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

We are given the state-space system.

vn+1=Avn+Bxnyn=Cvn+Dxn

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

zVzzv0=AVz+BXz

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

zIAVz=zv0+BXz

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

Vz=zzIA1v0+zIA1BXz

(5.11)

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

Yz=CVz+DXz=CzzIA1v0+zIA1BXz+DXz

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

Yz=CVz+DXz=CzIA1B+DXz

or

YzXz=CzIA1B+D

(5.12)

5.5    General Solution ...

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