ln[(s-1)/(s+1)] Laplace inverse的詳細過程?
回答 (1)
2016-06-07 7:44 am
✔ 最佳答案
令 F(s) = ln [ (s-1) / (s+1) ] = ln (s-1) - ln (s+1)dF/ds = 1/(s-1) - 1/(s+1)
ℒ⁻¹ { dF/ds } = ℒ⁻¹ { 1/(s-1) } - ℒ⁻¹ { 1/(s+1) }
- t * f(t) = e^t - e^(-t)
f(t) = [ e^t - e^(-t) ] / ( - t ) = [ e^(-t) - e^t ] / t
Ans: [ e^(-t) - e^t ] / t
收錄日期: 2021-05-02 14:09:50
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160606082936AAGLT6x