The steady-state error in many common feedbac control loops can be written in th

ID: 2994272 • Letter: T

Question

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)

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-->0 S * E(s)

= limit s-->0 S* (T*s + 1)*(1/s)/ [ (T*s + 1) +kp ]

= limit s-->0 (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-->0 S * E(s)

= limit s-->0 S* (T*s + 1)/ [ s(T*s + 1) +s K_p + K_i ]

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