Re: [理工] [工數]-拉氏兩題
※ 引述《redwing119 (翼迷)》之銘言:
: 1. L[(sinat)^2]
=0.5*L{1-cos(2at)}
=0.5*[1/s - s/s^2+(2a)^2]
: -1 s+a
: 2. L [ln(-----)]
: s+b
-1
= L{ln(s+a) - ln(s+b)}
-1
= (-1/t)L {1/(s+a) - 1/(s+b)}
= (-1/t)*[e^(-at) - e^(-bt)]
: THANKS...
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