Re: [理工] [工數]-Laplace
看板Grad-ProbAsk作者CRAZYAWIND (考試快到了!!)時間15年前 (2010/03/13 20:34)推噓14(14推 0噓 21→)留言35則, 12人參與討論串21/38 (看更多)
※ 引述《cccoco (危機感)》之銘言:
: 請問
: y'+y=u(t-1)-u(t-2)
: 的拉式轉換
: 1
: Y=---------(e^-s - e^-2s)
: (s+1)s
: 再來要怎麼處理呢?
: 謝謝喔!
部分分式
1 1 1 1
y(t) = (── - ───)e^-s -(── - ───)e^-2s
s s+1 s s+1
剩下的在inverse回來就好了
y(s) = [1- e^-(t-1)]u(t-1) -[1 -e^-(t-2)]u(t-2)
我趁這題我在問一下很新的一個題目
y" - 5y' +6y = u(t-1)-u(t-2) 初值等於0
求y(1)≦ y(t) ≦ y(1.5)
今天中興99電機的工數題目= =
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◆ From: 59.105.159.190
※ 編輯: CRAZYAWIND 來自: 59.105.159.190 (03/13 20:35)
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※ 編輯: CRAZYAWIND 來自: 59.105.159.190 (03/13 20:48)
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