[工數] 拉普拉斯轉換

[kax0205 (說好的數學史呢?)]

f(t)={    cos(t) for 0<=t<2π
        2-sin(t) for t>= 2π

求拉普拉斯轉換後的f(t)=?

答案是 s/(s+1)+(2/s- s/(s^2+1) - 1/(s^2+1) ) e^(-2πs)

敢問各位大大解法or破題方式QAQ


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→ darrenmm    :f(t)= cost*[u(t)-u(t-2π)]+(2-sint)*[u(t-2π)]
→ darrenmm    :然後轉換就好了
→ ntustjc     :heaviside??