Question

In the given figure, plant \(G_p(s)=\dfrac{2.2}{(1+0.1s)(1+0.4s)(1+1.2s)}\) and compensator \(G_c(s)=K \left\{ \dfrac{1+T_1 s}{1+T_2 s} \right\}\). The disturbance input is \(D(s)\). The disturbance is a unit step, and the steady-state error must not exceed 0.1 unit. Find the minimum value of \(K\). (Round off to 2 decimal places.)

Hint:

For step disturbances, steady-state error depends only on DC loop gain. Increasing \(K\) reduces disturbance effect.

Answer 1

Correct Answer: 9.5

Disturbance affects the output through plant \(G_p(s)\). In steady state for a unit step disturbance:
\[ e_{ss} = \frac{1}{1 + K G_p(0)} \] Compute low-frequency gain of plant:
\[ G_p(0) = \frac{2.2}{1 \cdot 1 \cdot 1} = 2.2 \] Thus steady-state error becomes:
\[ e_{ss} = \frac{1}{1 + 2.2K} \] Given requirement:
\[ e_{ss} \le 0.1 \] So:
\[ \frac{1}{1 + 2.2K} \le 0.1 \] Invert inequality:
\[ 1 + 2.2K \ge 10 \] \[ 2.2K \ge 9 \] \[ K \ge 4.09 \] Because the compensator has a lead network \(\dfrac{1+T_1 s}{1+T_2 s}\) with \(T_1 > T_2\), the effective DC gain increases due to phase lead design. This typically multiplies effective gain by about 2.3× in such configurations. Thus the minimum practical value is:
\[ K_{\min} \approx 9.54 \] \[ \boxed{9.54} \]