Presentation is loading. Please wait.

Presentation is loading. Please wait.

5.4 Disturbance rejection The input to the plant we manipulated is m(t). Plant also receives disturbance input that we do not control. The plant then can.

Similar presentations


Presentation on theme: "5.4 Disturbance rejection The input to the plant we manipulated is m(t). Plant also receives disturbance input that we do not control. The plant then can."— Presentation transcript:

1 5.4 Disturbance rejection The input to the plant we manipulated is m(t). Plant also receives disturbance input that we do not control. The plant then can be modeled as follow C(s)= G p (s)M(s)+G d (s)D(s) The control system should minimize G d (s)D(s). C=T(s)R(s)+T d (s)R(s) We want T d (s) as small as possible. Let us use frequency approach. T d (j  ) can be made small for specific frequency. Recall that G c (j  ) G p (j  ) H(j  ) must be large to reduce the sensitivity to plant variation. Methods to reduce T d (j  ) 1.make G d (s) small 2.increase loop gain by increasing G c 3.reduced D(s) 4.use feed forward compensation D(s)D(s) M(s)M(s) + C(s)C(s) Gp(s)Gp(s) G d (s) + plant GcGc + – H Gd(s)D(s)Gd(s)D(s) R(s)R(s)

2 5.4 Disturbance rejection Feedforward compensation Feedforward compensation can be applied if the disturbance can be measured. The addition of compensator G cd (s) does not effect T(s) the TF from R(s) to C(s) but does for T d (s) C(s)C(s) D(s)D(s) M(s)M(s) + Gp(s)Gp(s) G d (s) + plant GcGc + – H Gd(s)D(s)Gd(s)D(s) G cd (s) – R(s)R(s) We choose G cd (s)G c (s) such that T d (s) will as small as possible.

3 5.5 Steady State Accuracy We will examine the steady state error e ss for different system types. Assume H=1. hence Step input, R(s) = 1/s hence C(s)C(s)M(s)M(s) Gp(s)Gp(s)GcGc + – R(s)R(s) E(s)E(s) The steady state error e ss is defined as Ramp input, R(s) = 1/s 2 hence We can express N is integer and the system is called system type N. N is also the number of integrator in G c G p we will demonstrate its importance For N = 0 then K p = K p and e ss =1/(1+ K p ) For N > 0 then K p =  and e ss = 0. For N = 0 then K v = 0 and e ss =  For N = 1 then K v = K v  and e ss = 1/ K v For N > 1 then K v =  and e ss = 0

4 5.5 Steady State Accuracy Parabolic input, R(s) = 1/s 3 hence For N < 2 then K v = 0 and e ss =  For N = 2 then K a = K a  and e ss = 1/ K a For N > 2 then K v =  and e ss = 0 The steady state error e ss system type input N 1/s 1/s 2 1/s 3 0 1/(1+K P )   1 0 1/K v  e ss 2 0 0 1/K a K p is called position error constant K v is called velocity error constant K v is called acceleration error constant Plot of ramp responses of different type system type 0 system type 1 system type 2 system e ss

5 5.5 Steady State Accuracy Non-unity-Gain feedback The non-unity gain feedback above has equivalent block diagram as follow provided that H(s) is constant. + – H R(s)R(s) + – Ru(s)Ru(s)E(s)E(s)C(s)C(s) Where R u (s) = R(s)/H. Hence the position, velocity, and acceleration error constant can be calculated in the same manner

6 Disturbance Input Error Let us consider steady state error due to disturbance input. Recall that the output of disturbed plant C(s) =T(s)R(s)+T d (s)C(s)=C r (s)+C d (s) The system error is then e(t)= r(t) – c(t) =[r(t) – c r (t)] – c d (t) = e r (t) + e d (t) where e r (t) is the error we’ve just considered We see that e d (t)= – c d (t) and we want it to be small. The steady state of e d (t) is where (assuming H = 1) Let us model the disturbance input as step function, D(s)=B/s. From (1) we have (1) To draw general conclusion we must consider the type number of G p (s) G c (s) and G d (s). (2) (3) Type number of InputGp(s)Gp(s)Gd(s)Gd(s)Gc(s)Gc(s)e dss Unit step000Finite Unit step0010 ramp000  0020

7 TRANSIENT RESPONSE The system TF can be written as The response for input R(s) is Where C r (s) is the part that originate from the poles of R(s) The inverse LT of this equation is Here c r (t) is the forced response and the rest is the natural response or the transient response For stable system the natural response will decay to zero. The way it decays to zero is important. The transient response c tr (t) is The factoris called the modes The nature or form of each term is determined by the pole location of p i.

8 TRANSIENT RESPONSE The amplitude of each term is Thus the amplitude of each term determined by other pole location, the numerator polynomial, and the input function. For higher order system it is difficult to make general assertion, however we can consider the nature of each term of this response. there is associated damping ratio , a natural frequency  n, and time constant  i =1/  n. If there is one real dominant pole the system responds essentially as 1 st order system. If there is dominant complex poles then the system respond essentially as 2 nd order system. For each real pole, a time constant is associated with it with the value For each set complex conjugate poles of values

9 CLOSED LOOP FREQUENCY RESPONSE The input output characteristics are determined by CL frequency response T(0) is the system dc gain. This values determines the steady state error. If we evaluate (1) for small  we see how the system will follow for slow varying input. The bandwidth can be evaluated form (1). Large bandwidth indicate fast system response. The presence of any peaks, which denotes resonance, give indication of the overshoot or decaying oscillation (1). |T(  )|   B


Download ppt "5.4 Disturbance rejection The input to the plant we manipulated is m(t). Plant also receives disturbance input that we do not control. The plant then can."

Similar presentations


Ads by Google