Find the value of “K” to limit steady state error to 10, when input to system is {1+10t+20t2}. Here ‘t’ is the time.

G(S)H(S)=$\frac{k}{(S^2 (S+2)(S+3))}$, r(t)=1+10t+20t^2

=1+10t+40$\frac{t_2}{2}$

Input is the combination of all the three type, step of magnitude $A_1$=1, Ramp of magnitude $A_2$=10 and parabolic of magnitude $A_3$=40.

Position error $k_p$=\lim_(S\to 0)$G(S) H(S) =$\lim_(S\to 0)\frac{k}{(S^2 (S+2)(S+3))}$=$\frac{k}{0}k_p$=∞ Velocity error,$k_v$=$\lim_(S\to 0)$ S G(S) H(S)

=$\lim_(S\to 0)$ $\frac{(S.k)}{(S^2 (S+2)(S+3))}$

=$\frac{k}{0}$

$k_v$=∞

Acceleration error, $k_a$=$\lim_(S\to 0)$ $S^2$G(S) H(S)

=$\lim_(S\to 0)$ $\frac{(S^2 k)}{(S^2 (S+2)(S+3))}$

=$\frac{k}{((2)(3))}$

=$\frac{k}{6}$

$k_a$=$\frac{k}{6}$

$e_ss$ =$e_ss1$+$e_ss2$+$e_ss3$=$\frac{A_1}{(1+k_p )}$+$\frac{A_2}{k_v}$ +$\frac{A_3}{k_a}$

=$\frac{1}{(1+∞)}$+$\frac{10}{∞}$+$\frac{40}{\frac{k}{6})}$

=0+0+$\frac{240}{k}$

Since, $e_ss$=10

10=$\frac{240}{k}$

k=$\frac{240}{10}$

k=24