Worked examples — Pole placement — Ackermann's formula

Recall the machine we are feeding: $K=\left[\begin{array}{cccc}0& \cdots & 0& 1\end{array}\right]{\mathcal{C}}^{-1}\alpha (A),\mathcal{C}=\left[\begin{array}{cccc}B& AB& \cdots & A^{n-1}B\end{array}\right].$ Here $A$ is the $n\times n$ system matrix, $B$ the input column, $\alpha (s)$ the desired characteristic polynomial (built from the poles you WANT), and $\mathcal{C}$ the controllability matrix. Every letter here was earned in the parent note — we only USE them now.

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The scenario matrix

Every Ackermann problem lands in one of these cells. The examples below are labelled by cell so you can see the whole space is covered.

Cell	What makes it special	Covered by
A. Stable → faster	open loop already stable, we just want speed	Ex 1
B. Unstable → stabilise	a pole with $\text{Re}>0$ we must drag left	Ex 2
C. Complex target	desired poles are $\sigma \pm j\omega$ (oscillation shaping)	Ex 3
D. Degenerate: uncontrollable	$\det \mathcal{C}=0$, formula BREAKS	Ex 4
E. Zero desired coefficient	a pole exactly at $s=0$ (marginal)	Ex 5
F. Higher order $n=3$	selector row really matters, hand-guessing fails	Ex 6
G. Real-world word problem	cart / satellite phrased in English	Ex 7
H. Exam trap	wrong polynomial / wrong sign temptation	Ex 8

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Cell A — Stable system, want it faster

Forecast: since the system is already stable, do you expect a big gain or a small one? Guess the sign of $k_1,k_2$.

1. Desired polynomial. $\alpha (s)=(s+4)(s+5)=s^2+9s+20$, so $a_1=9,a_0=20$. Why this step? The coefficients $9,20$ literally ARE the target dynamics — see Characteristic polynomial.
2. Controllability matrix. $AB=\left[\begin{array}{c}-1\\ -3\end{array}\right]$, so $\mathcal{C}=\left[\begin{array}{cc}1& -1\\ 1& -3\end{array}\right]$, $\det =-3+1=-2\ne 0$. Why? Non-zero determinant ⟹ controllable ⟹ Ackermann is legal.
3. Invert. ${\mathcal{C}}^{-1}=\frac{1}{-2}\left[\begin{array}{cc}-3& 1\\ -1& 1\end{array}\right]=\left[\begin{array}{cc}1.5& -0.5\\ 0.5& -0.5\end{array}\right]$. Why? This is the "return ticket" from canonical coordinates back to ours.
4. $\alpha (A)$. $A$ is diagonal, so $\alpha (A)=\text{diag}(\alpha (-1),\alpha (-3))$. Compute $\alpha (-1)=1-9+20=12$, $\alpha (-3)=9-27+20=2$. So $\alpha (A)=\left[\begin{array}{cc}12& 0\\ 0& 2\end{array}\right]$. Why? For a diagonal matrix, a polynomial acts on each eigenvalue separately — the cleanest way to see $\alpha (A)\ne 0$.
5. Assemble. ${\mathcal{C}}^{-1}\alpha (A)=\left[\begin{array}{cc}1.5& -0.5\\ 0.5& -0.5\end{array}\right]\left[\begin{array}{cc}12& 0\\ 0& 2\end{array}\right]=\left[\begin{array}{cc}18& -1\\ 6& -1\end{array}\right]$. Then $K=[01]\left[\begin{array}{cc}18& -1\\ 6& -1\end{array}\right]=[6-1]$. Why? Multiplying the two pieces completes the formula, and the selector row $[01]$ picks off the bottom row — the only row that matters, because in canonical coordinates only the last state is actuated.

Verify: $BK=\left[\begin{array}{c}1\\ 1\end{array}\right][6-1]=\left[\begin{array}{cc}6& -1\\ 6& -1\end{array}\right]$, so $A-BK=\left[\begin{array}{cc}-7& 1\\ -6& -2\end{array}\right]$. Trace $=-9=-(4+5)$ ✓, $\det =14+6=20=4\cdot 5$ ✓. Char poly $s^2+9s+20$ — poles at $-4,-5$. Notice $k_2=-1$: pushing against the already-stable second mode isn't always "positive."

