20 例:一系统的开环传递函数(M (0.5s+1)(0.04s+1) 求:r(=1(t)及时的稳态误差 解 e, R(s= lim s (0.5s+1)(0.04s+1) R(s s→>01+G(s)H(S s→>0(0.5s+1)(0.04s+1)+20 r(t)=1(t)时,R(s)=1/s ssIm~(0.5s+1)(004s+1) 0.05 s→>0(0.5s+1)(0.04s+1)+20s21 r(t)=t时,R(s)=1/s2 S (0.5s+1)(0.04s+1)1 50(0.5+1)(0.04s+1)+20s25 例: 一系统的开环传递函数 求：r(t)=1(t)及t时的稳态误差 解: (0.5 1)(0.04 1) 20 ( ) ( ) + + = s s G s H s ( ) (0.5 1)(0.04 1) 20 (0.5 1)(0.04 1) ( ) lim 1 ( ) ( ) 1 lim 0 0 R s s s s s R s s G s H s e s s s s s + + + + + = + = → → 0.05 21 1 1 (0.5 1)(0.04 1) 20 (0.5 1)(0.04 1) lim 0 • =  + + + + + = → s s s s s e s s s s r(t) = 1(t) 时, R(s)=1/s r(t) = t 时, R(s)=1/s2 • =  + + + + + = → 2 0 1 (0.5 1)(0.04 1) 20 (0.5 1)(0.04 1) lim s s s s s e s s ss