单位阶跃输入时的动态误差级数: c (t) (二)=(1-2 R(二) 0 T 2T 3T 4T 5t t E(x)=①2(x)R()=(1-z)(,-)=1 e(0)=1,e(1)=e(2)=…=e(m)=0 ΦD()=1-Φ()=z1 C(z)=Φ(z)R(x)= 7十x十z° C(1)=C(2)=C(3) 1 11-Φ2(z) D(z)G(z)Φ(z) D(x)= G(z)(1-z-) 返回单位阶跃输入时的动态误差级数： 1 ( ) (1 ) e z z −  = − 1 1 ( ) 1 R z z − = − ) 1 1 1 ( ) ( ) ( ) (1 )( 1 1 = − =  = − − − z E z z R z z e e(0) = 1, e(1) = e(2) == e(n) = 0 = + + + − =  = − − − − − 1 2 3 1 1 1 ( ) ( ) ( ) z z z z z C z z R z  C(1) = C(2) = C(3) == 1 ( )(1 ) ( ) 1 1 − − − = G z z z D z T 2T 3T 4T 5T t C （ t ） 1 0 ( ) 1 ( ) ( ) 1 ( ) z z G z D z e e  −  =  返回 1 ( ) 1 ( ) e z z z −  = −  =