L[tf(t)]=1/[s(s^2-4)]
find out L[e^(-t)f(2t)]
laplace transform question
2009-12-26 5:37 am
回答 (1)
2009-12-26 6:56 am
✔ 最佳答案
L{ f(t) }=F(s) => -F'(s)=L{ tf(t) }=1/[s(s^2-4)] =>F(s)=(1/8)[2lns-ln(s^2-4)]L{ f(2t)}= ∫[0~∞] exp(-st)f(2t)dt= (1/2)∫[0~∞] exp(-0.5s t)f(t) dt
= (1/2) F(s/2)= (1/16)[2ln(s/2)-ln(s^2/4 - 4)]
L{ exp(-t) f(2t) }= (1/16){ 2ln[(s+1)/2] - ln[(s+1)^2/4 - 4] }
=(1/16) { 2ln(s+1)- ln[(s+1)^2-16] }
收錄日期: 2021-05-04 00:46:57
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20091225000016KK06293