## 5.4 Solution of the State-Space Equations in the *z*-Domain

We are given the state-space system.

$\begin{array}{c}v\left(n+1\right)=Av\left(n\right)+Bx\left(n\right)\\ y\left(n\right)=Cv\left(n\right)+Dx\left(n\right)\end{array}$

By taking the *z*-transform on the previous state equations, we will get

$zV\left(z\right)-zv\left(0\right)=AV\left(z\right)+BX\left(z\right)$

Notice that *z* is a scalar and cannot be subtracted from **A**. Therefore, we write the previous equations as

$\left(zI-A\right)V\left(z\right)=zv\left(0\right)+BX\left(z\right)$

where **I** is the identity matrix. Solving the previous equation for the states we get

$V\left(z\right)=z{\left(zI-A\right)}^{-1}v\left(0\right)+{\left(zI-A\right)}^{-1}BX\left(z\right)$ |
(5.11) |

**v**(*n*) is the inverse transform of the previous equation. For the output **y**(*n*), we have

$\begin{array}{c}Y\left(z\right)=CV\left(z\right)+DX\left(z\right)\\ =C\left[z{\left(zI-A\right)}^{-1}v\left(0\right)+{\left(zI-A\right)}^{-1}BX\left(z\right)\right]+DX\left(z\right)\end{array}$

This is a good place to try to find **H**(*z*) from the state-space equations. With **v**(0) = 0, we have

$\begin{array}{c}Y\left(z\right)=CV\left(z\right)+DX\left(z\right)\\ =\left[C{\left(zI-A\right)}^{-1}B+D\right]X\left(z\right)\end{array}$

or

$\frac{Y\left(z\right)}{X\left(z\right)}=\left[C{\left(zI-A\right)}^{-1}B+D\right]$ |
(5.12) |

## 5.5 General Solution ...

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