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Cell B — Unstable pole, stabilise

Forecast: the unstable pole is at $+1.73$. To reach $-3$ we drag it across zero — expect a sizeable gain on the state that "feeds" the instability.

1. Desired polynomial. $\alpha (s)=(s+3{)}^2=s^2+6s+9$, so $a_1=6,a_0=9$. Why? These coefficients encode the target: both poles at $-3$, no oscillation.
2. Controllability. $AB=\left[\begin{array}{c}1\\ 0\end{array}\right]$, $\mathcal{C}=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$, $\det =-1$, ${\mathcal{C}}^{-1}=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$. Why? $\det =-1\ne 0$ ⟹ controllable ⟹ Ackermann is legal; here $\mathcal{C}$ happens to be its own inverse.
3. $\alpha (A)$. $A^2=\left[\begin{array}{cc}0& 1\\ 3& 0\end{array}\right]\left[\begin{array}{cc}0& 1\\ 3& 0\end{array}\right]=\left[\begin{array}{cc}3& 0\\ 0& 3\end{array}\right]$. So $\alpha (A)=A^2+6A+9I=\left[\begin{array}{cc}3& 0\\ 0& 3\end{array}\right]+\left[\begin{array}{cc}0& 6\\ 18& 0\end{array}\right]+\left[\begin{array}{cc}9& 0\\ 0& 9\end{array}\right]=\left[\begin{array}{cc}12& 6\\ 18& 12\end{array}\right]$. Why? Use the DESIRED polynomial — the trap in mistake #2 of the parent. See Cayley–Hamilton theorem.
4. Assemble. ${\mathcal{C}}^{-1}\alpha (A)=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}12& 6\\ 18& 12\end{array}\right]=\left[\begin{array}{cc}18& 12\\ 12& 6\end{array}\right]$. $K=[01]\cdot \left[\begin{array}{cc}18& 12\\ 12& 6\end{array}\right]=[126]$. Why? The product ${\mathcal{C}}^{-1}\alpha (A)$ transports the canonical answer back into our coordinates; the selector row $[01]$ then reads off the bottom row where the gains live.

Verify: $A-BK=\left[\begin{array}{cc}0& 1\\ 3& 0\end{array}\right]-\left[\begin{array}{cc}0& 0\\ 12& 6\end{array}\right]=\left[\begin{array}{cc}0& 1\\ -9& -6\end{array}\right]$. Char poly $s^2+6s+9$ ✓, roots $-3,-3$. The big $k_1=12$ overpowers the $+3$ that was feeding the instability — exactly as forecast. See Eigenvalues and system stability.

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Cell C — Complex target poles (shape the oscillation)

Forecast: complex poles $\sigma \pm j\omega$ come from $s^2+2|\sigma |s+({\sigma }^2+{\omega }^2)$. Predict the two coefficients before computing.

Figure — the target poles in the complex $s$-plane. The two yellow crosses are our desired poles $-1\pm 2j$. The blue arrow measures the real part ($-1$, the decay rate) and the red arrow measures the imaginary part ($+2$, the wobble frequency). The green half-plane is where poles must sit for stability. Reading this picture tells you the polynomial coefficients before any algebra: the sum of decay pieces gives the $s^1$ coefficient, the squared distance from the origin gives the constant.

1. Desired polynomial. $\alpha (s)=(s+1{)}^2+2^2=s^2+2s+5$, so $a_1=2,a_0=5$. Why? A complex pair $\sigma \pm j\omega$ always factors to $s^2-2\sigma s+({\sigma }^2+{\omega }^2)$; with $\sigma =-1,\omega =2$ that is $s^2+2s+5$. The picture's arrows are exactly $\sigma$ and $\omega$.
2. Controllability. $AB=\left[\begin{array}{c}1\\ 0\end{array}\right]$, $\mathcal{C}=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$, ${\mathcal{C}}^{-1}=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$. Why? $\det \mathcal{C}=-1\ne 0$ ⟹ controllable, so the inverse in Ackermann exists; this permutation matrix is again self-inverse.
3. $\alpha (A)$. $A^2=0$ (the double integrator is nilpotent), so $\alpha (A)=0+2A+5I=\left[\begin{array}{cc}5& 2\\ 0& 5\end{array}\right]$. Why? Substitute the matrix $A$ into the DESIRED polynomial; because $A^2$ vanishes, only the $2A$ and $5I$ terms survive — the cleanest possible $\alpha (A)$.
4. Assemble. ${\mathcal{C}}^{-1}\alpha (A)=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}5& 2\\ 0& 5\end{array}\right]=\left[\begin{array}{cc}0& 5\\ 5& 2\end{array}\right]$. $K=[01]\left[\begin{array}{cc}0& 5\\ 5& 2\end{array}\right]=[52]$. Why? The product returns to our coordinates and the selector row $[01]$ picks the bottom row where the gains live.

