Diagonalizing basis

Changing basis in the state equation

As with all basis transformations, the basis transformation we seek can be written \[\begin{aligned} \bm{x} = P \bm{x}' \quad \Rightarrow \bm{x}' = P^{-1} \bm{x}, \end{aligned}\] where \(P\) is the transformation matrix, \(\bm{x}\) is a representation of the state vector in the original basis, and \(\bm{x}'\) is a representation of the state vector in the new basis.²

Substituting this transformation into the standard linear state-model equations yields the model $$\begin{align} \dot{\bm{x}}' &= \underbrace{P^{-1} A P}_{A'} \bm{x}' + \underbrace{P^{-1} B}_{B'} \bm{u} \\ \bm{y} &= \underbrace{C P}_{C'} \bm{x}' + \underbrace{D}_{D'} \bm{u}. \end{align}$$

Modal and eigenvalue matrices

Let a state equation have matrix \(A\) with \(n\) distinct eigenvalues \((\lambda_i)\) and eigenvectors \((\bm{m}_i)\). Let the eigenvalue matrix \(\Lambda\) be defined as \[\begin{aligned} \Lambda = \begin{bmatrix} \lambda_1 & & & 0 \\ & \lambda_2 & & \\ & & \ddots & \\ 0 & & & \lambda_n \end{bmatrix}.\end{aligned}\]

Furthermore, let the modal matrix \(M\) be defined as \[\begin{aligned} M = \left[ \begin{array}{cccc} \rule[-1ex]{0.5pt}{2.5ex}& \rule[-1ex]{0.5pt}{2.5ex}& & \rule[-1ex]{0.5pt}{2.5ex}\\ \bm{m}_{1} & \bm{m}_{2} & \cdots & \bm{m}_{n} \\ \rule[-1ex]{0.5pt}{2.5ex}& \rule[-1ex]{0.5pt}{2.5ex}& & \rule[-1ex]{0.5pt}{2.5ex} \end{array} \right] \end{aligned}\]

Diagonalization of the state equation

Let the modal matrix \(M\) be the transformation matrix for our state-model. Then³ \(\bm{x}' = M^{-1} \bm{x}\).

The state equation becomes \[\begin{aligned} \dot{\bm{x}}' &= M^{-1} A M \bm{x}' + M^{-1} B \bm{u}. \end{aligned}\] The eigenproblem implies that \[\begin{aligned} A \begin{bmatrix} \bm{m}_1 & \bm{m}_2 & \cdots & \bm{m}_n \end{bmatrix} &= \begin{bmatrix} \bm{m}_1 & \bm{m}_2 & \cdots & \bm{m}_n \end{bmatrix} \Lambda \quad\Rightarrow \\ A M &= M \Lambda \quad\Rightarrow \\ M^{-1} A M &= M^{-1} M \Lambda \\ &= \Lambda.\end{aligned}\] That is, \(A' = \Lambda\)! Recall that \(\Lambda\) is diagonal; therefore, we have diagonalized the state-space model. In full-form, the diagonalized model is

$$\begin{align} \label{ssresp_eq:diagonalized_state_space} \dot{\bm{x}}' &= \underbrace{\Lambda}_{A'} \bm{x}' + \underbrace{M^{-1} B}_{B'} \bm{u} \\ \bm{y} &= \underbrace{C M}_{C'} \bm{x}' + \underbrace{D}_{D'} \bm{u}. \end{align}$$

Computing the state transition matrix

Recall our definition of the state transition matrix \(\Phi(t) = e^{A t}\). Directly applying this to the diagonalized system of , $$\begin{align} \Phi'(t) &= e^{\Lambda t} \\ &= \begin{bmatrix} e^{\lambda_1 t} & & & 0 \\ & e^{\lambda_2 t} & & \\ & & \ddots & \\ 0 & & & e^{\lambda_n t} \end{bmatrix}. \end{align}$$ In this last equality, we have used the diagonal property of the state transition matrix, defined in .

Recall that the free response solution to the state equation is given by the initial condition and state transition matrix, so $$\begin{align} \bm{x}_\text{fr}'(t) &= \Phi'(t) \bm{x}'(0) \\ &= x_1'(0) e^{\lambda_1 t} + x_2'(0) e^{\lambda_2 t} + \cdots + x_n'(0) e^{\lambda_n t} \end{align}$$ where the initial conditions are \(\bm{x}'(0) = M^{-1} \bm{x}(0)\). We have completely decoupled each state’s free response, one of the remarkable qualities of the diagonalized system.

At this point, one could simply solve the diagonalized system for \(\bm{x}'(t)\), then convert the solution to the original basis with \(\bm{x}(t) = M \bm{x}'(t)\).

Sometimes, we might prefer to transform the diagonalized-basis state transition matrix into the original basis. The following is a derivation of that transformation.

Beginning with the free response solution in the diagonalized-basis and transforming the equation into the original basis, we find an expression for the original state transition matrix, as follows. \[\begin{aligned} \bm{x}_\text{fr}'(t) &= \Phi'(t) \bm{x}'(0) \quad\Rightarrow \\ M^{-1} \bm{x}_\text{fr}(t) &= \Phi'(t) M^{-1} \bm{x}(0) \quad\Rightarrow \\ \bm{x}_\text{fr}(t) &= \underbrace{M \Phi'(t) M^{-1}}_{\Phi(t)} \bm{x}(0).\end{aligned}\] This last expression is just the free response solution in the original basis, so we can identify \[\begin{aligned} \label{ssresp_eq:state_transition_pref} \Phi(t) = M \Phi'(t) M^{-1}. \end{aligned}\] This is a powerful result. is the preferred method of deriving the state transition matrix for a given system. The eigenvalues give \(\Phi'\) and the eigenvectors give \(M\).

Example 7.2

For the state equation $$\begin{align*} \dot{\bm{x}} = \begin{bmatrix} -2 & 2 \\ 2 & -3 \end{bmatrix} \bm{x} + \begin{bmatrix} 1 \\ -1 \end{bmatrix} \bm{u} \end{align*}$$ find the state’s free response to initial condition $\bm{x}(0) = \begin{bmatrix} 2 & -1 \end{bmatrix}^{\top}$.

Then just use and $$\begin{align*} \bm{x}_\text{fr}(t) = \Phi(t) \bm{x}(0). \end{align*}$$