§36.4 动态误差糸数法(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(S E(s C(s) R(S)E(s C(s) 解.②r(t)=2+t/2 S(s+1 s(10s+1) 2 r"(t)=0 25[e(t ①2(S)= E(s) (t) R(S)1+ en(t) s(10s+1) s(10s+1) 10 10s2+s+1 e- (t) s+9s 2-19s5 ea (t) t(min) 10 25 e,2(t)=Cr+C1r+C2r"=0+r+9"=65+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