推 q920110:硬拆成(s^2+2^(1/2)as+s^2)*(s^2-2^(1/2)as+a^2) 10/08 13:17
→ ziizi:q大(s^2+2^(1/2)as+s^2)應該要是(s^2+2^(1/2)as+a^2)吧? 10/09 00:15
→ ziizi:謝謝Q大!不過這是怎麼想到的阿?因為這個也不是配方的樣子 10/09 00:17
推 Ifanwei:我也有想到這個步驟但是好像也無用阿 10/09 00:31
→ Ifanwei:拆成這樣後有什麼辦法可以得到最後解答? Z大有繼續算完嗎? 10/09 00:31
→ ziizi:分母化成這兩式相乘，想拆開成兩個分式...可是仍卡住 10/09 16:13
※ 編輯: ziizi 來自: 106.107.13.179 (10/09 16:22)
用部分分式的作法
s As+B Cs+D
F(s) = ------- = -------------- + -------------
s^4+a^4 s^2+√2as+a^2 s^2-√2as+a^2
| s -1
As+B = F(s)(s^2+√2as+a^2)| = -------------------- = ------
|s^2=-√2as-a^2 -√2as-a^2-√2as+a^2 2√2a
| s 1
Cs+D = F(s)(s^2-√2as+a^2)| = ------------------- = -----
|s^2=√2as-a^2 √2as-a^2+√2as+a^2 2√2a
又s^2+√2as+a^2 = s^2+√2as + a^2/2 + a^2/2 = (s+√2a/2)^2+(√2a/2)^2
s^2-√2as+a^2 = s^2-√2as + a^2/2 + a^2/2 = (s-√2a/2)^2+(√2a/2)^2
F(s) = F1(s) + F2(s)
-1/(2√2a) - √2a/2 / (2a^2)
F1(s) = ----------------------- = -----------------------
(s+√2a/2)^2+(√2a/2)^2 (s+√2a/2)^2+(√2a/2)^2
1/(2√2a) √2a/2 / (2a^2)
F2(s) = ----------------------- = -----------------------
(s-√2a/2)^2+(√2a/2)^2 (s-√2a/2)^2+(√2a/2)^2
-1 -1 -1
L (F(s)) = L (F1(s)) + L (F2(s))
-√2at/2 √2at/2
= e sin(√2at/2)u(t)/(-2a^2) + e sin(√2at/2)u(t)/(2a^2)
-√2at/2 √2at/2
= sin(√2at/2)u(t)/a^2 * [ -e /2 + e /2 ]
= sinh(√2at/2)sin(√2at/2)u(t)/a^2
熟練的話其實寫起來很快
不過前提是要先想到分母要那樣拆
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推 ziizi:謝謝e大的回覆,了解:D 10/12 13:35
推 cha0417:這題超麻煩又只有5分....= = 10/12 19:29
推 gnsh:@@ 10/12 20:05