Answer :

10.45 (MTA)

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)

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

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

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

$\mathrm{K\; \ge \; 9.54}$

Minimum value of K is 9.54