Ackermann's formula

Consider a linear continuous-time invariant system with a state-space representation

$$\backslash begin\{align\}$$

\mathbf\dot{x}(t) &= \mathbf{Ax}(t) + \mathbf{Bu}(t) \\

\mathbf{y}(t) &= \mathbf{Cx}(t)

\end{align}

where {{math|x}} is the state vector, {{math|u}} is the input vector, and {{math|A, B, C}} are matrices of compatible dimensions that represent the dynamics of the system. An input-output description of this system is given by the transfer function

$$\backslash begin\{align\}$$

G(s) &= \mathbf{C}(s \mathbf{I} - \mathbf{A})^{-1} \mathbf{B} \\[4pt]

&= \mathbf{C}\ \frac{\operatorname{adj}(s \mathbf{I} - \mathbf{A})}{\det(s \mathbf{I} - \mathbf{A})}\ \mathbf{B}.

\end{align}

where {{math|det}} is the determinant and {{math|adj}} is the adjugate.

Since the denominator of the right equation is given by the characteristic polynomial of {{math|A}}, the poles of {{mvar|G}} are eigenvalues of {{math|A}} (note that the converse is not necessarily true, since there may be cancellations between terms of the numerator and the denominator). If the system is unstable, or has a slow response or any other characteristic that does not specify the design criteria, it could be advantageous to make changes to it. The matrices {{math|A, B, C}}, however, may represent physical parameters of a system that cannot be altered. Thus, one approach to this problem might be to create a feedback loop with a gain {{math|k}} that will feed the state variable {{math|x}} into the input {{math|u}}.

If the system is controllable, there is always an input {{math|u(t)}} such that any state {{math|x{{sub|0}}}} can be transferred to any other state {{math|x(t)}}. With that in mind, a feedback loop can be added to the system with the control input {{math|1=u(t) = r(t) − kx(t)}}, such that the new dynamics of the system will be

$$\backslash begin\{align\}$$

\mathbf\dot{x}(t) &= \mathbf{Ax}(t) + \mathbf{B}[\mathbf{r}(t) - \mathbf{kx}(t)] \\[2pt]

&= [\mathbf{A} - \mathbf{Bk}] \mathbf{x}(t) + \mathbf{Br}(t), \\[4pt]

\mathbf{y}(t) &= \mathbf{Cx}(t).

\end{align}

In this new realization, the poles will be dependent on the characteristic polynomial {{math|Δ{{sub|new}}}} of {{math|A − Bk}}, that is

$$\backslash Delta\_\backslash text\{new\}(s)\; =\; \backslash det\backslash bigl(s\; \backslash mathbf\{I\}\; -\; (\backslash mathbf\{A\}\; -\; \backslash mathbf\{Bk\})\; \backslash bigr).$$

Computing the characteristic polynomial and choosing a suitable feedback matrix can be a challenging task, especially in larger systems. One way to make computations easier is through Ackermann's formula. For simplicity's sake, consider a single input vector with no reference parameter {{math|r}}, such as

$$\backslash begin\{align\}$$

\mathbf{u}(t) &= -\mathbf{k}^{\rm T} \mathbf{x}(t) \\[2pt]

\mathbf\dot{x}(t) &= \mathbf{Ax}(t) - \mathbf{Bk}^{\rm T} \mathbf{x}(t),

\end{align}

where {{math|k{{sup|T}}}} is a feedback vector of compatible dimensions. Ackermann's formula states that the design process can be simplified by only computing the following equation:

$$\backslash mathbf\{k\}^\{\backslash rm\; T\}\; =\; \backslash begin\{bmatrix\}$$

0 & \cdots & 0 & 1

\end{bmatrix} \, \mathcal{C}^{-1} \Delta_\text{new}(\mathbf{A}),

in which {{math|Δ{{sub|new}}(A)}} is the desired characteristic polynomial evaluated at matrix {{math|A}}, and $\backslash mathcal\{C\}$ is the controllability matrix of the system.

