§3.6.4 动态误差糸数法(3) 例1两系统如图示,要求在4分钟内误差不超过6m,应选用哪个系统? 已知:r()=2+12/4 R(s)E(sI C(s) R(S)E(S) 解①.r(t)=2+t/ S(s+1 (10s+1) "(t)=1/2 25 (t)=0 s+ EO Φ1(s) 20-… R(S) 1+ 15 s(s+1 (S+1) s2+s+1 10 =C+cs+cas t. s-S+∴ ea (t) t(min en(r)=Cr+C;r'+C2r"=2+t/20 10 20 5 30§3.6.4 动态误差系数法(3) 例1 两系统如图示,要求在4分钟内误差不超过6m,应选用哪个系统? 已知： ( 1) 1 1 1 ( ) ( ) ( ) 1 + +  = = s s R s E s s e 解 ①. ( ) 2 4 2 r t = t + t r(t) = 2 + t 2 r (t) = 1 2 r (t) = 0 = C0 + C1s + C2 s 2 + (1 )[ ] 2 0 1 2 s + s 2 = + s + s 2 C + C s + C s + = C0 + C1s + C2 s 2 + C3 s 3 + C0 s + C1s 2 + C2 s 3 + C0 s 2 + C1s 3 + 1 ( 1) 2 + + + = s s s s = s − s 3 + es1 (t) = C0r + C1r + C2r  = 2 + t 2 C0 + (C0+C1 )s + (C0+C1 + C2 )s 2 + 比较系数：     = − = = = 1 0 1 0 3 2 1 0 C C C C