Answer:
The attached system shows that there’s an integrator between the point where disturbance enters the system and error measuring element. A any time when R(s)=0 then
[tex]\frac {C(s)}{D(s)}=\frac {G(s)}{1+G_c(s)G(s)}[/tex] and considering that [tex]E(s)=D(s)-G_c(s)C(s)[/tex] then
[tex]\frac {E(s)}{D(s)}=1-(\frac {C(s)}{D(s)})G_c(s)[/tex]
[tex]\frac {E(s)}{D(s)}=1-(\frac {G(s)}{1+G_c(s)D(s)})G_c(s)[/tex]
[tex]\frac {E(s)}{D(s)}=\frac {1}{1+G_c(s)G(s)}[/tex]
[tex]E(s)=\frac {D(s)}{1+G_c(s)G(s)}[/tex]
For ramp disturbance d(t)=at
[tex]D(s)=\frac {a}{s^{2}}[/tex] therefore, the steady state error is given by
[tex]e(\infty)= \lim_{s \to 0} s E(s)[/tex]
[tex]e(\infty)= \lim_{s \to 0} s [\frac {D(s)}{1+G_c(s)G(s)}][/tex]
[tex]e(\infty)= \lim_{s \to 0} s [\frac {a}{s^{2}+s^{2}G_c(s)G(s)}][/tex]
[tex]e(\infty)= \lim_{s \to 0} s [\frac {a}{s+sG_c(s)G(s)}][/tex]
[tex]e(\infty)= \lim_{s \to 0} s [\frac {a}{sG_c(s)G(s)}][/tex]
Whenever [tex]G_c(s)[/tex] has a double intergrator, the error [tex]e(\infty)[/tex] becomes zero