問題1.3 ラプラス逆変換

14. 4{exp(-2s) - 2exp(-5s)}/s
= 4 exp(-2s)/s - 8 exp(-5s)/s
= 4L{u(t-2)} - 8 L{u(t-5)}
= L{4u(t-2) - 8u(t-5)}

15. exp(-3s)/s^3
= exp(-3s) 2!/s^3・1/2!
= 1/2・exp(-3s)2!/s^3
= 1/2 L{(t-3)^2 u(t-3)}
= L{(t-3)^2/2 u(t-3)}

16. exp(-3s) / (s-1)^3
= exp(-3s) 2!/(s-1)^3・1/2!
= 1/2・exp(-3s)2!/(s-1)^3
= 1/2 L{(t-3)^2 exp(t-3)}
= L{(t-3)^2 exp(t-3) / 2}

17. 3(1 - exp(-πs)) / (s^2 + 9)
= 3/(s^2 + 3^2) - 3exp(-πs) / (s^2 + 3^2)
= L{sin(3t)} - L{u(t-π) cos3(t-π)}
= L{sin(3t)} + L{u(t-π)sin(3t)}
= L{(1 + u(t-π))sin(3t)}

18. exp(-2πs) / (s^2 + 2s + 2)
= exp(-2πs) /{(s+1)^2 + 1}
= L{u(t-2π) exp(-(t-2π)) sin ω(t-2π)}
= L{exp(2π - t) u(t - 2π) sin ω(t-2π)}

19. s exp(-2s) / (s^2 + π^2)
= exp(-2s) s/(s^2 + π^2)
= L{u(t-2) cos π(t-2)}
= L{u(t-2) cosπt cos2π + sinπt sin 2π}
= L{u(t-2) cosπt}