西业2量学 §3.6.4 动态误差数法(4 例1两系统如图示,要求在4分钟内系统不超过6m应选用哪个系统? 已知:r(t)=2+t2/4 R()E(「11c)RE C(s) 解.②r(t)=2+t/2 s(s+1) (10s+1 (t)=1/2 r"(t)=0 25[e(t E(s) 20 (t) R(S) en(t) 1+ 10s+1) s(10s+1) 10 10s2+S+1 e- (t) s+9s2-19s3+ e a (t) t(min) 25 e,2(t)=Cr+C1r'+C2r"=0+r+9r"=6.5+t/2§3.6.4 动态误差系数法(4) 例1 两系统如图示,要求在4分钟内系统不超过6m应选用哪个系统? 已知： (10 1) 1 1 1 ( ) ( ) ( ) 2 + +  = = s s R s E s s e 解. ② ( ) 2 4 2 r t = t + t r (t) = 2+ t 2 r (t) = 1 2 r (t) = 0 s 2 3 s + s + 10s 2 3 9s −10s 2 3 4 9s + 9s + 90s 10 1 (10 1) 2 + + + = s s s s = s + 9s 2 −19s 3 + e t C r C r C r s = +  +  2 0 1 2 ( ) 3 4 −19s − 90s 2 1+ s + 10s = 0+ r  + 9r  = 6.5+ t 2 2 + 9s 3 −19s 2 s + 10s