Topic: Control Systems
The voltage V in the circuit given below is :
- A. $\frac{u-1}{u}$
- B. $\frac{u-1}{u^{2}}$
- C. $\frac{u^{2}}{u-1}$
- D. $\frac{1}{u-1}$
Answer: D. $\frac{1}{u-1}$
Explanation: The closed-loop transfer function is $\frac{C(s)}{R(s)} = \frac{G(s)}{1+G(s)H(s)} = \frac{K}{s(s+1)(s+4) + K(s+u)}$.
For a unit step input $R(s) = \frac{1}{s}$, the steady-state output is $c(\infty) = \lim_{s \to 0} s C(s) = \lim_{s \to 0} \frac{K}{s(s+1)(s+4) + K(s+u)} = \frac{K}{Ku} = \frac{1}{u}$.
The steady-state error is $e_{ss} = 1 - c(\infty) = 1 - \frac{1}{u} = \frac{u-1}{u}$.
The sensitivity of $e_{ss}$ with respect to $u$ is $S_u^{e_{ss}} = \frac{u}{e_{ss}} \frac{\partial e_{ss}}{\partial u}$.
$\frac{\partial e_{ss}}{\partial u} = \frac{\partial}{\partial u} (1 - u^{-1}) = \frac{1}{u^2}$.
$S_u^{e_{ss}} = \frac{u}{(u-1)/u} \times \frac{1}{u^2} = \frac{u^2}{u-1} \times \frac{1}{u^2} = \frac{1}{u-1}$.