终端短路线段z=1z.g(2zD Z=IX 0 终端开路线段z=-/zcg(2z 4 4 Z=IX 0 22 终端短路线段 Z=jX Z Zg 0 4 g l  < 4 g l  = 4 g l  > l l ) 2 Z jZ tg( l g g   = 终端开路线段 ) 2 Z jZ ctg( l g g   = − Z=jX Z 0 4 g l  < 4 g l  = 4 g l  > l l Zg