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Solving LCCDEs by Unilateral Z-Transform

Due to its time delay property, the unilateral z-transform is a powerful tool for solving LCCDEs with arbitrary initial conditions.

Example: A system is described by this LCCDE

\begin{displaymath}y[n]+3y[n-1]=x[n]=\alpha u[n] \end{displaymath}

Taking unilateral z-transform of the DE, we get

\begin{displaymath}Y(z)+3Y(z)z^{-1}+3y[-1]=X(z)=\frac{\alpha}{1-z^{-1}} \end{displaymath}

When neither $y[-1]$ nor $x[n]$ is zero, we have

\begin{displaymath}Y(z)+3Y(z)z^{-1}+3\beta=X(z)=\frac{\alpha}{1-z^{-1}} \end{displaymath}

Solving this algebraic equation in z-domain for $Y(z)$ we get

\begin{displaymath}Y(z)=\frac{\alpha}{(1+3z^{-1})(1-z^{-1})}-\frac{3\beta}{1+3z^{-1}} \end{displaymath}

The first term is the particular solution caused by the input alone and the second term is the homogeneous solution caused by the initial condition alone. The $Y(z)$ can be further written as

\begin{displaymath}Y(z)=\frac{1}{1+3z^{-1}}(\frac{3}{4}\alpha-3\beta)
+\frac{\alpha}{4}\frac{1}{1-z^{-1}}
\end{displaymath}

and in time domain, we have the general solution

\begin{displaymath}y_g[n]=[(\frac{3}{4}\alpha-3\beta)(-3)^n+\frac{\alpha}{4}]u[n]
=y_h[n]+y_p[n] \end{displaymath}

which is the sum of both the homogeneous and particular solutions.

Note that bilateral z-transform can also be used to solve LCCDEs. However, as bilateral z-transform does not take initial condition into account, it is always implicitly assumed that the system is initially at rest. If this is not the case, unilateral z-transform has to be used.


next up previous
Next: About this document ... Up: Z_Transform Previous: Unilateral Z-Transform
Ruye Wang 2014-10-28