Range of K for stability (Jury test)

Step 1: Characteristic equation

For unity feedback, $1 + G(z) = 0$:

$$(z - 0.3679)(z - 1) + K(0.3679z + 0.2642) = 0$$ $$P(z) = z^2 + (0.3679K - 1.3679)z + (0.3679 + 0.2642K) = 0$$

So $n = 2$, $a_0 = 1$, $a_1 = 0.3679K - 1.3679$, $a_2 = 0.3679 + 0.2642K$.

Step 2: Jury conditions for a second-order system

For $n = 2$ there are only three conditions: $|a_2| < a_0$, $P(1) > 0$, and $P(-1) > 0$. No table rows are needed.

Condition 1: $|a_2| < a_0$

$$|0.3679 + 0.2642K| < 1$$ $$\Rightarrow\; -1 < 0.3679 + 0.2642K < 1 \;\Rightarrow\; -5.178 < K < \frac{1 - 0.3679}{0.2642} = 2.3925$$

Condition 2: $P(1) > 0$

$$P(1) = 1 + 0.3679K - 1.3679 + 0.3679 + 0.2642K = 0.6321K > 0 \;\Rightarrow\; K > 0$$

Condition 3: $(-1)^2P(-1) > 0$

$$P(-1) = 1 - 0.3679K + 1.3679 + 0.3679 + 0.2642K = 2.7358 - 0.1037K > 0 \;\Rightarrow\; K < 26.38$$

Step 3: Combine

The intersection of all three is

$$0 < K < 2.3925$$

Step 4: Check at the critical gain

At $K = 2.3925$: $P(z) = z^2 - 0.4877z + 1 = 0$, which gives $z = 0.244 \pm j0.970$ with $|z| = 1.000$. The poles lie on the unit circle, so the system is marginally stable and oscillates.

This $G(z)$ is the ZOH equivalent of $\dfrac{K}{s(s+1)}$ with $T = 1$ s. The continuous system is stable for every $K > 0$, but sampling limits the stable gain to $K < 2.39$.

Answer: The closed-loop system is stable for $0 < K < 2.3925$.