[理工] [工數] 拉氏轉換
題目是 y''+ 4y = f(t) , y(0)= y'(0) = 0
f(t)= 0 , 0<t<5 and f(t)=t+2 , t>5
答案是 1/4 u(t-5) [t+2 - 1/2 sin(t-5) - 7 cos2(t-5)]
我重複算了好幾次答案都是
1/4 u(t-5) [t-4 - 1/2 sin2(t-5) - 7 cos2(t-5)]
我的答案和解答插在 它是 t+2而我是t-4 還有sin那邊 我是sin2t而它是sint
想請問是否答案錯誤??
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◆ From: 61.231.244.148
推
06/17 22:51, , 1F
06/17 22:51, 1F
我想請問是怎麼算出來的?
我先列出我的算式好了
f(t)= u(t-5) (t+2)
拉普拉式轉換
(s^2+4)Y = (1/s^2 + 7 / s) e ^ -5s
Y = 1/(s^2+4) * (1/s^2 + 7 / s) e ^ -5s
之後把 1/s^2 (s^2+4) 拆成 1/4(1/s^2 -1/s^2+4) e^-5s
1/s* (s^2+4) 拆成 1/4(1/s - 7s/s^2+4)e^-5s
逆轉換後就會變成
y =1/4 u(t-5)[(t-5) - 1/2 sin2(t-5)] + 1/4 u(t-5)[1 - 7cos2(t-5) ]
整理後得 1/4 u(t-5) [t-4 - 1/2 sin2(t-5) - 7 cos2(t-5)]
是哪裡有錯嗎?
※ 編輯: gn0524 來自: 61.231.244.148 (06/17 23:11)
※ 編輯: gn0524 來自: 61.231.244.148 (06/17 23:16)
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