We will now solve the same examples which were used to understand the Jury's test.

Example 1 The characteristic equation: $P(z)=z^{3}-1.3z^{2}-0.08z+0.24=0$

Transforming P(z) into w -domain:

$\displaystyle Q(w) = \left [ \frac {w+1}{w-1} \right ]^{3}-1.3\left [ \frac {w+1}{w-1} \right ]^{2}-0.08\left [ \frac {w+1}{w-1} \right ]+0.24=0$

or, $\displaystyle \;\; Q(w) = 0.14 w^3 - 1.06 w^2 - 5.1 w -1.98 = 0$

We can now construct the Routh array as

There is one sign change in the first column of the Routh array. Thus the system is unstable with one pole at right hand side of the w -plane or outside the unit circle of z -plane.

Example2: The characteristic equation: $P(z)= z^{4}-1.2z^{3}+0.07z^{2}+0.3z-0.08=0$

Transforming P(z) into w -domain:

$\displaystyle Q(w) = \left [ \frac {w+1}{w-1} \right ]^{4} - 1.2 \left [\frac {... ...ft [ \frac {w+1}{w-1} \right ]^{2}+0.3\left [ \frac {w+1}{w-1} \right ]-0.08=0$

or, $\displaystyle \;\; Q(w) = 0.09w^4 + 1.32 w^3 +5.38 w^2 +7.32 w +1.89 = 0$

We can now construct the Routh array as

All elements in the first column of Routh array are positive. Thus the system is stable.