I am totally a noob about this subject :D
Here is a system :
If the open loop transfer function is equal to "-1", the output "Vout(s)" goes to infinity. Nevertheless it is said that a system is unstable only if the the tranfer function "1 + the open loop transfer function" has at least one zero in the right half plane ? Why it would be stable if the zero is in the left right plane ?
Thank you very much and have a nice day ! :D
I will drop the $(s)$ just to simplify notation.
First, the statement "1 + the open loop transfer function" is not correct if you have a $\beta$ block at the bottom. Usually, the block diagram has $\beta(s) = 1$ and then this statement is true. If we let $H = \frac{N_H}{D_H}$, $G = \frac{N_G}{D_G}$, and $\beta = 1$, i.e., $N_{X}$ is the numerator of $X$ and $D_X$ is the denominator of $X$. Then, the close loop transfer function is, \begin{align} \frac{V_{out}}{V_{ref}} &= \frac{HG}{1+HG} \\ &= \frac{\frac{N_H}{D_H}\frac{N_G}{D_G}}{1+\frac{N_H}{D_H}\frac{N_G}{D_G}} \\ &= \frac{\frac{N_H N_G}{D_H D_G}}{\frac{D_{H}D_{G}+N_H N_G}{D_{H}D_{G}}} \\ &= \frac{N_H N_G}{D_{H}D_{G}+N_H N_G} \qquad(*) \end{align}
The "1+open loop transfer funcion" $= 1+HG = \frac{D_{H}D_{G}+N_H N_G}{D_{H}D_{G}}$. Notice that the zeros of this transfer function is equal to the poles of the closed loop system, i.e., denominator of (*). If the a zero of $1+HG$ is in the open right half plane then the pole of $\frac{V_{out}}{V_{ref}}$ is in the open right half plane and the system is unstable. Similarly, if all the zeros of $1+HG$ are in the open left half plane then the poles of $\frac{V_{out}}{V_{ref}}$ are in the open left half plane and the system is stable.