Example 3:

Consider the system shown in Figure 1. Find out the range of K for which the system is stable.
Solution:

$\displaystyle G(z)$ $\displaystyle =$ $\displaystyle \frac {K(0.084z^2+0.17z+0.019)}{(z^3-1.5z^2+0.553z-0.05)}$

**Figure 1:** Figure for Example 3
$\begin{figure}\centering \input{m3l2f1.pstex_t}\end{figure}$

Characteristic equation:

$\displaystyle 1+\frac {K(0.084z^2+0.17z+0.019)}{(z^3-1.5z^2+0.553z-0.05)}=0$

$\displaystyle P(z) = z^3+(0.084K-1.5)z^2+(0.17K+0.553)z+(0.019K-0.05) = 0$

Transforming P(z) into w -domain:

$\displaystyle Q(w) = \left [ \frac {w+1}{w-1} \right ]^{3}+(0.084K-1.5)\left [ ... ...1} \right ]^{2}+(0.17K+0.553)\left [ \frac {w+1}{w-1} \right ]+(0.019K-0.05)=0$

or, $\displaystyle \;\; Q(w) = (0.003+0.27K) w^3 + (1.1-0.11K) w^2 +(3.8-0.27K) w + (3.1+0.07K) = 0$

We can now construct the Routh array as