862. LAPLACE变换 16/48圆 所以 [ (Rep>0). 同理 2[ pn+1) (Rep>0). 例4求e门和[e],s为常数 解:在Rep>Res的上半平面上, sto-pt dt (p-s)t dt 所以 (Rep>res). P 类似地 dt te p-s ● First●Prev●Next●Last● Go Back● Full Screen●cose●Quit
• First • Prev • Next • Last • Go Back • Full Screen • Close • Quit §6.2. LAPLACE C 16/48 ¤± L [t] = 1 p 2 , (Rep > 0). Ón L [t n ] = n! p (n + 1), (Rep > 0). ✿ ~¯¦ L e st Ú L te st , s ~ê© )µ3 Rep > Res þ²¡þ§ L e st = Z ∞ 0 e ste −ptdt = Z ∞ 0 e −(p−s)tdt = − 1 p − s e −(p−s)t ∞ 0 = 1 p − s . ¤± L e st = 1 p − s , (Rep > Res). aq/ L te st = Z ∞ 0 te ste −ptdt = Z ∞ 0 te −(p−s)td = − 1 p − s Z ∞ 0 td e −(p−s)t
62. LAPLACE变换 1748则 te dt 0 (p-s)2 (Rep >res). 同理 例5求[f(),其中f()是存在 Laplace变换的任意函数 解:将 Laplace变换的定义式(62-3),两边分别对p求导 df(t) (t)e"p f(t)dt, P 从而 tf(t)(-1 df(p) d P ● First●Prev●Next●Last● Go Back● Full Screen●cose●Quit
• First • Prev • Next • Last • Go Back • Full Screen • Close • Quit §6.2. LAPLACE C 17/48 = − 1 p − s te −(p−s)t ∞ 0 − Z ∞ 0 e −(p−s)tdt = 1 (p − s) 2 , (Rep > Res). Ón L t n e st = n! (p − s) n+1 . ✿ ~°¦ L [t f(t)]§Ù¥ f(t) ´3 Laplace C?¿¼ê© )µò Laplace C½Âª(6.2-3)§ü>©Oé p¦§ d ¯f(t) dp = Z ∞ 0 (−t)e −pt f(t)dt, l t f(t) ; (−1)d ¯f(p) dp .