# Control Systems/Eigenvalue Assignment for MIMO Systems

The design of control laws for MIMO systems are more extensive in comparison to SISO systems because the additional inputs ($q>1$ ) offer more options like defining the Eigenvectors or handling the activity of inputs. This also means that the feedback matrix K for a set of desired Eigenvalues of the closed-loop system is not unique. All presented methods have advantages, disadvantages and certain limitations. This means not all methods can be applied on every possible system and it is important to check which method could be applied on the own considered problem.

# Parametric State Feedback

A simple approach to find the feedback matrix K can be derived via parametric state feedback (in German: vollständige modale Synthese). A MIMO system

${\dot {x}}(t)=Ax(t)+Bu(t)$ with input vector

$u(t)=(u_{1}(t),u_{2}(t),\cdots ,u_{q}(t))=-K~x(t)$ input matrix $B\in \mathbb {R} ^{p\times q}$ and feedback matrix $K\in \mathbb {R} ^{q\times p}$ is considered. The Eigenvalue problem of the closed-loop system

${\dot {x}}(t)=(A-B~K)~x(t)=A_{CL}~x(t)$ is noted as

$A_{CL}~{\tilde {v}}_{i}=(A-B~K)~{\tilde {v}}_{i}={\tilde {\lambda }}_{i}~{\tilde {v}}_{i}$ where ${\tilde {\lambda }}_{i}\in \mathbb {C}$ denote the assigned Eigenvalues and ${\tilde {v}}_{i}\in \mathbb {C} ^{p}$ denote the Eigenvectors of the closed-loop system. Next, new parameter vectors $\phi _{i}=K{\tilde {v}}_{i}$ are introduced and assigned and the Eigenvalue problem is recasted as



$B~K~{\tilde {v}}_{i}=B~\phi _{i}=(A-{\tilde {\lambda }}_{i}~I)~{\tilde {v}}_{i}.$ ## Controller synthesis

1. From Equation  one defines the Eigenvector with

${\tilde {v}}_{i}=(A-{\tilde {\lambda }}_{i}~I)^{-1}~B~\phi _{i}$ 2. The new parameter vectors $\phi _{i}$ are concatenated as

$\Phi =[\phi _{1},\phi _{2},\cdots ,\phi _{p}]=K[{\tilde {v}}_{1},{\tilde {v}}_{2},\cdots ,{\tilde {v}}_{p}],$ where the feedback matrix K can be noted as

$K=\Phi ~[{\tilde {v}}_{1},{\tilde {v}}_{2},\cdots ,{\tilde {v}}_{p}]^{-1}.$ 3. Finally, the Eigenvector definition is used to hold the full description of the feedback matrix with

$K=[\phi _{1},\phi _{2},\cdots ,\phi _{p}]~[(A-{\tilde {\lambda }}_{1}~I)^{-1}~B~\phi _{1},\cdots ,(A-{\tilde {\lambda }}_{p}~I)^{-1}~B~\phi _{p}]^{-1}.$ The parameter vectors are defined arbitrarily but have to be linear independent.

## Example

Consider the dynamical system

${\begin{bmatrix}{\dot {x}}_{1}(t)\\{\dot {x}}_{2}(t)\end{bmatrix}}={\begin{bmatrix}3&1\\4&3\end{bmatrix}}~{\begin{bmatrix}{x}_{1}(t)\\{x}_{2}(t)\end{bmatrix}}+{\begin{bmatrix}1&2\\3&4\end{bmatrix}}~{\begin{bmatrix}u_{1}(t)\\u_{2}(t)\end{bmatrix}}$ which is unstable due to positive Eigenvalues $\lambda _{i}\in \{1,5\}$ . A feedback matrix K should be found to reach a stable closed-loop system with Eigenvalues ${\tilde {\lambda }}_{1}=-4,{\tilde {\lambda }}_{2}=-2$ .

