Re: [理工] [工數]-拉式轉換
※ 引述《smallprawn (水中瑕)》之銘言：
: L{(cos t)u(t-π/2)}=L{-(sin(t-π/2)u(t-π/2)}
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: 為什麼第二步不是 L{cos(t-π/2)u(t-π/2)}
π
cos(t- --- ) = sint
2
π
sin(t - --- ) = -cost
2
: 除了尤拉判斷以外.他的判斷依據又可以是什麼?!
: 如果變成L{(cos t)u(t-3π/2)}=L{(_____(t-3π/2))u(t-3π/2)}
: 或者L{(sin t)u(t-π/2)}=L{(_____(t-π/2))u(t-π/2)}
不要想太多@@
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