工程數學 (急) 20點

1.以微分方式求出轉換1-1.t*sin(wt)L{sin(wt)} = w/(s^2+w^2)L{t*sin(wt)} = -{w/(s^2+w^2)}'= 2ws/(s^2+w^2)^2
1-2.(1/2)t^2*cos(wt), w=π/2L{cos(wt)} = s/(s^2+w^2)L{t*cos(wt)} = -{s/(s^2+w^2)}'= [(s^2+w^2)-s*2s]/(s^2+w^2)^2= (w^2-s^2)/(s^2+w^2)^2L{t^2*cos(wt)} = -{(w^2-s^2)/(s^2+w^2)^2}'= -[-(s^2+w^2)2s - (w^2-s^2)*2(s^2+w^2)*2s]/(s^2+w^2)^4= [2s + 4s(w^2-s^2)]/(s^2 + w^2)^3= 2s(2w^2 + 1 - 2s^2)/(s^2 + w^2)^3=> L{t^2*cos(wt)}/2 = s(2w^2 + 1 - 2s^2)/(s^2 + w^2)^3
2.轉換式的積分: F(s)' = [ln(s^2) + 1/(s-1)^2]'= [2*ln(s) + 1/(s -1)^2]'= 2/s - 2(s-1)/(s - 1)^4= 2/s - 2/(s-1)^3=> f(t) = 2 - t^2*e^t
L{f(t)/t} = L{(2-t^2*e^t)/t} = ∫F(s)ds= ∫[2/s - 2/(s-1)^3]ds= ln(s^2) + 1/(s-1)^2=> f(t)/t = (2-t^2*e^t)/t = 2/t - t*e^t
3.摺積方式計算轉換Left = L{∫cos(wr)*1*dr}= L{sin(wr)/w}.....r=0~t= L{sin(wt)/w - 0}= L{sin(wt)/w}= w/(s^2 + w^2)w= 1/(s^2 + w^2)Right = L{f(t)*g(t)}= L{cos(wt)*1}= F(s)*G(s)= [s/(s^2 + w^2)]*(1/s)= 1/(s^2 + w^2)= Left
4.摺積方法求反轉換摺積只能正算不能反算回去

2015-01-08 07:13:21 補充：
H(s) = 2πs/(s^2+π^2)^2 = [2πs^2/(s^2+π^2)^2]*(1/s)


L{sin(πt)}/π = 1/(s^2 + π^2)

=> L{t*sin(πt)}/π

= -[1/(s^2 + π^2)]'

= 2s/(s^2 + π^2)^2.....(1)


L{cos(πt)} = s/(s^2 + π^2)

=> L{t*cos(πt)}

= -{s/(s^2 + π^2)}'

= -{[(s^2 + π^2) - 2s^2]/(s^2 + π^2)^2

= (s^2 - π^2)/(s^2 + π^2)^2.....(2)

2015-01-08 07:14:35 補充：
L{sin(πt)/π + t*cos(πt)}

= 1/(s^2 + π^2) + (s^2 - π^2)/(s^2 + π^2)^2

= [(s^2 + π^2) + (s^2 + π^2)]/(s^2 + π^2)^2

= 2s^2/(s^2 + π^2)

2015-01-08 07:15:36 補充：
Right = L{∫[sin(πr)/π + r*cos(πr)] * 1 * dr}

= L{∫sin(πr)dr/π + ∫r*d(sin(πr))/π}

= L{-cos(πr)/π^2 + r*sin(πr)/π - ∫sin(πr)dr/π}.....Partial Int.

= L{-cos(πr)/π^2 + r*sin(πr)/π + cos(πr)/π^2}

= L{r*sin(πr)/π}.....r=0~t

= L{t*sin(πt)}

= 2s/(s^2 + π^2)^2.....by (1)

2015-01-08 07:16:26 補充：
Left = L{[sin(πt)/π + t*cos(πt)]*1}

= L{f(t)*g(t)}

= F(s)*G(s)

= [2s^2/(s^2 + π^2)]*(1/s).....by (2)

= 2s/(s^2 + π^2)

= Right



=> h(t) = sin(πt)/π + t*cos(πt)

「以微分方式求出轉換」

求出什麼轉換？？？