1. The parameter vectors are defined as $\phi _{1}={\begin{bmatrix}1\\0\end{bmatrix}}$ and $\phi _{2}={\begin{bmatrix}0\\1\end{bmatrix}}$ 2. The resulting Eigenvectors are

${\tilde {v}}_{1}=(A-{\tilde {\lambda }}_{1}~I)^{-1}~B~\phi _{1}=\left({\begin{bmatrix}3&1\\4&3\end{bmatrix}}-{\begin{bmatrix}-4&0\\0&-4\end{bmatrix}}\right)^{-1}{\begin{bmatrix}1&2\\3&4\end{bmatrix}}~{\begin{bmatrix}1\\0\end{bmatrix}}\approx {\begin{bmatrix}0.09\\0.38\end{bmatrix}}$ and

${\tilde {v}}_{2}=(A-{\tilde {\lambda }}_{2}~I)^{-1}~B~\phi _{2}=\left({\begin{bmatrix}3&1\\4&3\end{bmatrix}}-{\begin{bmatrix}-2&0\\0&-2\end{bmatrix}}\right)^{-1}{\begin{bmatrix}1&2\\3&4\end{bmatrix}}~{\begin{bmatrix}0\\1\end{bmatrix}}\approx {\begin{bmatrix}0.29\\0.57\end{bmatrix}}.$ 3. The feedback matrix is calculated with

$K=[\phi _{1},\phi _{2}]~[(A-{\tilde {\lambda }}_{1}~I)^{-1}~B~\phi _{1},(A-{\tilde {\lambda }}_{2}~I)^{-1}~B~\phi _{2}]^{-1}\approx {\begin{bmatrix}1&0\\0&1\end{bmatrix}}~{\begin{bmatrix}0.09&0.29\\0.38&0.57\end{bmatrix}}^{-1}\approx {\begin{bmatrix}-9.68&4.92\\6.45&-1.53\end{bmatrix}}.$ More precise rounding leads to a feedback matrix

$K={\begin{bmatrix}-10&5\\6.6{\overline {1}}&-1.5{\overline {5}}\end{bmatrix}}$ # Singular Value Decomposition and Diagonalization

If the state matrix $A\in \mathbb {R} ^{p\times p}$ of system

${\dot {x}}(t)=A~x(t)+B~u(t)$ is diagonalizable, which means the number of Eigenvalues and Eigenvectors are equal, then the transform

$x=M~x_{M}$ can be used to yield

$M~{\dot {x}}_{M}(t)=AM~x_{M}(t)+B~u(t)$ and further

${\dot {x}}_{M}(t)=M^{-1}AM~x_{M}(t)+M^{-1}~B~u(t).$ Transformation matrix M contains the Eigenvectors $v_{i}\in \mathbb {C} ^{p}$ as

$M=[v_{1},v_{2},\cdots ,v_{p}]$ which leads to a new diagonal state matrix

$A_{M}=M^{-1}~A~M={\begin{bmatrix}\lambda _{1}\\&\lambda _{2}\\&&\ddots \\&&&\lambda _{p}\end{bmatrix}}$ consisting of Eigenvalues $\lambda _{i}\in \mathbb {C}$ , and new input

$u_{M}(t)=M^{-1}~B~u(t)={\begin{bmatrix}u_{M,1}\\u_{M,2}\\\cdots \\u_{M,p}\end{bmatrix}}.$ The control law for the new input $u_{M}$ is designed as

$u_{M}(t)=-K_{M}x_{M}(t)=-{\begin{bmatrix}K_{M,1}\\&K_{M,2}\\&&\ddots \\&&&K_{M,p}\end{bmatrix}}~{\begin{bmatrix}x_{M,1}(t)\\x_{M,2}(t)\\\cdots \\x_{M,p}(t)\end{bmatrix}}$ and the closed-loop system in new coordinates is noted as

${\dot {x}}_{M}(t)=A_{M}~x_{M}(t)+u_{M}(t)=(A_{M}-K_{M})~x_{M}(t)={\begin{bmatrix}\lambda _{1}-K_{M,1}\\&\lambda _{2}-K_{M,2}\\&&\ddots \\&&&\lambda _{p}-K_{M,p}\end{bmatrix}}~{\begin{bmatrix}x_{M,1}(t)\\x_{M,2}(t)\\\cdots \\x_{M,p}(t)\end{bmatrix}}$ Feedback matrix $K_{M}$ can be used to influence or shift each Eigenvalue directly.

