How can the Poles of the Root Locus be negative? My understanding of drawing a root locus diagram is that stability requires all roots of the characteristic polynomial of the open loop transfer function to lie in the negative real part of the plane. In other words, it must have roots $(s+a)(s+b)...(s+n)$ where $s$ is a negative number.

My question is that the Laplace Transform requires that the real part of $s$ be positive, so wouldn't the possibility of negative real parts of $s$ be a contradiction?

Thanks.
If the denominator polynomial is given by

$$Q(s)=(s-s_0)(s-s_1)\ldots (s-s_{n-1})\tag{1}$$

then the corresponding time domain function is made up of a sum of scaled exponentials of the form $e^{s_it}u(t)$, where $u(t)$ is the unit step function. These terms decay only if $\textrm{Re}\{{s_i}\}&lt;0$, so for stability the zeros of the denominator polynomial (i.e., the poles of the system function) must lie in the left half-plane.

Let $h(t)$ be the impulse response of a causal LTI system. The transfer function is the Laplace transform of $h(t)$:

$$H(s)=\int_{0}^{\infty}h(t)e^{-st}dt\tag{2}$$

This integral only converges in a region $\textrm{Re}\{s\}&gt;c$ ("region of convergence"), with some real-valued constant $c$. That constant depends on the properties of $h(t)$. If $h(t)$ is a weighted sum of exponentials $e^{s_it}u(t)$ then $c$ equals the most positive real part of the complex numbers $s_i$:

$$c=\max_i\big\{\textrm{Re}\{s_i\}\big\}\tag{3}$$

This makes sure that the damping by the term $e^{-st}$ in $(2)$ is enough to make $h(t)e^{-st}$ decay sufficiently such that the integral converges.

For stability we require $c&lt;0$, i.e., all poles lie in the left half-plane, and the imaginary axis of the $s$-plane $s=j\omega$ lies inside the region of convergence.