Maximum K for poles left of −0.5

Result: all poles lie left of $s=-0.5$ when $K > 3.75$. There is no finite maximum; $K = 3.75$ is the limiting value.

To check that all poles lie to the left of $s=-0.5$, shift the imaginary axis to $s=-0.5$ by putting $s = z - 0.5$. Then apply the Routh criterion in $z$. Unity feedback is assumed.

Characteristic equation

$$\begin{aligned} s(s+2)(s+3) + K(s+1) &= 0 \\ s^3 + 5s^2 + (6+K)s + K &= 0 \end{aligned}$$

Shift: $s = z - 0.5$

$$\begin{aligned} (z-0.5)(z+1.5)(z+2.5) &= z^3 + 3.5z^2 + 1.75z - 1.875 \\ K(z-0.5+1) &= Kz + 0.5K \end{aligned}$$ $$z^3 + 3.5z^2 + (1.75+K)z + (0.5K - 1.875) = 0$$

Routh array in z

Conditions

• $z^1$ row: $3K + 8 > 0 \Rightarrow K > -2.67$ (true for $K>0$)
• $z^0$ row: $0.5K - 1.875 > 0 \Rightarrow K > 3.75$

So all closed-loop poles lie to the left of $s=-0.5$ when

$$K > 3.75$$

Comment on "maximum K"

For this system there is no finite maximum. As $K\to\infty$, one pole moves to the zero at $-1$, and the other two follow asymptotes at $\pm90^\circ$ from the centroid $\frac{(0-2-3)-(-1)}{2}=-2$, which is left of $-0.5$. The limiting (boundary) value is $K=3.75$. At $K=3.75$ the poles are $s=-0.5$ and $s=-2.25\pm j1.56$, so one pole is exactly at $-0.5$.

Answer: all poles lie left of $s=-0.5$ for $K > 3.75$; the boundary value is $K = 3.75$.