【精选】拉普拉斯变换与拉普拉斯逆变换的常用结论与经典公式

设函数 f ( t ) f(t) f(t)是定义在 [ 0 , + ∞ ) [0,+\infty) [0,+∞)上的实值函数，如果对于复参数 s = β + j w s=\beta+\text{j}w s=β+jw，积分 F ( s ) = ∫ 0 + ∞ f ( t ) e − s t   d t (1) F(s)=\int_0^{+\infty}f(t)e^{-st}\,dt\tag{1} F(s)=∫0+∞​f(t)e−stdt(1) 在复平面 s s s的某一个区域内收敛，则称 F ( s ) F(s) F(s)为 f ( t ) f(t) f(t)的拉普拉斯变换（简称拉氏变换），记为 F ( s ) = L [ f ( t ) ] F(s)=\mathscr{L}[f(t)] F(s)=L[f(t)]，该函数被称为像函数。

已知函数 f ( t ) f(t) f(t)经过拉普拉斯变换后得到 F ( s ) F(s) F(s)，则原函数 f ( x ) f(x) f(x)可由 F ( s ) F(s) F(s)经过拉普拉斯逆变换得到: f ( t ) = L − 1 [ F ( s ) ] = 1 2 π j ∫ β − j ∞ β + j ∞ F ( s ) e s t   d s (2) f(t)=\mathscr{L}^{-1}[F(s)]=\frac{1}{2\pi j}\int_{\beta-j\infty}^{\beta+j\infty}F(s)e^{st}\,ds\tag{2} f(t)=L−1[F(s)]=2πj1​∫β−j∞β+j∞​F(s)estds(2) 记 f ( t ) = L − 1 [ F ( s ) ] f(t)=\mathscr{L}^{-1}[F(s)] f(t)=L−1[F(s)]， f ( t ) f(t) f(t)被称为原像函数，此过程称为拉普拉斯逆变换（简称拉氏逆变换）。

已知函数 f ( t ) = δ ( t ) f(t)=\delta (t) f(t)=δ(t)，因此该函数经过拉普拉斯变换将得到像函数：

已知 δ ( t ) \delta(t) δ(t)的定义如下： δ ( t ) = { A − τ 2 ⩽ t ⩽ τ 2 0 t > ∣ τ 2 ∣    , 且 A τ = 1 , τ → 0 + \delta(t)=\left\{ \begin{aligned} A && -\frac{\tau}{2}\leqslant t\leqslant\frac{\tau}{2} \\ 0 && t> \left|\frac{\tau}{2} \right | \\ \end{aligned}\,\,,且A\tau=1,\tau\to0^+ \right. δ(t)=⎩ ⎨ ⎧​A0​​−2τ​⩽t⩽2τ​t> ​2τ​ ​​,且Aτ=1,τ→0+

F ( s ) = ∫ 0 + ∞ f ( t ) e − s t   d t = ∫ 0 + ∞ δ ( t ) e − s t d t = ∫ − ∞ + ∞ δ ( t ) e − j w t d t = A ∫ − τ 2 τ 2 e − j w t d t = A − j w ( e − j w τ 2 − e j w τ 2 ) = 2 A w sin ⁡ w τ 2 = lim ⁡ τ → 0 + 2 sin ⁡ w τ 2 w τ = 1 \begin{aligned} F(s)&=\int_0^{+\infty}f(t)e^{-st}\,dt\\&=\int_0^{+\infty}\delta(t)e^{-st}dt\\ &=\int_{-\infty}^{+\infty}\delta(t)e^{-jwt}dt\\&=A\int_{-\frac{\tau}{2}}^{\frac{\tau}{2}}e^{-jwt}dt\\ &=\frac{A}{-jw}\left(e^{-jw\frac{\tau}{2}}-e^{jw\frac{\tau}{2}}\right)\\&=\frac{2A}{w}\sin\frac{w\tau}{2}\\ &=\lim\limits_{\tau\to0^+}\frac{2\sin\frac{w \tau}{2}}{w\tau}\\&=1 \end{aligned} F(s)​=∫0+∞​f(t)e−stdt=∫0+∞​δ(t)e−stdt=∫−∞+∞​δ(t)e−jwtdt=A∫−2τ​2τ​​e−jwtdt=−jwA​(e−jw2τ​−ejw2τ​)=w2A​sin2wτ​=τ→0+lim​wτ2sin2wτ​​=1​ 证毕

