# Quick question about a closed loop control system

Using above loop, we see that

$\dfrac{y(s)}{r(s)} = \dfrac{p(s)c(s)}{1+p(s)c(s)}$

and

$\dfrac{e(s)}{r(s)} = \dfrac{1}{1+p(s)c(s)}$

(well known results)

Why is that:

If the close loop system is stable, then $1+p(s)c(s) = 0$ ?

Shouldn't it be if the closed loop poles of $\dfrac{y(s)}{r(s)}$ are in open left hand plane?

Thanks

The poles of the TF are the roots of the characteristic equation, $\small 1+P(s)C(s)=0$. For a stable system all of these roots must be in the left half s-plane.