Question

If $G\left(s\right)$ is a stable transfer function, then $F\left(s\right)=\frac{1}{G\left(s\right)}$ is always a stable transfer function.

• A. True
• B. False

Answer 1

The correct option is B False

T.F of $G\left(s\right)=\frac{(s+Z_1)(s+Z_2)}{(s+P_1)(s+P_2)}$
T.F of $F\left(s\right)=\frac{(s+P_1)(s+P_2)}{(s+Z_1)(s+Z_2)}$
The condition for stability is that none of the pole of $G\left(s\right)$ should beonthe right half of s-plane, but $G\left(s\right)$ may have zeros in the right half of s-plane. These zeros become pole of $F\left(s\right)$. Therefore the $F\left(s\right)$ need not be stable (False).