Example 2: The characteristic equation is

$P(z)= z^{4}-1.2z^{3}+0.07z^{2}+0.3z-0.08=0$
Thus, $a_{0}=1$ $a_{1}=-1.2$ $a_{2}=0.07$ $a_{3}=0.3$ $a_{4}=-0.08$

We will now check the stability conditions.

1. $\vert a_{n}\vert = \vert a_{4}\vert = 0.08 < a_{0}=1 \Rightarrow$ First condition is satisfied.

2. $P(1)=1-1.2+0.07+0.3-0.08+0.09>0 \Rightarrow$ Second condition is satisfied.

3. $P(-1)= 1+1.2+0.07-0.3-0.08=1.89>0 \Rightarrow$ Third condition is satisfied.

Next we will construct the Jury Table.

Jury Table

$b_{1}=-0.0756$

Rest of the elements are also calculated in a similar fashion. The elements are $b_{1}=-0.0756$ $b_{0}=-0.204$ $c_{2}=0.946$ $c_{1}=-1.184$ $c_{0}=0.315$ . One can see
$\vert b_{3}\vert=0.9936>\vert b_{0}\vert=0.204$
$\vert c_{2}\vert=0.946>\vert c_{0}\vert=0.315$

All criteria are satisfied. Thus the system is stable.