K and T for roots left of s=−a

To make all roots lie to the left of $s=-a$, shift the imaginary axis to $s=-a$ by putting $s = z - a$. Then apply the Routh criterion to the new polynomial in $z$ (relative stability).

Characteristic equation

$$\begin{aligned} 1 + \frac{K}{s(1+Ts)} &= 0 \\ Ts^2 + s + K &= 0 \end{aligned}$$

Shift the axis: $s = z - a$

$$\begin{aligned} T(z-a)^2 + (z-a) + K &= 0 \\ Tz^2 + (1-2aT)z + (Ta^2 - a + K) &= 0 \end{aligned}$$

Routh array in z

For no sign change (all roots left of $z=0$, i.e. left of $s=-a$):

1. $T > 0$
2. $1-2aT > 0 \Rightarrow T < \dfrac{1}{2a}$
3. $Ta^2 - a + K > 0 \Rightarrow K > a - a^2T = a(1-aT)$

$$0 < T < \frac{1}{2a}, \qquad K > a(1-aT)$$

Physical meaning: the sum of the closed-loop roots is $-\frac{1}{T}$, so both roots can be left of $-a$ only if $\frac{1}{T} > 2a$. The gain must be large enough that the slower real root moves past $-a$.

Example: for $a=1$ and $T=0.25$: $K > 1(1-0.25)=0.75$. With $K=1$, $0.25s^2+s+1=0$ gives $s = -2, -2$, both left of $-1$.