This proof is based on Encyclopedia of Life Support Systems entry on Pole Placement Control.{{Cite book|title=Control systems, robotics and automation|last=Ackermann|first=J. E.|date=2009|publisher=Eolss Publishers Co. Ltd|others=Unbehauen, Heinz.|isbn=9781848265905|location=Oxford|chapter=Pole Placement Control|oclc=703352455}} Assume that the system is controllable. The characteristic polynomial of $\backslash mathbf\{A\}\_\{\backslash rm\; CL\}\; :=\; (\backslash mathbf\{A\}\; -\; \backslash mathbf\{Bk\}^\{\backslash rm\; T\})$ is given by

$$\backslash Delta(\backslash mathbf\{A\}\_\{\backslash rm\; CL\})\; =\; (\backslash mathbf\{A\}\_\{\backslash rm\; CL\})^n\; +\; \backslash sum\_\{k=0\}^\{n-1\}\; \backslash alpha\_k\; \backslash mathbf\{A\}\_\{\backslash rm\; CL\}^\{k\}$$

Calculating the powers of {{math|A{{sub|CL}}}} results in

$$\backslash begin\{align\}$$

(\mathbf{A}_{\rm CL})^0 =&\ (\mathbf{A} - \mathbf{Bk}^{\rm T})^0 = \mathbf{I} \\[4pt]

(\mathbf{A}_{\rm CL})^1 =&\ (\mathbf{A} - \mathbf{Bk}^{\rm T})^1 = \mathbf{A} - \mathbf{Bk}^{\rm T} \\[4pt]

(\mathbf{A}_{\rm CL})^2 =&\ (\mathbf{A} - \mathbf{Bk}^{\rm T})^2 \\[2pt]

&= \mathbf{A}^2 - \mathbf{ABk}^{\rm T} - \mathbf{Bk}^{\rm T} \mathbf{A} + (\mathbf{Bk}^{\rm T})^2 \\[2pt]

&= \mathbf{A}^2 - \mathbf{ABk}^{\rm T} - (\mathbf{Bk}^{\rm T})[\mathbf{A} - \mathbf{Bk}^{\rm T}] \\[2pt]

&= \mathbf{A}^2 - \mathbf{ABk}^{\rm T} - \mathbf{Bk}^{\rm T} \mathbf{A}_{\rm CL} \\[4pt]

\vdots \ & \\[4pt]

(\mathbf{A}_{\rm CL})^n =&\ (\mathbf{A}-\mathbf{Bk}^{\rm T})^n \\[2pt]

&= \mathbf{A}^n - \mathbf{A}^{n-1} \mathbf{Bk}^{\rm T} - \mathbf{A}^{n-2} \mathbf{Bk}^{\rm T} \mathbf{A}_{\rm CL}- \ldots -\mathbf{Bk}^{\rm T} \mathbf{A}_{\rm CL}^{n-1}

\end{align}

Replacing the previous equations into {{math|Δ(A{{sub|CL}})}} yields

$$\backslash begin\{align\}$$

\Delta(\mathbf{A}_{\rm CL}) &= \overbrace{(\mathbf{A}^n-\mathbf{A}^{n-1} \mathbf{Bk}^{\rm T}-\mathbf{A}^{n-2} \mathbf{Bk}^{\rm T} \mathbf{A}_{\rm CL}-\ldots-\mathbf{Bk}^{\rm T} \mathbf{A}_{\rm CL}^{n-1})}^{(\mathbf{A}_{\rm CL})^n}

+ \overbrace{\ldots + \alpha_2(\mathbf{A}^2-\mathbf{ABk}^{\rm T}-\mathbf{Bk}^{\rm T} \mathbf{A}_{\rm CL}) + \alpha_1 (\mathbf{A}-\mathbf{Bk}^{\rm T})+\alpha_0\mathbf{I}}^{\sum_{k=0}^{n-1} \alpha_k \mathbf{A}_{\rm CL}^k} \\[4pt]

