Re: [理工] [工數]-拉氏
※ 引述《wil0829ly (汪汪)》之銘言:
: 1.
: √(s-a)
已知
-1Γ(n+1) n
L ─── = t
n+1
s
把 s軸平移 換進去
-1 Γ(n+1) n at
L ──── = t e
(s-a)^n+1
n = -3/2
-1 Γ(-1/2) -3/2 at
L ─────── = t e
(s-a)^(-1/2)
-1 1 -3/2 at 1
L ─────── = t e ─────
(s-a)^(-1/2) Γ(-1/2)
Γ(-1/2) ≒ 2.363 多
: 2.
: s
: arc cot(──)
: π
-1 -1 s
L cot (──)
π
1/π
L tf(t) = ─────────
1 + (s/π)^2
π
= ────────
s^2 + π^2
tf(t) = sin πt
sin πt
f(t) = ─────
t
: 這兩題的拉氏逆轉換怎麼求
: 請高手幫忙一下@@
: 感恩
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