Verify: $A-BK=\left[\begin{array}{cc}0& 1\\ -5& -2\end{array}\right]$. Char poly $s^2+2s+5$ ✓. Discriminant $4-20=-16<0$ ⟹ complex roots $-1\pm 2j$ as designed. Compare with LQR — optimal pole placement alternative if you'd rather let a cost function choose $\omega ,\sigma$ for you.

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Cell D — Degenerate: an uncontrollable system

Forecast: $B$ only touches the first state. Can a push on state 1 ever influence state 2 (they're decoupled)? Guess whether Ackermann even runs.

1. Controllability first. $AB=\left[\begin{array}{c}-1\\ 0\end{array}\right]$, so $\mathcal{C}=\left[\begin{array}{cc}1& -1\\ 0& 0\end{array}\right]$, $\det =0$. Why check first? If $\det \mathcal{C}=0$ the inverse in Ackermann does not exist — STOP.
2. Interpret. The second row of $\mathcal{C}$ is all zeros: the $-2$ mode is unreachable. Its pole is fixed at $-2$ no matter what $K$ you choose. Why? A zero row means no combination of $B,AB$ ever excites state 2 — that direction is invisible to the actuator.

Verify: try any $K=[k_1{k}_2]$. $A-BK=\left[\begin{array}{cc}-1-k_1& -k_2\\ 0& -2\end{array}\right]$ — upper triangular, so eigenvalues are $-1-k_1$ (movable) and $-2$ (frozen). We literally cannot place the second pole. This is the "It always works" trap (mistake #3) made concrete. The system is stable here, so it's stabilisable; if that frozen pole were $+2$ we'd be stuck with instability. See Controllability and the controllability matrix.

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Cell E — A desired pole exactly at the origin

Forecast: a pole at $s=0$ means the constant term $a_0$ of $\alpha$ is... which value? Predict it before step 1.

1. Desired polynomial. $\alpha (s)=(s-0)(s+5)=s^2+5s+0$, so $a_1=5,a_0=0$. Why? The zero constant term is the whole point of this cell — a root at $s=0$ forces $a_0=0$.
2. Controllability. $AB=\left[\begin{array}{c}1\\ -3\end{array}\right]$, $\mathcal{C}=\left[\begin{array}{cc}0& 1\\ 1& -3\end{array}\right]$, $\det =-1$, ${\mathcal{C}}^{-1}=\left[\begin{array}{cc}3& 1\\ 1& 0\end{array}\right]$. Why? $\det =-1\ne 0$ ⟹ controllable ⟹ inverse exists ⟹ Ackermann is legal.
3. $\alpha (A)$. $A^2=\left[\begin{array}{cc}0& 1\\ -2& -3\end{array}\right]\left[\begin{array}{cc}0& 1\\ -2& -3\end{array}\right]=\left[\begin{array}{cc}-2& -3\\ 6& 7\end{array}\right]$. Then $\alpha (A)=A^2+5A+0\cdot I=\left[\begin{array}{cc}-2& -3\\ 6& 7\end{array}\right]+\left[\begin{array}{cc}0& 5\\ -10& -15\end{array}\right]=\left[\begin{array}{cc}-2& 2\\ -4& -8\end{array}\right]$. Why include $0\cdot I$? To hammer home that $a_0=0$ drops the identity term — a common slip.
4. Assemble. ${\mathcal{C}}^{-1}\alpha (A)=\left[\begin{array}{cc}3& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}-2& 2\\ -4& -8\end{array}\right]=\left[\begin{array}{cc}-10& -2\\ -2& 2\end{array}\right]$. $K=[01]\left[\begin{array}{cc}-10& -2\\ -2& 2\end{array}\right]=[-22]$. Why? The product transports back to our coordinates; the selector row reads off the bottom-row gains.

