推 momo210723:很佩服 有人能用ptt把這些符號打出來 而且我也看的懂的 04/25 19:40
-2t t 2τ
f(t)= e ∫ e cos(3τ)dτ
0
(sol)
s
1. L[cos3t] = ---------
(s^2)+9
s-2
2. L[e^(2t) cos(3t)] = --------------
[(s-2)^2] +9
t s-2
3. L[∫ e^(2τ) cos(3τ)dτ] = -----------------
0 s{[(s-2)^2] +9}
-2t t 2τ s
4. L[e ∫ e cos(3τ)dτ] = ----------------
0 (s+2) [(s^2)+9]
set L[f(t)] = F(s) 取I.E.之L.T.
s
-> F(s) = ----------------
(s+2) [(s^2)+9]
-1
L t -2(t-τ)
---> f(t) = ∫ e cos(3τ) dτ
0
2τ
-2t e τ=t
= e [ ----------- (2cos(3τ)+3sin(3τ))]
(2^2)+(3^2) τ=0
2t
-2t e 1
= e [---- (2cos(3t)+3sin(3t)) - ---- (2) ]
13 13
1 -2t
= ---- (2cos(3t)+3sin(3t) - 2e )
13
※ 引述《asweknow ( )》之銘言：
: http://140.134.4.2/~admission/fcu_exam/pm99/015.pdf
: 請問第二題的laplace要怎麼求解
: 右邊那一串取laplace應該會很醜吧
: 謝謝
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 163.22.18.57