In the last step, the new input is transformed backwards to original coordinates to yield the original feedback matrix K. The new input is defined by

$u_{M}(t)=M^{-1}~B~u(t)$ and

$u_{M}(t)=-K_{M}~x_{M}(t)=-K_{M}~M^{-1}~x(t).$ From these formulas one gains the identity

$M^{-1}~B~u(t)=-K_{M}~M^{-1}~x(t)$ and further

$u(t)=-B^{-1}~M~K_{M}~M^{-1}~x(t)=-K~x(t).$ Therefore, the feedback matrix is found as

$K=B^{-1}~M~K_{M}~M^{-1}.$ ## Requirements

This controller design is applicable only if the following requirements are guaranteed.

• State matrix A is diagonalizable.
• The number of states and inputs are equal $p=q$ .
• Input matrix $B\in \mathbb {R} ^{p\times p}$ is invertible.

## Example

Consider the dynamical system

${\begin{bmatrix}{\dot {x}}_{1}(t)\\{\dot {x}}_{2}(t)\end{bmatrix}}={\begin{bmatrix}2&1\\1&2\end{bmatrix}}~{\begin{bmatrix}{x}_{1}(t)\\{x}_{2}(t)\end{bmatrix}}+{\begin{bmatrix}1&2\\2&1\end{bmatrix}}~{\begin{bmatrix}u_{1}(t)\\u_{2}(t)\end{bmatrix}}$ which is unstable due to positive Eigenvalues $\lambda _{i}\in \{1,3\}$ . The Eigenvectors are

${\tilde {v}}_{1}={\begin{bmatrix}-{\frac {\sqrt {2}}{2}}\\{\frac {\sqrt {2}}{2}}\end{bmatrix}}$ and

${\tilde {v}}_{2}={\begin{bmatrix}{\frac {\sqrt {2}}{2}}\\{\frac {\sqrt {2}}{2}}\end{bmatrix}}.$ Thus, the transformation matrix is noted as

$M={\begin{bmatrix}-{\frac {\sqrt {2}}{2}}&{\frac {\sqrt {2}}{2}}\\{\frac {\sqrt {2}}{2}}&{\frac {\sqrt {2}}{2}}\end{bmatrix}}$ and the state matrix in new coordinates is derived as

$A_{M}=M^{-1}~A~M={\begin{bmatrix}1&0\\0&3\end{bmatrix}}.$ The desired Eigenvalues of the closed-loop system are ${\tilde {\lambda }}_{1}=-5$ and ${\tilde {\lambda }}_{2}=-1$ , so feedback matrix is found with

$\lambda _{1}-K_{M,1}=1-K_{M,1}={\tilde {\lambda }}_{1}=-5\quad \Rightarrow K_{M,1}=1+5=6$ and

$\lambda _{2}-K_{M,2}=3-K_{M,1}={\tilde {\lambda }}_{2}=-1\quad \Rightarrow K_{M,2}=3+1=4$ and thus one holds

$K_{M}={\begin{bmatrix}6&0\\0&4\\\end{bmatrix}}.$ Finally, the feedback matrix in original coordinates are calculated by

$K=B^{-1}~M~K_{M}~M^{-1}={\begin{bmatrix}1&2\\2&1\end{bmatrix}}^{-1}~{\begin{bmatrix}-{\frac {\sqrt {2}}{2}}&{\frac {\sqrt {2}}{2}}\\{\frac {\sqrt {2}}{2}}&{\frac {\sqrt {2}}{2}}\end{bmatrix}}~{\begin{bmatrix}6&0\\0&4\\\end{bmatrix}}~{\begin{bmatrix}-{\frac {\sqrt {2}}{2}}&{\frac {\sqrt {2}}{2}}\\{\frac {\sqrt {2}}{2}}&{\frac {\sqrt {2}}{2}}\end{bmatrix}}^{-1}={\frac {1}{3}}{\begin{bmatrix}-7&11\\11&-7\\\end{bmatrix}}.$ # Sylvester Equation

