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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

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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.

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