Hard Laplace Transform

2011-07-20 9:55 pm

回答 (1)

2011-07-22 5:57 am
✔ 最佳答案
(a) cost * (1/3)sin(3t)
=∫[0~t] cos(t-x) (1/3)sin(3x) dx
=(1/6)∫[0~t] sin(2x+t)+ sin(4x-t) dx
=(-1/6) [ (1/4) cos(4x-t) + (1/2) cos(2x+t) ] sub. for x=0~ t
=(-1/6) [ (3/4) cos(3t) - (3/4) cos(t) ]
=(1/8) [cos(t)- cos(3t)]
(b) By the method of Laplace transform.
s^2 Y - s - 0 + 9Y= - e^(-πs) s/(s^2+1) + e^(-3πs)
(Note: u_π(t) cos(t)= -u_π(t) cos(t-π) )
Y=s/(s^2+9)+ e^(-3πs)/(s^2+9) - e(-πs) s/[(s^2+1)(s^2+9)]
y= cos(3t)- (1/3)u_3π(t) sin(3t)- (1/8) u_π(t) [-cos(t)+cos(3t)]


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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110720000016KK05063

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