1. 阿克曼公式
设有如下系统
{ x ˙ = A x + B u y = C x \begin{cases} \dot x = Ax + Bu \\ y = Cx \end{cases} {
x˙=Ax+Buy=Cx显然，通过矩阵A能够得到其特征多项式
φ A ( λ ) = λ n + a n − 1 λ n − 1 + ⋯ + a 1 λ + a 0 \varphi _A ( \lambda ) = \lambda ^n + a_{n-1} \lambda^{n-1} + \cdots + a_1 \lambda + a_0 φA(λ)=λn+an−1λn−1+⋯+a1λ+a0通过将控制量设为模态反馈控制（详见模态反馈控制一文）
u = − K x u = - Kx u=−Kx系统演变为
x ˙ = ( A − B K ) x \dot x = \left( A - BK \right) x x˙=(A−BK)x假设期望系统的期望特征多项式为
φ w ( λ ) = λ n + γ n − 1 λ n − 1 + ⋯ + γ 1 λ + γ 0 \varphi _w ( \lambda ) = \lambda ^n + \gamma_{n-1} \lambda^{n-1} + \cdots + \gamma_1 \lambda + \gamma_0 φw(λ)=λn+γn−1λn−1+⋯+γ1λ+γ0那么有阿克曼公式
K = [ 0 0 ⋯ 0 1 ] ⋅ [ B A B ⋯ A n − 1 B ] − 1 ⋅ φ w ( A ) (1) K = \left[ \begin{matrix} 0 & 0 & \cdots & 0 & 1 \end{matrix} \right] \cdot \left[ \begin{matrix} B & AB & \cdots & A^{n-1} B \end{matrix} \right]^{-1} \cdot \varphi _w (A) \tag{1} K=[00⋯01]⋅[BAB⋯An−1B]−1⋅φw(A)(1)其中
φ w ( A ) = A n + γ n − 1 A n − 1 + ⋯ + γ 1 A + γ 0 I (2) \varphi _w (A) = A ^n + \gamma_{n-1} A^{n-1} + \cdots + \gamma_1 A + \gamma_0 I \tag{2} φw(A)=An+γn−1An−1+⋯+γ1A+γ0I(2)利用式(1)和(2)，可以计算出模态反馈控制中所需的增益矩阵 K K K。
2. 举例
例：设系统的柯西形式为
A = [ 0 1 − 2 − 3 ] , B = [ 0 1 ] A = \left[ \begin{matrix} 0 & 1 \\ -2 &-3 \end{matrix} \right], \quad B = \left[ \begin{matrix} 0 \\ 1 \end{matrix} \right] A=[0−21−3],B=[01]试确定 K 1 , K 2 K_1, K_2 K1,K2，使得系统的期望特征多项式为 φ w ( s ) = s 2 + 4 s + 3 \varphi _w (s) = s^2 + 4s + 3 φw(s)=s2+4s+3。
系统为2阶，即 n = 2 n=2 n=2，故可以先计算 A B AB AB与 A 2 A^2 A2：
A B = [ 0 1 − 2 − 3 ] ⋅ [ 0 1 ] = [ 1 − 3 ] AB = \left[ \begin{matrix} 0 & 1 \\ -2 &-3 \end{matrix} \right] \cdot \left[ \begin{matrix} 0 \\ 1 \end{matrix} \right] = \left[ \begin{matrix} 1 \\ -3 \end{matrix} \right] AB=[0−21−3]⋅[01]=[1−3] A 2 = [ 0 1 − 2 − 3 ] ⋅ [ 0 1 − 2 − 3 ] = [ − 2 − 3 6 7 ] A^2 = \left[ \begin{matrix} 0 & 1 \\ -2 &-3 \end{matrix} \right] \cdot \left[ \begin{matrix} 0 & 1 \\ -2 &-3 \end{matrix} \right] = \left[ \begin{matrix} -2 & -3 \\ 6 & 7 \end{matrix} \right] A2=[0−21−3]⋅[0−21−3]=[−26−37]期望的多项式为 s 2 + 4 s + 3 s^2 + 4s + 3 s2+4s+3，可以得到 γ 0 = 3 , γ 1 = 4 \gamma_0 = 3, \gamma_1 = 4 γ0=3,γ1=4。则
φ w ( A ) = A 2 + γ 1 A + γ 0 I = [ − 2 − 3 6 7 ] + 4 [ 0 1 − 2 − 3 ] + 3 [ 1 0 0 1 ] = [ 1 1 − 2 − 2 ] \begin{aligned} \varphi _w (A) &= A ^2 + \gamma_1 A + \gamma_0 I \\ &= \left[ \begin{matrix} -2 & -3 \\ 6 & 7 \end{matrix} \right] + 4 \left[ \begin{matrix} 0 & 1 \\ -2 &-3 \end{matrix} \right] + 3 \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \\ &= \left[ \begin{matrix} 1 & 1 \\ -2 & -2 \end{matrix} \right] \end{aligned} φw(A)=A2+γ1A+γ0I=[−26−37]+4[0−21−3]+3[1001]=[1−21−2]那么，根据式(1)有
K = [ 0 1 ] ⋅ [ B A B ] − 1 ⋅ φ w ( A ) = [ 0 1 ] ⋅ [ 0 1 1 − 3 ] − 1 ⋅ [ 1 1 − 2 − 2 ] = [ 1 1 ] \begin{aligned} K &= \left[ \begin{matrix} 0 & 1 \end{matrix} \right] \cdot \left[ \begin{matrix} B & AB \end{matrix} \right]^{-1} \cdot \varphi _w (A) \\ &= \left[ \begin{matrix} 0 & 1 \end{matrix} \right] \cdot \left[ \begin{matrix} 0 & 1 \\ 1 & -3 \end{matrix} \right]^{-1} \cdot \left[ \begin{matrix} 1 & 1 \\ -2 & -2 \end{matrix} \right] \\ &= \left[ \begin{matrix} 1 & 1 \end{matrix} \right] \end{aligned} K=[01]⋅[BAB]−1⋅φw(A)=[01]⋅[011−3]−1⋅[1−21−2]=[11]即：负反馈通路中的矩阵 K K K为
K = [ 1 1 ] K = \left[ \begin{matrix} 1 & 1 \end{matrix} \right] K=[11]相应的控制量为
u = − K x = − x 1 − x 2 u = -K x = -x_1 - x_2 u=−Kx=−x1−x2