Answer:
(E) 1 or 2
Explanation:
Steady state error for unit step input is given by [tex]e_{ss}=\frac{1}{1+k_p}[/tex] where [tex]K_P[/tex] is position error coefficient the value of [tex]K_P[/tex] is given by [tex]k_p=\lim_{s\rightarrow 0}G(s)[/tex] it will be only finite when the type of the open loop system will be 0 when the of the system will be 1 or 2 then [tex]k_p[/tex] becomes infinite and steady state error will become zero so the option (E) will be the correct answer
