反laplace轉換3/S(S^2+9)，請求解題?
回答 (1)
2015-12-12 11:18 pm
￡{3/S(S^2+9)} = ?
= ￡{a/s + bs/(s^2+9) + c/(s^2+9)}
= ￡{1/3s - s/3(s^2+9)} ;;; Note
= 1/3 - 1/3 * cos(3t)
= (1 - cos(3t))/3
Note:
3 = a(s^2+9) + bs^2 + cs
= (a+b)s^2 + cs + 9a
=> a = 1/3
=> b = -a = -1/3
=> c = 0
= ￡{a/s + bs/(s^2+9) + c/(s^2+9)}
= ￡{1/3s - s/3(s^2+9)} ;;; Note
= 1/3 - 1/3 * cos(3t)
= (1 - cos(3t))/3
Note:
3 = a(s^2+9) + bs^2 + cs
= (a+b)s^2 + cs + 9a
=> a = 1/3
=> b = -a = -1/3
=> c = 0
收錄日期: 2021-04-30 20:01:15
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20151212090438AAeHLsv