Similarity Transformations

A similarity transformation of a state-space system is a linear change of state variable coordinates:

$$\underline{x}(n) \isdef \mathbf{E}\tilde{\underline{x}}(n)$$

where

• $\underline{x}(n) =$ original state vector
• $\tilde{\underline{x}}(n) =$ state vector in new coordinates
• $\mathbf{E}=$ any invertible (one-to-one) matrix
  (linear transformation)

Substituting $\underline{x}(n) = \mathbf{E}\tilde{\underline{x}}(n)$ gives

$$\begin{eqnarray*} \mathbf{E}\tilde{\underline{x}}(n+1) & = & \mathbf{A}\, \mathbf{E}\tilde{\underline{x}}(n) + \mathbf{B}\underline{u}(n) \nonumber \\ \underline{y}(n) & = & \mathbf{C}\mathbf{E}\tilde{\underline{x}}(n) + \mathbf{D}\underline{u}(n) \end{eqnarray*}$$

Premultiplying the first equation above by $\mathbf{E}^{-1}$ gives

$$\begin{eqnarray*} \tilde{\underline{x}}(n+1) & = &\left(\mathbf{E}^{-1}\mathbf{A}\mathbf{E}\right) \tilde{\underline{x}}(n) + \left(\mathbf{E}^{-1}\mathbf{B}\right) \underline{u}(n) \nonumber \\ \underline{y}(n) & = & \left(\mathbf{C}\mathbf{E}\right) \tilde{\underline{x}}(n) + \mathbf{D}\underline{u}(n) \end{eqnarray*}$$

Define the transformed system matrices by

$$\begin{eqnarray*} \tilde{\mathbf{A}}&=& \mathbf{E}^{-1}\mathbf{A}\mathbf{E}\nonumber \\ \tilde{\mathbf{B}}&=& \mathbf{E}^{-1}\mathbf{B}\nonumber \\ \tilde{\mathbf{C}}&=& \mathbf{C}\mathbf{E}\nonumber \\ \tilde{\mathbf{D}}&=& \mathbf{D} \end{eqnarray*}$$

We can now write

$$\begin{eqnarray*} \tilde{\underline{x}}(n+1) & = &\tilde{\mathbf{A}}\tilde{\underline{x}}(n) + \tilde{\mathbf{B}}\underline{u}(n) \nonumber \\ \underline{y}(n) & = & \tilde{\mathbf{C}}\tilde{\underline{x}}(n) + \mathbf{D}\underline{u}(n) \end{eqnarray*}$$

The transformed system describes the same system in new state-variable coordinates

Let's verify that the transfer function has not changed:

$$\begin{eqnarray*} {\tilde{\mathbf{H}}}(z) &=& \tilde{\mathbf{D}}+ \tilde{\mathbf{C}}(z\mathbf{I}- \tilde{\mathbf{A}})^{-1}\tilde{\mathbf{B}}\\ &=& \mathbf{D}+ (\mathbf{C}\mathbf{E}) \left(z\mathbf{I}- \mathbf{E}^{-1}\mathbf{A}\mathbf{E}\right)^{-1}(\mathbf{E}^{-1}\mathbf{B})\\ &=& \mathbf{D}+ \mathbf{C}\left[\mathbf{E}\left(z\mathbf{I}- \mathbf{E}^{-1}\mathbf{A}\mathbf{E}\right)\mathbf{E}^{-1}\right]^{-1} \mathbf{B}\\ &=& \mathbf{D}+ \mathbf{C}\left(z\mathbf{I}- \mathbf{A}\right)^{-1} \mathbf{B}= \mathbf{H}(z) \end{eqnarray*}$$

• Since the eigenvalues of $\mathbf{A}$ are the poles of the system, it follows that the eigenvalues of $\tilde{\mathbf{A}}=\mathbf{E}^{-1}\mathbf{A}\mathbf{E}$ are the same. In other words, eigenvalues are unaffected by a similarity transformation.

• The transformed Markov parameters, $\tilde{\mathbf{C}}\tilde{\mathbf{A}}^n \tilde{\mathbf{B}}$ , are also unchanged since they are given by the inverse $z$ transform of the transfer function ${\tilde{\mathbf{H}}}(z)$ . However, it is also easy to show this by direct calculation.