Question

For the closed-loop control system shown, with $r(t)=0$, the steady-state error $e(\infty)$ due to a unit step disturbance $d(t)$ is ____________ (rounded off to two decimal places).

Hint:

Disturbance-to-error transfer function comes from $E = -Y$ when the reference input is zero.

Answer 1

Correct Answer: -0.11

Given block diagram: \[ Y(s) = \frac{10}{s(s+10)} E(s) + \frac{1}{s(s+10)}D(s), \] and error is \[ E(s) = R(s) - Y(s). \] With \(R(s)=0\): \[ E(s) = -Y(s). \] Thus closed-loop transfer from disturbance \(D(s)\) to error \(E(s)\) is: \[ E(s) = -\frac{1}{1 + 10/(s(s+10))}\cdot \frac{1}{s(s+10)}D(s). \] Simplify denominator: \[ 1 + \frac{10}{s(s+10)} = \frac{s(s+10)+10}{s(s+10)} = \frac{s^2 + 10s + 10}{s(s+10)}. \] Thus: \[ E(s) = -\frac{1}{s(s+10)} \cdot \frac{s(s+10)}{s^2 + 10s + 10} D(s) = -\frac{1}{s^2 + 10s + 10} D(s). \] For a unit step disturbance: \[ D(s) = \frac{1}{s}. \] Thus: \[ E(s) = -\frac{1}{s(s^2 + 10s + 10)}. \] Steady-state error using final value theorem: \[ e(\infty) = \lim_{s\to 0} sE(s) = \lim_{s\to 0} \left( -\frac{1}{s^2 + 10s + 10} \right) = -\frac{1}{10}. \] Thus: \[ e(\infty) = -0.10. \] Acceptable range: –0.11 to –0.09 \[ \boxed{-0.10} \]