Re: [理工] inverse laplace
※ 引述《ziizi (ziizi)》之銘言：
s
----------
-1 4 4
L { s + a }
針對這題我是使用複變函數來解
可是解完就花很多時間了...
想問板上高手們遇到這題會怎麼處理呢?
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 1.170.186.180
推
10/08 13:17,
10/08 13:17
→
10/09 00:15,
10/09 00:15
→
10/09 00:17,
10/09 00:17
推
10/09 00:31,
10/09 00:31
→
10/09 00:31,
10/09 00:31
→
10/09 16:13,
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
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 140.121.146.175
推
10/12 13:35, , 1F
10/12 13:35, 1F
推
10/12 19:29, , 2F
10/12 19:29, 2F
推
10/12 20:05, , 3F
10/12 20:05, 3F
討論串 (同標題文章)
完整討論串 (本文為第 3 之 3 篇)：