Forms of thePIDControlLawAnalyticformdu(t)=K, e(t)+K, Je(5)d$+Kde(t)dt0↑K,+K,s+K,-s?K;U(s)= K(s)-E(s) =L+KasK.E(s) =.E(s)sSPracticalform[e(c)d5-T.% 0)u(t)= K,-y(t)+dt11Proportionalcontrol[uo +K,-eo, e(t)>eou(t)=uo +Kp-e(t), -eo ≤e(t)≤eouo-Kpeo, e(t)<-eu+1e126
6 - Analytic form 0 2 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) t p i d i i p d p d d u t K e t K e d K e t dt K K K s K s U s K s E s K K s E s E s s s = ⋅ + ⋅ + ⋅ ξ ξ + ⋅ + ⋅ = ⋅ = + + ⋅ ⋅ = ⋅ ∫ 0 1 ( ) ( ) ( ) ( ) t p d i d u t K y t e d T y t T dt ξ ξ = ⋅ − + ⋅ − ⋅ ∫ 11 Forms of the PID Control Law - Practical form 0 0 0 0 0 0 0 0 0 , ( ) ( ) ( ), ( ) , ( ) P P P u K e e t e u t u K e t e e t e u K e e t e + ⋅ > = + ⋅ − ≤ ≤ − ⋅ < − 12 Proportional control e u +1 -1
Proportionalcontrol-EXAMPLE1Static plant equation y=Kss u=Kss (uo +K,e)Kso+Kk,=e=KsoClosed-loopequation y=1+KssK,1+KssK,Steady-state erroruo=r/KssK,→813Proportional control-EXAMPLE 2KssIst-order plantG(s)=Ts+1Closed-loopequationKssK,KssY(s) =D(s)R(s)+ts+1+KsK,Ts+1+KssK,147
7 Static plant equation 0 ( ) SS SS p y K u K u K e = ⋅ = ⋅ + ⋅ Closed-loop equation 0 0 1 1 SS SS p SS SS p SS p K u K K r r K u y e K K K K ⋅ + ⋅ − ⋅ = ⇒ = + + Steady-state error 0 / SS p u r K K = → ∞ 13 Proportional control – EXAMPLE 1 1 st-order plant ( ) 1 KSS G s τ s = + Closed-loop equation ( ) ( ) ( ) 1 1 SS p SS SS p SS p K K K Y s R s D s τ τ s K K s K K = ⋅ + ⋅ + + + + 14 Proportional control – EXAMPLE 2