This method is taken from the online resource

Consider the closed-loop system

${\dot {x}}(t)=A~x(t)+B~u(t)=(A-B~K)~x(t)=A_{CL}~x(t)$ with input $u(t)=-K~x(t)$ and closed-loop state matrix $A_{CL}=A-B~K$ . The desired closed-loop Eigenvalues ${\tilde {\lambda }}_{i}\in \mathbb {C}$ can be chosen real- or complex-valued as ${\tilde {\lambda }}_{i}=\alpha _{i}\pm j\beta _{i}$ and the matrix of the desired Eigenvalues is noted as

$\Lambda ={\begin{bmatrix}\alpha _{1}&\beta _{1}\\-\beta _{1}&\alpha _{1}\\&&\ddots \\&&&{\tilde {\lambda }}_{i}\\&&&&\ddots \end{bmatrix}}$ The closed-loop state matrix $A_{CL}$ has to be similar to $\Lambda$ as

$A_{CL}=A-B~K\sim \Lambda$ which means that there exists a transformation matrix $M\in \mathbb {R} ^{p\times p}$ such that

$M^{-1}~A_{CL}~M=M^{-1}~(A-B~K)~M=\Lambda$ holds and further



$A~M-M~\Lambda =B~K~M.$ An arbitrary Matrix $G=K~M$ is introduced and Equation  is seperated in a Sylvester equation

[Sylvester]

$A~M-M~\Lambda =B~G$ and a feedback matrix formula

$K=G~M^{-1}.$ ## Algorithm

1. Choose an arbitrary matrix $G\in \mathbb {R} ^{q\times p}$ .

2. Solve the Sylvester equation for M (numerically).

3. Calculate the feedback matrix K.

## Remarks

• State matrix A and the negative Eigenvalue matrix $-\Lambda$ shall not have common Eigenvalues.
• For some choices of G the computation could fail. Then another G has to be chosen.

## Example

Consider the dynamical system

${\begin{bmatrix}{\dot {x}}_{1}(t)\\{\dot {x}}_{2}(t)\end{bmatrix}}={\begin{bmatrix}2&1\\1&2\end{bmatrix}}~{\begin{bmatrix}{x}_{1}(t)\\{x}_{2}(t)\end{bmatrix}}+{\begin{bmatrix}1&2\\2&1\end{bmatrix}}~{\begin{bmatrix}u_{1}(t)\\u_{2}(t)\end{bmatrix}}$ which is unstable due to positive Eigenvalues $\lambda _{i}\in \{1,3\}$ . The complex-valued Eigenvalues ${\tilde {\lambda }}_{1,2}=-2\pm 1j$ are desired for the closed-loop system. So, the eigenvalue matrix is noted as

$\Lambda ={\begin{bmatrix}-2&1\\-1&-2\end{bmatrix}}.$ Matrix G is chosen as

$G={\begin{bmatrix}1&0\\0&1\end{bmatrix}}.$ and Sylvester equation

$A~M-M~\Lambda ={\begin{bmatrix}2&1\\1&2\end{bmatrix}}~M-M~{\begin{bmatrix}-2&1\\-1&-2\end{bmatrix}}={\begin{bmatrix}1&2\\2&1\end{bmatrix}}=B~G$ is noted. The Sylvester equation is solved numerically and the transformation matrix is computed as

$M={\begin{bmatrix}0.0307692&0.446154\\0.430769&0.246154\end{bmatrix}}.$ Finally, the feedback matrix is found as

$K=G~M^{-1}={\begin{bmatrix}1&0\\0&1\end{bmatrix}}~{\begin{bmatrix}0.0307692&0.446154\\0.430769&0.246154\end{bmatrix}}^{-1}={\frac {1}{3}}{\begin{bmatrix}-4&7.25\\7&-0.5\end{bmatrix}}.$ 