# 8.4: Linear Transformation of State Variables

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### Linear State Transformation

This section describes a general procedure to transform a state variable model into an alternate model using state variables that are linear combinations of the original variables.

Consider the general state variable model of a SISO system, described as:

$\dot{x}(t)=Ax(t)+bu(t) \nonumber$

$y(t)=c^{T} x(t) \nonumber$

Define a bilinear transformation of the state variable vector, $$x(t)$$, by multiplying with a constant invertible matrix $$P$$, resulting in a new set of state variables, $$z(t)$$:

$z=Px,\; \; \; x=P^{-1} z \nonumber$

Substitute the above relations in the state and output equations:

$P^{-1} \dot{z}=AP^{-1} z+bu, \;\;y=c^{T} P^{-1} z \nonumber$

Multiplying on the left by $$P$$ results in a new state variable model:

$\dot{z}=PAP^{-1} z+Pbu, \;\;y=c^{T} P^{-1} z \nonumber$

The two models share the same transfer function, i.e.,

$G(s)=c^{T} P^{-1} (sI-PAP^{-1} )^{-1} Pb=c^{T} (sI-A)^{-1} b \nonumber$

Next, we explore the possibility to impart a desired structure to the system matrix through linear transformation of the state variables.

First, define $$\bar{A}=PAP^{-1} ,\, \, \bar{b}=Pb$$; the first equation is alternatively expressed as: $$\bar{A}P=PA$$.

Then, given $$A,\, b$$, and the desired structure of $$\bar{A},\, \bar{b}$$, the above relations may be used to define the transformation matrix, $$P$$. In particular, the transformation to the controller and modal forms is explored below.

### Transformation into Controller Form

A necessary condition to obtain a linear transformation matrix $$P$$ to convert a state variable model into controller form is that the following $$n\times n$$ controllability matrix has full rank:

$M_{\rm C} =[b,\; Ab,\ldots ,\;A^{n-1} b] \nonumber$

The controllability matrix is guaranteed to be of full rank if the transfer function description of the model has the same order, $$n$$, as the number of state variables used to describe the system, i.e., there are no pole-zero cancelations in the transfer function, $$G\left(s\right)={\bf c}^T(s{I-A})^{-1}{\bf b}$$.

Next, assume that the controllability matrix, $$M_{\rm C}$$, of pair $$\left(A,b\right)$$ is of full rank.  Let $$M_{\rm CF}$$ denote the controllability matrix of the controller form representation $$\left(\bar{ A}, \bar{ b}\right)$$; then, $$M_{\rm CF}$$ contains the coefficients of the characteristic polynomial and can be written by inspection.

The required transformation matrix is defined by:

$Q=P^{-1} =M_{\rm C} M_{\rm CF}^{-1} \nonumber$

##### Example $$\PageIndex{1}$$

The state and output equations for a small DC motor model are given as:

$\frac{\rm d}{\rm dt} \left[\begin{array}{c} {i_a } \\ {\omega } \end{array}\right]=\left[\begin{array}{cc} {-100} & {-5} \\ {5} & {-10} \end{array}\right] \left[\begin{array}{c} {i_a} \\ {\omega } \end{array}\right]+\left[\begin{array}{c} {100} \\ {0} \end{array}\right]V_a ,\, \, \, \omega =\left[\begin{array}{cc} {0} & {1} \end{array}\right]\left[\begin{array}{c} {i_a} \\ {\omega } \end{array}\right] \nonumber$

The controllability matrix for the model is formed as: $$M_{\rm C} =[b,\; Ab]=\left[\begin{array}{cc} {100} & {-10^{4} } \\ {0} & {500} \end{array}\right]$$.

As the controllability matrix is of full rank, the dc motor model is controllable.

From the transfer function description, $$G(s)=\frac{500}{s^{2} +110s+1025}$$, the controller form realization is obtained as:

$\dot{x}=\left[\begin{array}{cc} {0} & {1} \\ {-1025} & {-110} \end{array}\right]x+\left[\begin{array}{c} {0} \\ {1} \end{array}\right]V_a ,\, \, \, \omega =\left[\begin{array}{cc} {500} & {0} \end{array}\right]x \nonumber$

The controllability matrix for the controller form realization is given as: $$M_{\rm CF} =\left[\begin{array}{cc} {0} & {1} \\ {1} & {-110} \end{array}\right]$$.

The state transformation matrix for the model is obtained as:

$Q=P^{-1} =\left[\begin{array}{cc} {1000} & {100} \\ {500} & {0} \end{array}\right],\quad P=\left[\begin{array}{cc} {0} & {0.002} \\ {0.01} & {-0.02} \end{array}\right] \nonumber$

Indeed, $$\overline{A}=PAQ$$ matches the system matrix in the controller form.

