Discrete Systems and Digital Signal Processing with MATLAB, 2nd Edition

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 ...