GATE 2021 EE – Question 57
In the given figure, plant $G_p(s)=\dfrac{2.2}{(1+0.1s)(1+0.4s)(1+1.2s)}$ and compensator $G_c(s)=K\left(\dfrac{1+T_1s}{1+T_2s}\right)$. The external disturbance input is $D(s)$. It is desired that when the disturbance is a unit step, the steady-state error should not exceed 0.1 unit. The minimum value of K is ________. (Round off to 2 decimal places.)
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Correct answer: 9.50 to 9.60
Explanation
The disturbance enters at the plant input, so $\frac{E}{D}=-\frac{G_p}{1+G_cG_p}$. At DC, $G_p(0)=2.2$ and $G_c(0)=K$, so the steady-state error is $\frac{2.2}{1+2.2K}\le0.1$, which needs $1+2.2K\ge22$ and $K\ge9.545\approx9.55$.