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.)

1. Mathematical Formulation:
From the given block diagram, the closed-loop transfer function relationship for the output $C\left(s\right)$ is expressed as:
$C\left(s\right)=\left[E\right(s\left)G_c\right(s)+D(s\left)\right]G_p\left(s\right)$

Here, the error signal is given by:
$E\left(s\right)=R\left(s\right)-C\left(s\right)$

To find the response due to the disturbance $D\left(s\right)$, we assume the reference input $R\left(s\right)=0$, which gives:
$E\left(s\right)=-C\left(s\right)$

Substituting $C\left(s\right)=-E\left(s\right)$ into the output equation:
$-E\left(s\right)=\left[E\right(s\left)G_c\right(s)+D(s\left)\right]G_p\left(s\right)$
$-E\left(s\right)=E\left(s\right)G_c\left(s\right)G_p\left(s\right)+D\left(s\right)G_p\left(s\right)$
$E\left(s\right)[1+G_c(s\left)G_p\right(s\left)\right]=-D\left(s\right)G_p\left(s\right)$

Solving for the error transfer function:
$E\left(s\right)=\frac{-D\left(s\right)G_p\left(s\right)}{1+G_c\left(s\right)G_p\left(s\right)}$

2. Steady-State Error Calculation:
For a unit step disturbance input, $D\left(s\right)=\frac{1}{s}$. Using the Final Value Theorem, the steady-state error is:
$e_{ss}=\underset{s\to 0}{\lim }sE\left(s\right)$
$e_{ss}=\underset{s\to 0}{\lim }s\left[\frac{-\frac{1}{s}{G}_p\left(s\right)}{1+G_c\left(s\right)G_p\left(s\right)}\right]=\frac{-\underset{s\to 0}{\lim }{G}_p\left(s\right)}{1+\underset{s\to 0}{\lim }{G}_c\left(s\right)G_p\left(s\right)}$

Substitute the limits of the system components as $s\to 0$:
$\underset{s\to 0}{\lim }{G}_p\left(s\right)=\underset{s\to 0}{\lim }\frac{2.2}{(1+0.1s)(1+0.4s)(1+1.2s)}=2.2$
$\underset{s\to 0}{\lim }{G}_c\left(s\right)=\underset{s\to 0}{\lim }K\left(\frac{1+T_{1}s}{1+T_{2}s}\right)=K$

Thus, the magnitude of the steady-state error is:
$\left|e_{ss}\right|=\frac{2.2}{1+2.2K}$

3. Solving for the Minimum Value of K:
We require the steady-state error magnitude to not exceed 0.1:
$\frac{2.2}{1+2.2K}\le 0.1$
$2.2\le 0.1(1+2.2K)$
$22\le 1+2.2K$
$2.2K\ge 21$
$K\ge \frac{21}{2.2}\approx 9.55$

Therefore, the minimum value of $K$ is 9.55 (or 9.54 depending on rounding).