Nyquist plot of GH(z), as shown in Figure 3, intersects the negative real axis at -0.025K when $\omega = \dfrac{\omega _s }{2} = 31.4$ rad/sec.

$\phi$ can be computed as

$\displaystyle \phi = - (0 + 0.5 \times 1)180^0=-90^0$

It can be seen from Figure 3 that for $\phi$ to be -90°, (-1, j 0) point should be located at the left of -0.025K point. Thus for stability

$\begin{displaymath} \begin{array}{l} - 1 < - 0.025K \\ \\ \Rightarrow K < 40\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\ \end{array} \end{displaymath}$

If K > 40, (-1, j 0) will be at the right of point, hence making $\phi=90^o$ .

If $\phi=90^o$ , we get from (1)

$\displaystyle Z_{-1}= \frac{\phi}{180^o}+0+0.5=1$

Thus for K > 40, one of the closed loop poles will be outside the unit circle.

If K is negative we can still use the same Nyquist plot but refer (+1, j0 ) point as the critical point. $\phi$ in this case still equals +90° and the system is unstable. Hence the stable range of K is 0 ≤ K < 40

More details can be found in Digital Control Systems by B. C. Kuo.