# 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.

• Where did you read that the real part of 's' has to be positive? – Dan Szabo Sep 4 at 1:00
• From the definition of the Laplace Transform. If s is negative then the integral does not converge. – user2813684 Sep 4 at 1:04
• Could you edit your post to include the integral that does not converge when ‘s’ is negative? I think it would help answer your question. If you want to share a link to a reference figure, that would be as good if it would be easier. – Dan Szabo Sep 4 at 1:59
• en.m.wikipedia.org/wiki/_transform the section on convergence is a bit different than your assumption s can be negative – Stanley Pawlukiewicz Sep 4 at 2:08
• That actually makes more sense to me now. So s does not need need to be positive, but the coefficient on t in the exponential does, which then puts a condition upon the range of values that s can take, which may be either positive or negative? – user2813684 Sep 4 at 2:42

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 $$\textrm{Re}\{{s_i}\}<0$$, 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 $$\textrm{Re}\{s\}>c$$ ("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 $$c<0$$, 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.