已知函数 f ( t ) = u ( t ) f(t)=u(t) f(t)=u(t)，因此该函数经过拉普拉斯变换将得到像函数： F ( s ) = ∫ 0 + ∞ f ( t ) e − s t   d t = ∫ 0 + ∞ u ( t ) e − s t d t = − 1 s e − s t ∣ 0 + ∞ = 1 s F(s)=\int_0^{+\infty}f(t)e^{-st}\,dt=\int_0^{+\infty}u(t)e^{-st}dt=-\frac{1}{s}e^{-st}\big|_0^{+\infty}=\frac{1}{s} F(s)=∫0+∞​f(t)e−stdt=∫0+∞​u(t)e−stdt=−s1​e−st ​0+∞​=s1​ 证毕

已知函数 f ( t ) = t u ( t ) f(t)=tu(t) f(t)=tu(t)，因此该函数经过拉普拉斯变换将得到像函数： F ( s ) = ∫ 0 + ∞ f ( t ) e − s t   d t = ∫ 0 + ∞ t u ( t ) e − s t d t = − 1 s ∫ 0 + ∞ t d ( e − s t ) = − 1 s [ t e − s t ∣ 0 + ∞ − ∫ 0 + ∞ e − s t d t ] = 1 s 2 \begin{aligned} F(s)&=\int_0^{+\infty}f(t)e^{-st}\,dt\\ &=\int_0^{+\infty}tu(t)e^{-st}dt\\ &=-\frac{1}{s}\int_0^{+\infty}td(e^{-st})\\ &=-\frac{1}{s}\left[te^{-st}\big|_0^{+\infty}-\int_0^{+\infty}e^{-st}dt\right]\\&=\frac{1}{s^2} \end{aligned} F(s)​=∫0+∞​f(t)e−stdt=∫0+∞​tu(t)e−stdt=−s1​∫0+∞​td(e−st)=−s1​[te−st ​0+∞​−∫0+∞​e−stdt]=s21​​ 证毕

已知函数 f ( t ) = e a t u ( t ) f(t)=e^{at}u(t) f(t)=eatu(t)，因此该函数经过拉普拉斯变换将得到像函数： F ( s ) = ∫ 0 + ∞ f ( t ) e − s t   d t = ∫ 0 + ∞ e a t u ( t ) e − s t d t = 1 a − s e ( a − s ) t ∣ 0 + ∞ = 1 s − a F(s)=\int_0^{+\infty}f(t)e^{-st}\,dt=\int_0^{+\infty}e^{at}u(t)e^{-st}dt=\frac{1}{a-s}e^{(a-s)t}\big|_0^{+\infty}=\frac{1}{s-a} F(s)=∫0+∞​f(t)e−stdt=∫0+∞​eatu(t)e−stdt=a−s1​e(a−s)t ​0+∞​=s−a1​ 证毕

已知函数 f ( t ) = sin ⁡ w t f(t)=\sin wt f(t)=sinwt，因此该函数经过拉普拉斯变换将得到像函数： F ( s ) = L [ sin ⁡ w t ] = 1 2 j ( L [ e j w t ] − L [ e − j w t ] ) = 1 2 j ( 1 s − j w − 1 s + j w ) = w s 2 + w 2 \begin{aligned} F(s)&=\mathscr{L}[\sin wt]=\frac{1}{2j}\left(\mathscr{L}[e^{jwt}]-\mathscr{L}[e^{-jwt}]\right)\\&=\frac{1}{2j}\left(\frac{1}{s-jw}-\frac{1}{s+jw}\right)=\frac{w}{s^2+w^2} \end{aligned} F(s)​=L[sinwt]=2j1​(L[ejwt]−L[e−jwt])=2j1​(s−jw1​−s+jw1​)=s2+w2w​​ 证毕

