The steady-state error in many common feedbac control loops can be wri.docx
1. The steady-state error in many common feedbac control loops can be written in the form
E(s) = 1/(1 + K(s)*G(s)) * R(s)
where K(s) is the transfer function of the controller
G(s) = 1/(T*s + 1) transfer function of the system to be controlled
R(s) = 1/s (unit step function) is the input
For each case below find the steady state error, i.e. lim e(t) as t approaches infinity
1) K(s) = K_p, K_p > 0 is a constant (proportional control)
2) K(s) = K_p + K_i/s, where K_p, K_i > 0 are constants (proportional+integral control)
Solution
G(s) = 1/(T*s + 1)
R(s) = 1/s
1) k(s) =kp
![E(s) = 1/(1 + K(s)*G(s)) * R(s)
= R(s)/[ 1+ kp* 1/(T*s + 1) ]
=(T*s + 1)*R(s)/ [ (T*s + 1) +kp ]
=(T*s + 1)(1/s)/ [ (T*s + 1) +kp ]
steady state error = limit s-->0Â Â S * E(s)
= limit s-->0 S* (T*s + 1)*(1/s)/ [ (T*s + 1) +kp ]
= limit s-->0 Â Â (T*s + 1)/ [ (T*s + 1) +kp ]
= 1/(1+kp)
2) K(s) = K_p + K_i/s
E(s) =
E(s) = 1/(1 + K(s)*G(s)) * R(s)
= R(s)/[ 1+ k(s)* 1/(T*s + 1) ]
=(T*s + 1)*R(s)/ [ (T*s + 1) +k(s) ]
=(T*s + 1)(1/s)/ [ (T*s + 1) + K_p + K_i/s ]
= (T*s + 1)/ [ s(T*s + 1) +s K_p + K_i ]
steady state error = limit s-->0Â Â S * E(s)
= limit s-->0 S* (T*s + 1)/ [ s(T*s + 1) +s K_p + K_i ]](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)