CHAPTER 5 LAPLACE TRANSFORM Laft bgt a Lft

CHAPTER 5 拉普拉斯變換 (LAPLACE TRANSFORM)

(定理一) 拉普拉斯具有線性性質 L(af(t) + bg(t)) = a. L(f(t)) + b. L(g(t)) = a. F(S) + b. G(S) L(a 1 f 1 + a 2 f 2 + a 3 f 3 +…+ anfn ) = a 1 L(f 1) + a 2 L(f 2) + a 3 L(f 3)+…+ an. L(fn ) = a 1 F 1(S) + a 2 F 2(S) + a 3 F 3(S)+…+ an. Fn(S) Example 3 and 4(p 252 ~ p 253)

(CASE 2)重覆因子 (s-a)m G(s) = (s-a)m Y(s) = ( Am / (s – a)m) + ( Am-1 /( s – a)m-1) + ( Am-2 /( s – a)m-2) +…+ ( A 1 / (s – a)) y(t) = [Am/(m-1)!] tm-1 eat + [Am-1/(m-2)!] tm-2 eat +…+A 1 eat L(tneat) = n! / (s-a)n+1 Example 2

(CASE 4)重覆複數因子[(s-a)(s-a＊)]2 Y(s) 有 As + B / [(s-a)(s-a＊)]2 + Ms + N / (s-a)(s-a＊) 型態之部份分式 L[s / (s 2 + w 2)2] = (t/2 w)sinwt 或 L[s-α / ((s-α)2+ w 2)2 ] = (t / 2 w)eαtsinwt L[1 / (s 2 + w 2)2] = 1 / 2 w 3(sinwt - wtcoswt) 或 L[1 / ((s-α)2+ w 2)2] = eαt / 2 w 3(sunwt - wtcoswt) Example 4

有變換式 e-as. F(s) ie L[f(t - a)u(t - a)] = e-as. F(s) 或逆變式 f(t - a)u(t - a) = L-1[e-as. F(s)] 當 f(t - a) = 1 L[f(t - a)u(t - a)] = L[u(t - a)] = e-as / s Example 1 , 2

roblem 5 -1 f(t) = t 0 <= t <= 0 2 -t 1 <= t <= 2 (法一) L(f(t)) = ∫e-stf(t)dt = ∫te-stdt + ∫(2 -t)e-stdt = (1 – 2 e-s + e-2 s) / s 2

Example f(t) = t 0<= t <=1 1 t >= 1

= [(1/6)(1 / s) – (1/2)(1 / s+2) + (1/3)(1 / s+3)]e-s + [(1 / s+2) – (1 / s+3)]e-2 s + [(1 / s+2) –(1 / s+3) ] y(t) = [(1/6)u(t - 1) – (1/2)e-2(t -1)u(t-1) + (1/3)e-3(t-1)u(t-1)] + [e-2(t-2)u(t -2) – e-3(t-2)u(t-2)] + e-2 t –e-3 t = e-2 t – e-3 t + [(1/6) – (1/2)e-2(t-1) + (1/3)e-3(t-1)]u(t-1) + [e-2(t-2) – e-3(t-2)]u(t -2)

Example L(t 2 sint) = (-1)2 (d 2/ds 2) L (sint) = (d 2/ds 2)(1/( s 2+1)) = d/ds (-2 s/( s 2+1)2) =[ -2( s 2+1)2 + (2 s)2(s 2+1)(2 s)] / (s 2+1)4 ={ (s 2+1)[-2(s 2+1)+8 s 2] }/ (s 2+1) 4 =6 s 2 -2/( s 2+1)3 ∴ L(t y′) = (d 2/ds 2) (s Y – y(0)) = d/ds (Y+s(d. Y/ds)) = d. Y/ds + s( d 2 Y/ds 2 ) = s( d 2 Y/ds 2 ) + 2( d. Y/ds)

L (t 2 y〞) = (d 2/ds 2) (s 2 Y – sy(0)-y′(0)) = d/ds (2 s. Y + s 2 (d Y/ds) - y(0)) = 2 Y +2 s (d Y/ds) + s 2( d 2 Y/ds 2) = s 2( d 2 Y/ds 2) + 4 S(d Y/ds) +2 Y

Example t y〞+(t-1) y′+ y = 0 y(0) = 0 L (t y〞) = -2 s. Y - s 2( d. Y/ds) +y(0) L(t y′) = - Y – s ( d. Y/ds) → (s 2+s) Y′+ 3 s Y = 0 →(s+1) Y′+ 3 Y =0 ∴ d. Y/ds = (-3 / s+1 ) Y → d. Y/Y =( -3/s+1) ds

Ln Y = -3 ln(s+1) + c* = ln c (s+1)-3 ∴ Y= c (s+1)-3 = c/(s+1)3 y(t) = L -1 (Y(s)) = (c/2) t 2 e-t = k t 2 e-t

(積分之推度) ∫∫………∫ F(s)ds…ds = L {F(t)/tn} n Ex L {sin 2 t/t} = ? L (sin 2 t) = 2/(s 2+4) L {sin 2 t/t} = ∫ {2/ (S 2+4) } d. S = tan -1 S/2 | = π/2 - tan -1 S/2 = cos -1 S/2