Verify: $A-BK=\left[\begin{array}{cc}0& 1\\ -2& -3\end{array}\right]-\left[\begin{array}{cc}0& 0\\ -2& 2\end{array}\right]=\left[\begin{array}{cc}0& 1\\ 0& -5\end{array}\right]$. Char poly $s(s+5)=s^2+5s$ ✓. Determinant $=0$ correctly reflects the pole at the origin. A pole at $0$ means one direction neither grows nor decays — usable but marginal; avoid in flight hardware.

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Cell F — Higher order ($n=3$), where guessing dies

Forecast: in CCF the gains overwrite the bottom row directly (parent Step 1). So $K$ should equal the desired coefficients. Guess them from $(s+2{)}^3$.

1. Desired polynomial. $\alpha (s)=(s+2{)}^3=s^3+6s^2+12s+8$, so $a_2=6,a_1=12,a_0=8$. Why? Three equal poles at $-2$; expanding the cube gives these three coefficients directly.
2. Controllability. $AB=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]$, $A^{2}B=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$. $\mathcal{C}=\left[\begin{array}{ccc}0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right]$ (anti-diagonal), $\det =-1$, and ${\mathcal{C}}^{-1}=\mathcal{C}$ (self-inverse permutation). Why? $\det \ne 0$ ⟹ controllable; this anti-diagonal permutation reverses row order and equals its own inverse.
3. $\alpha (A)$. With this $A$: $A^2=\left[\begin{array}{ccc}0& 0& 1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$, $A^3=0$. So $\alpha (A)=A^3+6A^2+12A+8I=\left[\begin{array}{ccc}8& 12& 6\\ 0& 8& 12\\ 0& 0& 8\end{array}\right]$. Why? Powers of a nilpotent shift matrix vanish quickly — the polynomial fills an upper-triangular band.
4. Assemble. ${\mathcal{C}}^{-1}\alpha (A)=\mathcal{C}\alpha (A)$. Since $\mathcal{C}$ reverses row order, $\mathcal{C}\alpha (A)=\left[\begin{array}{ccc}0& 0& 8\\ 0& 8& 12\\ 8& 12& 6\end{array}\right]$. $K=[001]\cdot \text{(this)}=[8126]$. Why? The selector row $[001]$ picks the last row — which, in CCF, is exactly where the gains land.

Verify: because we were in CCF, $K=[a_0{a}_1{a}_2]=[8126]$ ✓ — the selector row just read off the bottom row, exactly the parent's Step 1 claim. Closed-loop bottom row becomes $[-8-12-6]$, giving char poly $s^3+6s^2+12s+8=(s+2{)}^3$ ✓. Try guessing this by eye — you can't; that's why we have the formula.

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Cell G — Real-world word problem

Forecast: "no oscillation" ⟹ real poles. "Time constant $0.5$ s" ⟹ pole at $-1/0.5=-2$. Guess the target poles before reading.

1. Translate spec to poles. Real, repeated, at $s=-2$ (critically damped, $\tau =0.5\text{ s}$). $\alpha (s)=(s+2{)}^2=s^2+4s+4$. Why? The decay $e^{\lambda t}$ has time constant $1/|\lambda |$; equal real poles = no wobble.
2. This is exactly parent Example 1, so $\mathcal{C}=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$, $\alpha (A)=\left[\begin{array}{cc}4& 4\\ 0& 4\end{array}\right]$. Why? Same plant, same steps — no need to redo controllability and $\alpha (A)$ from scratch.
3. Assemble. $K=[01]\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}4& 4\\ 0& 4\end{array}\right]=[44]$. Why? Selector row picks the bottom row of the transported $\alpha (A)$ — the gains.

Verify (with units!): the control law is $u=-4p-4v$. Units: $[k_1]={\text{s}}^{-2}$ (force-per-position), $[k_2]={\text{s}}^{-1}$ (force-per-velocity) — dimensionally a spring ($-4p$) plus a damper ($-4v$): we built an ideal spring-damper. Closed loop $s^2+4s+4$, poles $-2,-2$, settling time $\approx 4\tau =2\text{ s}$ ✓, no imaginary part ⟹ no oscillation ✓. The whole plan is just: pick poles → run Ackermann.

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Cell H — Exam trap (wrong polynomial / wrong sign)

Forecast: by Cayley–Hamilton theorem, $\chi (A)=0$. What gain does that give? And what does the wrong sign do to a correct gain?

