拉氏轉換,逆轉換問題

Q1:
L{sin(wt+δ)}=L{ sin(wt)cosδ+cos(wt)sinδ}= (wcosδ+s sinδ)/(s^2+w^2)

Q2: sin2tcos2t= 0.5 sin(4t)
L{ sin(2t)cos(2t)}= L{ 0.5sin(4t) }= 2/(s^2+16)

Q3:
L^(-1){ 4/( s^2 - 2s - 3 ) }=L^(-1) { 1/(s-3) - 1/(s+1) }= e^(3t) - e^(-t)

Q4:
L^(-1) { 3/(s^2+9) }= f(t)= sin(3t)
L^(-1){3(1-e^(-πs))/(s^2+9)}
=f(t)- u(t-π)f(t-π)
= sin(3t) - u(t-π) sin[3(t-π)]
= sin(3t) + u(t-π) sin(3t)

Q5:
L^(-1) { s/(s^2+π^2) } = f(t)= cos(πt)
L^(-1) { s e^(-2s)/(s^2+π^2) }
= u(t-2) f(t-2)
= u(t-2) cos[π(t-2)]
= u(t-2) cos(πt)

Q6:
L^(-1) { s/[ (s+0.5)^2+1 ] }
=L^(-1) { (s+0.5)/[(s+0.5)^2+1] - 0.5/[(s+0.5)^2+1] }
= e^(-0.5t) L^(-1) { (s-0.5)/(s^2+1)}
= e^(-0.5t) (cost - 0.5sint)

Note:(1) L {u(t-a)f(t-a)} = e^(-as) L{ f(t) }
(2) L {f(t)}=F(s), then L{e^(at)f(t) }= F(s-a)

我一開始轉錯了~~把cos2倍角轉成cos、sin的平方相減@@

1. sin(wt+δ) = sin(wt)cos(δ)+cos(wt)sin(δ) + linearity of Laplace transform
2. sin2tcos2t = 1/2 sin(4t)

這樣就太簡單了

第一題是不是有點怪怪的