&= (\mathbf{A}^n + \alpha_{n-1}\mathbf{A}^{n-1} + \ldots + \alpha_2 \mathbf{A}^2 + \alpha_1 \mathbf{A} + \alpha_0 \mathbf{I}) - (\mathbf{A}^{n-1}\mathbf{Bk}^{\rm T} + \mathbf{A}^{n-2}\mathbf{Bk}^{\rm T} \mathbf{A}_{\rm CL} + \ldots + \mathbf{Bk}^{\rm T}\mathbf{A}_{\rm CL}^{n-1}) + \ldots -\alpha_2 (\mathbf{ABk}^{\rm T} + \mathbf{Bk}^{\rm T} \mathbf{A}_{\rm CL}) - \alpha_1(\mathbf{Bk}^{\rm T}) \\[4pt]

&= \Delta(\mathbf{A}) - (\mathbf{A}^{n-1}\mathbf{Bk}^{\rm T} + \mathbf{A}^{n-2} \mathbf{Bk}^{\rm T} \mathbf{A}_{\rm CL} + \ldots + \mathbf{Bk}^{\rm T} \mathbf{A}_{\rm CL}^{n-1}) - \ldots - \alpha_2 (\mathbf{ABk}^{\rm T} + \mathbf{Bk}^{\rm T} \mathbf{A}_{\rm CL}) - \alpha_1 (\mathbf{Bk}^{\rm T})

\end{align}

Rewriting the above equation as a matrix product and omitting terms that {{math|k{{sup|T}}}} does not appear isolated yields

$$\backslash Delta(\backslash mathbf\{A\}\_\{\backslash rm\; CL\})\; =\; \backslash Delta(\backslash mathbf\{A\})\; -\; \backslash begin\{bmatrix\}$$

\mathbf{B} & \mathbf{AB} & \cdots & \mathbf{A}^{n-1}\mathbf{B}

\end{bmatrix}\begin{bmatrix}

\star \\

\vdots \\

\mathbf{k}^{\rm T}

\end{bmatrix}

From the Cayley–Hamilton theorem, {{math|1=Δ(A{{sub|CL}}) = 0}}, thus

\begin{bmatrix}

\mathbf{B} & \mathbf{AB} & \cdots & \mathbf{A}^{n-1}\mathbf{B}

\end{bmatrix}\begin{bmatrix}

\star \\

\vdots \\

\mathbf{k}^{\rm T}

\end{bmatrix} = \Delta(\mathbf{A})

Note that is the controllability matrix of the system. Since the system is controllable, $\backslash mathcal\{C\}$ is invertible. Thus,

\begin{bmatrix}

\star \\

\vdots \\

\mathbf{k}^{\rm T}

\end{bmatrix} = \mathcal{C}^{-1} \Delta(\mathbf{A})

To find {{math|k{{sup|T}}}}, both sides can be multiplied by the vector $$

\begin{bmatrix}

0 & \cdots & 0 & 1

\end{bmatrix}

giving

$$\backslash begin\{bmatrix\}$$

0 & \cdots & 0 & 1

\end{bmatrix}\begin{bmatrix}

\star \\

\vdots \\

\mathbf{k}^{\rm T}

\end{bmatrix} = \begin{bmatrix}

0 & \cdots & 0 & 1

\end{bmatrix} \, \mathcal{C}^{-1} \Delta(\mathbf{A})

Thus,

$$\backslash mathbf\{k\}^\{\backslash rm\; T\}\; =$$

\begin{bmatrix}

0 & \cdots & 0 & 1

\end{bmatrix} \, \mathcal{C}^{-1} \Delta(\mathbf{A})

Consider{{cite web|url=http://web.mit.edu/16.31/www/Fall06/1631_topic13.pdf |title=Topic #13 : 16.31 Feedback Control|website=Web.mit.edu |access-date=2017-07-06}}

\mathbf\dot{x} = \begin{bmatrix}

1 & 1 \\

1 & 2

\end{bmatrix}\mathbf{x} + \begin{bmatrix}

1 \\

0

\end{bmatrix} \mathbf{u}

We know from the characteristic polynomial of {{math|A}} that the system is unstable since

$$\backslash begin\{align\}$$

\det(s\mathbf{I} - \mathbf{A}) &= (s-1)(s-2) - 1 \\

&= s^2 - 3s + 2,

\end{align}

the matrix {{math|A}} will only have positive eigenvalues. Thus, to stabilize the system we shall put a feedback gain

$$\backslash mathbf\{k\}\; =\; \backslash begin\{bmatrix\}$$

k_1 & k_2 \end{bmatrix}.