1. Trap 1 — open-loop polynomial. $\chi (A)=A^2+A-2I$. Compute $A^2=\left[\begin{array}{cc}2& -1\\ -2& 3\end{array}\right]$, then $A^2+A-2I=\left[\begin{array}{cc}2& -1\\ -2& 3\end{array}\right]+\left[\begin{array}{cc}0& 1\\ 2& -1\end{array}\right]+\left[\begin{array}{cc}-2& 0\\ 0& -2\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$. Why? Cayley–Hamilton guarantees a matrix kills its OWN char poly. So $K=[01]{\mathcal{C}}^{-1}\cdot 0=[00]$ — no feedback at all, poles unchanged at $1,-2$ (still unstable!). This is mistake #2 of the parent, quantified.
2. Correct computation — Step 2a: desired polynomial. $\alpha (s)=(s+1{)}^2+1=s^2+2s+2$, so $a_1=2,a_0=2$. Why? Poles $-1\pm j$ give $\sigma =-1,\omega =1$, so $s^2-2\sigma s+({\sigma }^2+{\omega }^2)=s^2+2s+2$ — the DESIRED polynomial, not the open-loop one.
3. Step 2b: controllability. $AB=\left[\begin{array}{c}1\\ -1\end{array}\right]$, $\mathcal{C}=\left[\begin{array}{cc}0& 1\\ 1& -1\end{array}\right]$, $\det =-1$, ${\mathcal{C}}^{-1}=\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]$. Why? $\det \ne 0$ ⟹ controllable ⟹ Ackermann applies.
4. Step 2c: $\alpha (A)$. $A^2=\left[\begin{array}{cc}2& -1\\ -2& 3\end{array}\right]$, so $\alpha (A)=A^2+2A+2I=\left[\begin{array}{cc}2& -1\\ -2& 3\end{array}\right]+\left[\begin{array}{cc}0& 2\\ 4& -2\end{array}\right]+\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]=\left[\begin{array}{cc}4& 1\\ 2& 3\end{array}\right]$. Why? Substitute $A$ into the DESIRED polynomial — same computation as parent Example 2.
5. Step 2d: assemble. ${\mathcal{C}}^{-1}\alpha (A)=\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}4& 1\\ 2& 3\end{array}\right]=\left[\begin{array}{cc}6& 4\\ 4& 1\end{array}\right]$. $K=[01]\left[\begin{array}{cc}6& 4\\ 4& 1\end{array}\right]=[41]$. Why? The product returns to our coordinates; the selector row reads off the bottom-row gains $[41]$.
6. Trap 2 — wrong sign. If the friend uses $u=+Kx$ but keeps $K=[41]$, the closed loop is $A+BK=\left[\begin{array}{cc}0& 1\\ 6& 0\end{array}\right]$, char poly $s^2-6$, poles $\pm \sqrt{6}$ — more unstable than the start. Why? With $u=+Kx$ the closed loop is $A+BK$, not $A-BK$; the fix is to use $K=-[41]=[-4-1]$.

Verify: correct case $A-BK=\left[\begin{array}{cc}0& 1\\ 2& -1\end{array}\right]-\left[\begin{array}{cc}0& 0\\ 4& 1\end{array}\right]=\left[\begin{array}{cc}0& 1\\ -2& -2\end{array}\right]$, char poly $s^2+2s+2$, roots $-1\pm j$ ✓. Wrong-polynomial case: $\chi (A)=0$ ✓ so $K=0$, poles stay at $1,-2$. Wrong-sign case: $A+B[41]$ has char poly $s^2-6$ ✓ (unstable) — confirming both traps numerically.

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Recall Self-test across all cells

Which cell has $\det \mathcal{C}=0$, and what happens? ::: Cell D — Ackermann breaks; the unreachable pole is frozen. In CCF ($n=3$), what does $K$ equal directly? ::: The desired polynomial coefficients $[a_0{a}_1{a}_2]$ (Ex 6, $=[8126]$). A spec of "time constant $\tau$, no oscillation" maps to which poles? ::: Repeated real poles at $-1/\tau$ (Ex 7). What gain does plugging the OPEN-loop char poly give? ::: $K=0$ — because $\chi (A)=0$ by Cayley–Hamilton (Ex 8). A desired pole at $s=0$ forces which coefficient to zero? ::: The constant term $a_0=0$ (Ex 5).

See also Observer design and duality (Ackermann for observers) — the same formula, transposed, places observer poles.