作者Frobenius (i^(-i)= e^(π/2))
看板trans_math
标题Re: [转录][问题] 救救我呀
时间Wed Feb 7 01:48:35 2007
※ 引述《Frobenius (i^(-i)= e^(π/2))》之铭言：
: ※ 引述《Jasy (Transferers)》之铭言：
: -1 ( s + 1 )
: Ｌ { ─────────────── }
: ( s^2 + 1 )( s^2 + 4s + 13 )
: -1 1 1 1 s -1 3
: = Ｌ { ─ ──── + ─ ──── + ─ ─────────
: 10 s^2 + 1 20 s^2 + 1 15 [ (s+2)^2 + 3^2 ]
: -1 (s+2)
: + ─ ───────── }
: 20 [ (s+2)^2 + 3^2 ]
: = (1/10)sin(t) + (1/20)cos(t) - (1/15)e^(-2t)sin(3t) - (1/20)e^(-2t)cos(3t)
1 s 3 (s+2)
设 A ──── + B ──── + C ───────── + D ─────────
s^2 + 1 s^2 + 1 [ (s+2)^2 + 3^2 ] [ (s+2)^2 + 3^2 ]
(至於为什麽要这样设，因为这样比较好看出反转换)
( A + B s ) ( s^2 + 4s + 13 ) + ( 3 C + D(s+2) ) ( s^2 + 1 ) = s + 1 = 1 + s
令 s = i，(A + B i) ( -1 + 4i + 13 ) = (12A - 4B) + i(4A + 12B) = 1 + i
12A - 4B = 1 36A - 12B = 3
=>
4A + 12B = 1 +) 4A + 12B = 1
────────
40A = 4 => A = (1/10) => 12B = (6/10) => B = (1/20)
s^2 + 4s + 13 = 0 => s = -2 ± 3i
令 s = -2 + 3i，1 + s = 1 - 2 + 3i = -1 + 3i，s + 2 = 3i
s^2 = -5 - 12i，s^2 + 1 = -4 -12i，
( 3 C + 3 D i ) ( -4 -12i ) = (-12 C + 36 D) + i(-36 C - 12 D) = -1 + 3i
-12C + 36D = -1 -12C + 36D = -1
=>
-36C - 12D = 3 -) -12C - 4D = 1
─────────
40D = -2 => D = (-1/20)
=> -12C = (6/5) => C = (-1/10)
A = (1/10)，B = (1/20)，C = (-1/10)，D = (-1/20)
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◆ From: 140.122.225.109
1F：→ Frobenius:这题说不定用留数更快解决140.122.225.109 02/07 04:06
2F：→ Frobenius:用留数反而慢140.122.225.109 12/27 19:21