From Ackermann's formula, we can find a matrix {{math|k}} that will change the system so that its characteristic equation will be equal to a desired polynomial. Suppose we want $\backslash Delta\_\backslash text\{desired\}(s)\; =\; s^2\; +\; 11s\; +\; 30.$

Thus, $\backslash Delta\_\backslash text\{desired\}(\backslash mathbf\{A\})\; =\; \backslash mathbf\{A\}^2\; +\; 11\backslash mathbf\{A\}\; +\; 30\backslash mathbf\{I\}$ and computing the controllability matrix yields

$$\backslash begin\{align\}$$

\mathcal{C} &= \begin{bmatrix}

\mathbf{B} & \mathbf{AB}

\end{bmatrix} = \begin{bmatrix}

1 & 1 \\

0 & 1

\end{bmatrix} \\[4pt]

\implies \mathcal{C}^{-1} &= \begin{bmatrix}

1 & -1 \\

0 & 1

\end{bmatrix}

\end{align}

Also, we have that $\backslash mathbf\{A\}^2\; =\; \backslash left[\backslash begin\{smallmatrix\}$

2 & 3 \\

3 & 5

\end{smallmatrix}\right].

Finally, from Ackermann's formula

$$\backslash begin\{align\}$$

\mathbf{k}^{\rm T} &= \begin{bmatrix}

0 & 1

\end{bmatrix}\begin{bmatrix}

1 & -1\\

0 & 1

\end{bmatrix}\left(\begin{bmatrix}

2 & 3\\

3 & 5

\end{bmatrix} + 11\begin{bmatrix}

1 & 1\\

1 & 2

\end{bmatrix} + 30\mathbf{I} \right) \\[2pt]

&= \begin{bmatrix}

0 & 1

\end{bmatrix} \begin{bmatrix}

1 & -1\\

0 & 1

\end{bmatrix}\begin{bmatrix}

43 & 14\\

14 & 57

\end{bmatrix} \\[2pt]

&= \begin{bmatrix}

0 & 1

\end{bmatrix} \begin{bmatrix}

29 & -43\\

14 & 57

\end{bmatrix} \\[6pt]

&= \begin{bmatrix}

14 & 57

\end{bmatrix}

\end{align}

Ackermann's formula can also be used for the design of state observers. Consider the linear discrete-time observed system

$$\backslash begin\{align\}$$

\mathbf\hat{x}(n+1) &= \mathbf{A\hat{x}}(n) + \mathbf{Bu}(n) + \mathbf{L}[\mathbf{y}(n) - \mathbf\hat{y}(n)] \\

\mathbf\hat{y}(n) &= \mathbf{C\hat{x}}(n)

\end{align}

with observer gain {{math|L}}. Then Ackermann's formula for the design of state observers is noted as

\mathbf{L}^{\rm T} = \begin{bmatrix}

0 & 0 & \cdots & 1

\end{bmatrix}(\mathcal{O}^{\rm T})^{-1} \Delta_\text{new}(\mathbf{A}^{\rm T})

with observability matrix $\backslash mathcal\{O\}$. Here it is important to note, that the observability matrix and the system matrix are transposed: $\backslash mathcal\{O\}^\{\backslash rm\; T\}$ and {{math|A{{sup|T}}}}.

Ackermann's formula can also be applied on continuous-time observed systems.

• Full state feedback

{{Reflist}}

• Chapter about Ackermann's Formula on Wikibook of Control Systems and Control Engineering

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