{2} B] = 001010100, C^{- 1} = C (it is its own inverse) .$ Step 3 — $α (A)$ . Powers of this nilpotent $A$ : $A^{2} = 000000100$ , $A^{3} = 0$ . $α (A) = A^{3} + 3 A^{2} + 3 A + I = 100310331 .$ Step 4. Selector row $[001]$ times $C^{- 1}$ picks the top row of $α (A)$ (because $C^{- 1}$ reverses order here): $[001] C^{- 1} = [100], K = [100] α (A) = [133] .$ Check: for CCF the gains are $k_{i} = a_{i - 1} - a_{i - 1}^{'}$ with open-loop coefficients all zero, so $K = [a_{0} a_{1} a_{2}] = [133]$ ✓. Closed-loop char poly $s^{3} + 3 s^{2} + 3 s + 1 = (s + 1)^{3}$ ✓.

L4.2

Design spec: for a double integrator $A = [0010]$ , $B = [01]$ , you want a critically damped response with natural frequency $ω_{n} = 3$ . This means desired poles $μ = - ω_{n}, - ω_{n} = - 3, - 3$ . Find $K$ and state the resulting $A - B K$ . Compare with what LQR would give (conceptually).

Recall Solution

Translate the spec: critically damped $\Rightarrow$ repeated real poles; $ω_{n} = 3 \Rightarrow$ both at $- 3$ . So $α (s) = (s + 3)^{2} = s^{2} + 6 s + 9$ — identical to L2.1. Hence $K = [96]$ , and $A - B K = [0 - 9 1 - 6]$ . Conceptual compare: state feedback via Ackermann places poles exactly where you dictate; LQR instead minimises a cost $\int (x^{T} Q x + u^{T} R u) d t$ and lets the optimal poles fall out — you tune $Q, R$ rather than pole locations directly. Ackermann = "I choose the poles"; LQR = "I choose the trade-off, poles emerge."

Level 5 — Mastery

Prove, generalise, or exploit duality.

L5.1

Show that for any single-input system already in controllable canonical form (all open-loop coefficients $a_{i}^{'}$ ), Ackermann reduces to the direct rule $k_{i} = a_{i - 1} - a_{i - 1}^{'}$ . Do it for the $n = 2$ case $A_{c} = [0 - a_{0}^{'} 1 - a_{1}^{'}]$ , $B_{c} = [01]$ with desired $α (s) = s^{2} + a_{1} s + a_{0}$ .

Recall Solution

Step 1 — controllability of CCF. $A_{c} B_{c} = [1 - a_{1}^{'}]$ , $C = [01 1 - a_{1}^{'}]$ , $det = - 1$ , $C^{- 1} = [a_{1}^{'} 1 10]$ . Step 2 — the selector row shortcut. $[01] C^{- 1} = [01] [a_{1}^{'} 1 10] = [10]$ . So $K = [10] α (A_{c})$ = top row of $α (A_{c})$ . Step 3 — $α (A_{c})$ top row. With $A_{c}^{2} = [- a_{0}^{'} a_{0}^{'} a_{1}^{'} - a_{1}^{'} a_{1}^{'2} - a_{0}^{'}]$ , $α (A_{c}) = A_{c}^{2} + a_{1} A_{c} + a_{0} I .$ Top row: from $A_{c}^{2}$ : $(- a_{0}^{'}, - a_{1}^{'})$ ; from $a_{1} A_{c}$ : $(0, a_{1})$ ; from $a_{0} I$ : $(a_{0}, 0)$ . Sum: $top row = (a_{0} - a_{0}^{'}, a_{1} - a_{1}^{'}) .$ Conclusion: $K = [a_{0} - a_{0}^{'}, a_{1} - a_{1}^{'}]$ , i.e. $k_{1} = a_{0} - a_{0}^{'}, k_{2} = a_{1} - a_{1}^{'}$ — exactly the "overwrite the bottom row" rule from the parent's Step 1. Ackermann and the CCF shortcut agree. $■$

L5.2 (Duality)

For the same double integrator, an observer needs a gain $L$ placing the poles of $A - L C$ with $C = [10]$ . By duality, $L$ for $(A, C)$ is the transpose of the controller gain $K$ for the dual pair $(A^{T}, C^{T})$ . Place observer poles at $- 4, - 4$ and find $L$ .

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. $A^{T} C^{T} = [01]$ , $C = [C^{T} A^{T} C^{T}] = [1001] = I$ , so $C^{- 1} = I$ . Step 3. $(A^{T})^{2} = 0$ , so $α (A^{T}) = 8 A^{T} + 16 I = [168016]$ . Step 4. $K_{dual} = [01] I α (A^{T}) = [01] [168016] = [816]$ . Transpose back: $L = K_{dual}^{T} = [816]$ . 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 $L = [816]^{T}$