問題1.3 ラプラス変換

8. t^2 (0< t <1)
これを単位階段関数を用いて表すと
t^2 u(t) - t^2 u(t-1)
L{t^2 u(t) - t^2 u(t-1)}
= L{t^2 u(t)} - L{t^2 u(t-1)}
= L{t^2 u(t)} - L{(t-1)^2 u(t-1)} + 2L{(t-1) u(t-1)} + L{u(t-1)}
= 2!/s^3 * exp(0)/s - exp(-s) * 2!/s^3 + 2 exp(-s)* 1!/s^2 + exp(-s) / s
= 2/s^4 - 2 exp(-s) / s^3 + 2 exp(-s) / s^2 + exp(-s) / s
= 2s^(-4) - 2s^(-3) exp(-s) + 2s^(-2) exp(-s) + s^(-1) exp(-s)

9. sin ωt (0< t < π/ω)
これを単位階段関数を用いて表すと
u(t)sin ωt - u(t - π/ω)sin ωt
L{u(t)sin ωt - u(t - π/ω) sin ωt}
= L{u(t) sin ωt} + L{u(t - π/ω) sin ω(t-π)}
= ω/(s^2 + ω^2) + ω exp(-πs/ω)/{(s-π/ω)^2 + ω^2}
= {ω{(s - π/ω)^2 + ω^2} + ω exp(-πs/ω)(s^2 + ω^2)}/ (s^2+ω^2){(s-π/ω)^2 + ω^2}
= ω{1+exp(-πs/ω)}/(s^2 + ω^2)

10. 1- exp(-t) (0<t<2)
これを単位階段関数を用いて表すと
u(t) {1 - exp(-t)} - u(t-2){1 - exp(-t)}
L{u(t){1 - exp(-t)} - u(t-2){1 - exp(-t)}}
= L{u(t)} - L{u(t)exp(-t)} - L{u(t-2)} + L{u(t-2)exp(-(t-2))}
= exp(-0)/s - exp(0)/(s+1) - exp(-2s)/s + exp(-2)L{exp(-(t-2))u(t-2)}
= 1/s - 1/(s+1) - exp(-2s)/s + exp(-2)exp(-2x)/(s+1)
= (1 - exp(-2s)/s + (exp(-2s-2) - 1)/(s+1)

11. exp(t) (0<t<1)
これを単位階段関数を用いて表すと
exp(t)u(t) - exp(t)u(t-1)
L{exp(t) u(t) - exp(t) u(t-1)}
= L{exp(t) u(t)} - e L{exp(t-1) u(t-1)}
= exp(0) / (s-1) - e・exp(-s)/(s-1)
= (1 - exp(1-s))/(s-1)

12. sin t (2π < t < 4π)
これを単位階段関数を用いて表すと
u(t - 2π)sin(t) - u(t - 4π)sin(t)
L{u(t - 2π)sin(t) - u(t - 4π) sin(t)}
= L{u(t - 2π)sin(t - 2π)} - L{u(t - 4π)sin(t - 4π)}
= exp(-2πs) / (s^2 + 1) - exp(-4πs) / (s^2 + 1)
= {exp(-2πs) - exp(-4πs)} / (s^2 + 1)

13. 10cos(πt) (1<t<2)
これを単位階段関数を用いて表すと
u(t-1)・10cos(πt) - u(t-2)・10cos(πt)
 L{10u(t-1)cos(πt) - 10 u(t-2)cos(πt)}
= - 10 L{u(t-1)cosπ(t-1)} - 10 L{u(t-2)cosπ(t-2)}
= -10s exp(-s)/(s^2 + π^2) - 10s exp(-2s)/(s^2 + π^2)
= -10(exp(-s) + exp(-2s))/(s^2 + π^2)