In the given, figure, plant $G_p(s)=\frac{2.2}{(1+0.1s)(1+0.4s)(1+1.2s)}$ and compensator $G_c(s)=K\left(\frac{1+T_{1}s}{1+T_{2}s}\right)$. 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.)

Given,

$G_p(s)=\frac{2.2}{(1+0.1s)(1+0.4s)(1+1.2s)}$ ...(1)

$G_c(s)=K\left(\frac{1+T_{1}s}{1+T_{2}s}\right)$ ....(2)

Maximum steady state error due to external input = 0.1

For step external disturbance input D(s) = 1/s .....(3)

Assume R(s) = 0,

⇒ C(s) = - E(s)

Then the output expression is given as

C(s) = [E(s)Gc(s) + D(s)]Gp(s)

E(s) = E(s)Gc(s)Gp(s) + D(s)Gp(s)

Error function $E(s)=\frac{-D(s)G_p(s)}{1+G_c(s)G_p(s)}$ ....(4)

Steady state error is given by

$e_{ss}=\underset{s\to 0}{\lim }s.E(s)$

From equations (1), (2), (3)&(4)

$|e_{ss}|=\underset{s\to 0}{\lim }\frac{s.\frac{1}{s}{G}_p(s)}{1+G_c(s)G_p(s)}$

$e_{ss}\ge \frac{2.2}{1+2.2K}$

$\frac{2.2}{1+2.2K}\le 0.1$

K ≥ 9.54

Minimum value of K is 9.54