### Transformation into Modal Form

A matrix that has a full set of eigenvectors is diagonalizable by a linear transformation matrix when the eigenvectors of $$A$$ are selected as the columns of $$P^{-1}$$.

In the event when $$A$$ has complex eigenvalues, its eigenvectors are also complex. By placing the real and imaginary parts of the complex eigenvector into separate columns of $$P^{-1}$$, the resulting modal form matrix, $$\bar{A}=PAP^{-1} ,\, \, \bar{b}=Pb$$, has real entries.

##### Example $$\PageIndex{2}$$

The state and output equations for a mass–spring–damper model are given as:

$\frac{\rm d}{\rm dt} \left[\begin{array}{c} {x} \\ {v} \end{array}\right]=\left[\begin{array}{cc} {0} & {1} \\ {-2} & {-2} \end{array}\right]\, \left[\begin{array}{c} {x} \\ {v} \end{array}\right]+\left[\begin{array}{c} {0} \\ {1} \end{array}\right]u,\, \, \, y=\left[\begin{array}{cc} {1} & {0} \end{array}\right]\left[\begin{array}{c} {x} \\ {v} \end{array}\right]. \nonumber$

The eigenvalues of the system matrix are: $$-1\pm j1$$; the complex eigenvectors are: $$\left[\begin{array}{c} {1} \\ {1\pm j1} \end{array}\right]$$.

By choosing $$P^{-1} =\left[\begin{array}{cc} {1} & {0} \\ {-1} & {1} \end{array}\right]$$, the modal form system matrix is obtained as: $$\bar{A}=PAP^{-1} =\left[\begin{array}{cc} {-1} & {1} \\ {-1} & {-1} \end{array}\right]$$.

##### Example $$\PageIndex{3}$$

The state variable model of a small DC motor is given as:

$\frac{\rm d}{\rm dt} \left[\begin{array}{c} {i_{a} } \\ {\omega } \end{array}\right]=\left[\begin{array}{cc} {-100} & {-5} \\ {5} & {-10} \end{array}\right]\left[\begin{array}{c} {i_a } \\ {\omega } \end{array}\right]+\left[\begin{array}{c} {100} \\ {0} \end{array}\right]V_a , \;\; \omega =\left[\begin{array}{cc} {0} & {1} \end{array}\right]\left[\begin{array}{c} {i_a } \\ {\omega } \end{array}\right] \nonumber$

We use MATLAB ‘eig’ command to obtain: $$V=P^{-1} =\left[\begin{array}{cc} {-0.9985} & {0.0556} \\ {0.0556} & {-0.9985} \end{array}\right]$$.

The transformed state variable model has a diagonal structure.

$\left[\begin{array}{c} {\dot{x}_{1} } \\ {\dot{x}_{2} } \end{array}\right]=\left[\begin{array}{cc} {-99.72} & {0} \\ {0} & {-10.28} \end{array}\right]\left[\begin{array}{c} {x_{1} } \\ {x_{2} } \end{array}\right]+\left[\begin{array}{c} {-100.47} \\ {-5.6} \end{array}\right]V_a, \;\; \omega =\left[\begin{array}{cc} {0.056} & {-0.998} \end{array}\right]\, \left[\begin{array}{c} {x_{1} } \\ {x_{2} } \end{array}\right] \nonumber$

The decoupled system of equations can be easily integrated. Assuming a unit-step input, the state variables are solved as:

$x_{1} (t)=1.0075+(x_{10} -1.0075)e^{-t/99.72} \nonumber$

$x_{2} (t)=0.545+(x_{20} -0.545)e^{-t/10.28} \nonumber$

where $$x_{10} ,\; x_{20}$$ represent the initial conditions on state variables. The output is computed as:

$\omega (t)=0.056x_{1} (t)-0.998x_{2} (t) \nonumber$

The original state variables are recovered as: $$\left[ \begin{array}{c} i_a \\ \omega \end{array} \right]=P^{-1}\left[ \begin{array}{c} x_1 \\ x_2 \end{array} \right]$$, i.e.,

$i_a\left(t\right)=-0.9985x_1\left(t\right)+0.0556x_2\left(t\right) \nonumber$

$\omega \left(t\right)=0.0556x_1\left(t\right)-0.9985x_2\left(t\right) \nonumber$

This page titled 8.4: Linear Transformation of State Variables is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Kamran Iqbal.