Re: [理工] [工數]-傅立葉轉換
※ 引述《goodguychung (泡麵終結者)》之銘言：
: sin(t)sin(0.5t)
: x(t)=-----------------
: πt^2
: 請問這題該如何做傅立葉轉換?
1 |t|<1
令f(t) = { 0 |t|>1
1 |t|<1/2
g(t) = { 0 |t|>1/2
F{f(t)} = 2sinw /w
F{g(t)} = 2sin(w/2)/w
F{1/4π f(t)*g(t)} = sinwsin(w/2)/πw^2
用對偶性質
F{sin(-t)sin(-t/2)/π(-t)^2} = F{sintsin(t/2)/πt^2}
=2π(1/4π f(w)*g(w)) = 1/2f(w)*g(w)
1 w+0.5 w 3
───∫ dw = ── + ───
2 -1 2 4
1 w+0.5 1
───∫ dw = ───
2 w-0.5 2
1 1 w 3
───∫ dw = - ─── + ────
2 w-0.5 2 4
1/2w + 3/4 -3/2<w<-1/2
{1/2 -1/2<w<1/2
-1/2w +3/4 1/2<w<3/2
0 其他
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◆ From: 59.105.159.190
※ 編輯: CRAZYAWIND 來自: 59.105.159.190 (03/03 19:36)
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