→ suker :sinwx = (1/2i)*[e^(iwx)-e^(-iwx)] 03/27 21:16
→ suker :ζ<代表反拉不好打用這個表示>,ζsin(wx) = (1/2i)* 03/27 21:18
→ suker :[ζe^(iwx)-ζe^(-iwx)]=(1/2i)*[1/(s-iw)-1/(s+iw)] 03/27 21:20
→ suker :然後你在通分=w/(s^2+w^2) ; ζe^ax = 1/ s-a 03/27 21:21
→ soupbone :看懂了!!! 感謝^^ 03/27 21:53