理工] [工數]-拉氏!!!~~~~~~~~~~~~~~~~

[topee (eason)]

[u(t)-u(t-1)]t^2

拉氏轉換


    2     1       2    1
=  ----  ---  -  ---- ---e^-s
   s^3    s      s^3   s

     2       2
=  ----- -  ----e^-s
    s^4     s^4

這樣可行??


= e^-0s L[t^2]           - e^-s L[t^2]
              t=t-0                   t = t+1

= e^-0s L[t^2]   -  e^-s L[(t+1)^2]

    2                2        2      1
= -----e^-0s   -  [----- +  ----- + ---]e^-s
   s^3              s^3      s^2     s

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◆ From: 125.228.221.105
→ iyenn:公式...在看一下  L(f(t)u(t-a))=L(f(t+a))e^-as
→ iyenn:http://en.wikipedia.org/wiki/Laplace_transform
推 squallting:轉回來就知道了 (茶)
→ iyenn:還可以順便練習逆轉換是嗎=.=
→ iyenn:這轉的我有點昏了 1/s 是?= =
→ iyenn:我看懂原po的疑點了XDD
→ topee:那個平移公式 我有想過 可是套直套不出來!!
→ topee:1/s   我當成u(t)     想
→ futurexplor:因為有一個是非同步的押!
→ topee:哪為神人拿套出來 幫解一下``
→ iyenn:你的想法是錯的呀,相乘轉換過去不是相乘...
→ topee:平移怎麼套出來?
→ topee:[1-t^2]e^-s ?
→ iyenn:分開移...
※ 編輯: topee           來自: 125.228.221.105      (03/31 23:24)
→ topee:麻煩I大再檢查一次~
推 squallting:e^-0s = 1
→ topee:對厚... 是不是只有脈衝才要這樣補?