862. LAPLACE变换 1/48回 在p的某一区域内收敛,则由此积分确定了一个复变数的复值函数 f(t)- Laplace变换 f( f(t)e pdt. 由式(6,22)表示的 Fourier变换的逆变换,有 g(t)= G(o)eo de f(o +io)e do f(t)= f(o+io)e(otio)de 0+1o f(t)= f(pe dp. (6.2-4) 0-1oo 上式中p=σ+io,do=}dp 式(6,2-4)应理解为 0+10 f(t)= lim 2io→∞ f(p)el dp. d-10 ● First●Prev●Next●Last● Go Back● Full Screen●cose●Quit
• First • Prev • Next • Last • Go Back • Full Screen • Close • Quit §6.2. LAPLACE C 11/48 3 p ,«SÂñ§KddÈ©(½ ECêE¼ê ¯f(t) —Laplace C ¯f(p) = Z ∞ 0 f(t)e −ptdt. dª(6.2-2)L« Fourier C_C§k g(t) = Z ∞ −∞ G(ω)e iωtdω = Z ∞ −∞ 1 2π ¯f(σ + iω)e iωtdω f(t) = 1 2π Z ∞ −∞ ¯f(σ + iω)e (σ+iω)tdω f(t) = 1 2πi Z σ+i∞ σ−i∞ ¯f(p)e ptdp. (6.2-4) þª¥ p = σ + iω, dω = 1 i dp. ª(6.2-4)An) f(t) = 1 2πi lim ω→∞ Z σ+iω σ−iω ¯f(p)e ptdp.
62. LAPLACE变换 12/48回 式(62-4)称为 Laplace逆变换或 Laplace变换的反演_ Riemann-Mellin 反演公式 立 Laplace变换和 Laplace逆变换常记为 f(p)=if(t)ls (6.2-5) f(t)=If(p) (6.2-6) 或等价地 f(p)言∫(t), (6.2-7) f(t)与∫(p) (6.2-8) 至注意:原函数均应理解为fOH(),在逆变换中尤其要注意此点 62.1 Laplace变换的存在性 Laplace积分和变换的存在条件是 ●Fist●Prev·Next●Last● Go Back● Full Screen●cose●Quit
• First • Prev • Next • Last • Go Back • Full Screen • Close • Quit §6.2. LAPLACE C 12/48 ª(6.2-4)¡ Laplace _C½ Laplace Cü—Riemann-Mellin üúª© ~ Laplace C. . Ú. Laplace _. C. . ~. P. . ¯f(p) = L [f(t)], (6.2-5) f(t) = L −1 [ ¯f(p)]. (6.2-6) ½ d/ ¯f(p) : f(t), (6.2-7) f(t) ; ¯f(p). (6.2-8) ~ 5. ¿. µ. ¼. ê. þ. A. n. ). . f(t)H(t)§3. _. C. . ¥. c. Ù. . 5. ¿. d. :. © 6.2.1 Laplace C35∗ Laplace È©ÚC3^´µ
862. LAPLACE变换 13/48则 1在0≤t<∞的任一有限区间上,除了有有限个第一类间断点 外,函数及其导数处处连续; 存在常数M>0和σ≥0,使对任何t值(0≤t<∞),有 If(t)l Meo (6.2-9) 上式的意思是f(t)的增长速度不超过指数函数,这样的函数称 为指数级函数.o的下界称为收敛指标,用00表示.实际应用 中,大多数函数满足此充分条件 例 S例1求 Heaviside函数的 Laplace变换,即x[0 Laplace变换 解 cIH(]=H(p)=/e-p'dt=--e"pt=00 ● First●Prev●Next●Last● Go Back● Full Screen●cose●Quit
• First • Prev • Next • Last • Go Back • Full Screen • Close • Quit §6.2. LAPLACE C 13/48 1 3 0 ≤ t < ∞ ?k«mþ§Ø kk1amä: §¼ê9Ùê??ëY¶ 2 3~ê M > 0 Ú σ ≥ 0§¦é?Û t £0 ≤ t < ∞¤§k |f(t)| < Me σt . (6.2-9) þª¿g´ f(t) OÝØLê¼ê§ù¼ê¡ ê?¼ê©σ e.¡ÂñI§^ σ0 L«©¢SA^ ¥§õê¼ê÷vd¿©^© ~ ~ ✿ ~¬¦ Heaviside ¼ê Laplace C§=L [1] Laplace C© )µ L [H(t)] = H¯ (p) = Z ∞ 0 e −ptdt = − 1 p e −pt t=∞ t=0 ,
62. LAPLACE变换 1448回 仅当σ=Rep>0时, imtoo e-pr存在且为零,所以 H(p) (Rep >0). S例2求δ函数8()和8(t-)的 Laplace变换x[6()和 [6(t-τ) 解 z[6(t)] d(t) di d(te dt S(t)eP'dt= S(t)e pdi 2b(t-]=/6(-emdt 6(t-τepdt ● First●Prev●Next●Last● Go Back● Full Screen●cose●Quit
• First • Prev • Next • Last • Go Back • Full Screen • Close • Quit §6.2. LAPLACE C 14/48 ∵ e −pt = e −(σ+iω)t = e −σt , ∴ = σ = Rep > 0 §limt→∞ e −pt 3 "§¤± H¯ (p) = 1 p , (Rep > 0). ✿ ~¦ δ ¼ê δ(t) Ú δ(t − τ) Laplace C L [δ(t)] Ú L [δ(t − τ)]© )µ L [δ(t)] = Z ∞ 0 δ(t)e −ptdt = Z ∞ 0 − δ(t)e −ptdt = Z 0 + 0 − δ(t)e −ptdt = Z ∞ −∞ δ(t)e −ptdt = e −pt |t=0 = 1. L [δ(t − τ)] = Z ∞ 0 δ(t − τ)e −ptdt = Z ∞ 0 − δ(t − τ)e −ptdt
62. LAPLACE变换 15/48则 6(t-τepdt d(t - dt 即 z[6(t)=1,x[(t-t)=e 例3求x[l 解:在Rep>0的上半平面上 gIt] te p dt td(e pRo t(e pi)l P 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 15/48 = Z 0 + 0 − δ(t − τ)e −ptdt = Z ∞ −∞ δ(t − τ)e −ptdt = e −pt |t=τ = e −pτ . = L [δ(t)] = 1, L [δ(t − τ)] = e −pτ . ✿ ~®¦ L [t]© )µ3 Rep > 0 þ²¡þ§ L [t] = Z ∞ 0 te −ptdt = − 1 p Z ∞ 0 td e −pt = − 1 p t e −pt∞ 0 + 1 p Z ∞ 0 e −ptdt = 1 p Z ∞ 0 e −ptdt = 1 p 2 ,