已知函数 f ( t ) = cos ⁡ w t f(t)=\cos wt f(t)=coswt，因此该函数经过拉普拉斯变换将得到像函数： F ( s ) = L [ cos ⁡ w t ] = 1 2 ( L [ e j w t ] + L [ e − j w t ] ) = 1 2 ( 1 s − j w + 1 s + j w ) = s s 2 + w 2 \begin{aligned} F(s)&=\mathscr{L}[\cos wt]=\frac{1}{2}\left(\mathscr{L}[e^{jwt}]+\mathscr{L}[e^{-jwt}]\right)\\&=\frac{1}{2}\left(\frac{1}{s-jw}+\frac{1}{s+jw}\right)=\frac{s}{s^2+w^2} \end{aligned} F(s)​=L[coswt]=21​(L[ejwt]+L[e−jwt])=21​(s−jw1​+s+jw1​)=s2+w2s​​ 证毕

已知函数 f ( t ) = t m f(t)=t^m f(t)=tm，因此该函数经过拉普拉斯变换将得到像函数： F ( s ) = L [ t m ] = ∫ 0 + ∞ t m e − s t d t = − 1 s ∫ 0 + ∞ t m d ( e − s t ) = − 1 s t m e − s t ∣ 0 + ∞ + m s ∫ 0 + ∞ t m − 1 e − s t d t = m s L [ t m − 1 ] = m ( m − 1 ) s 2 L [ t m − 2 ]             ⋮ = m ! s m L [ t 0 ] = m ! s m + 1 \begin{aligned} F(s)&=\mathscr{L}[t^m]\\&=\int_0^{+\infty}t^me^{-st}dt\\&=-\frac{1}{s}\int_0^{+\infty}t^md(e^{-st})\\ &=-\frac{1}{s}t^me^{-st}\big|_0^{+\infty}+\frac{m}{s}\int_0^{+\infty}t^{m-1}e^{-st}dt\\ &=\frac{m}{s}\mathscr{L}[t^{m-1}]\\&=\frac{m(m-1)}{s^2}\mathscr{L}[t^{m-2}]\\ &\,\,\,\,\,\,\,\,\,\,\,\vdots\\ &=\frac{m!}{s^m}\mathscr{L}[t^0]\\&=\frac{m!}{s^{m+1}} \end{aligned} F(s)​=L[tm]=∫0+∞​tme−stdt=−s1​∫0+∞​tmd(e−st)=−s1​tme−st ​0+∞​+sm​∫0+∞​tm−1e−stdt=sm​L[tm−1]=s2m(m−1)​L[tm−2]⋮=smm!​L[t0]=sm+1m!​​ 证毕

设 α , β \alpha,\beta α,β为常数，且有 L [ f ( t ) ] = F ( s ) , L [ g ( t ) ] = G ( s ) \mathscr{L}[f(t)]=F(s),\mathscr{L}[g(t)]=G(s) L[f(t)]=F(s),L[g(t)]=G(s)，则有 L [ α f ( t ) + β g ( t ) ] = ∫ 0 + ∞ ( α f ( t ) + β g ( t ) ) e − s t d t = α ∫ 0 + ∞ f ( t ) e − s t d t + β ∫ 0 + ∞ g ( t ) e − s t d t = α F ( s ) + β G ( s ) \begin{aligned} \mathscr{L}[\alpha f(t)+\beta g(t)]&=\int_0^{+\infty}\left(\alpha f(t)+\beta g(t)\right)e^{-st}dt\\ &=\alpha \int_0^{+\infty}f(t)e^{-st}dt+\beta \int_0^{+\infty}g(t)e^{-st}dt\\ &=\alpha F(s)+\beta G(s) \end{aligned} L[αf(t)+βg(t)]​=∫0+∞​(αf(t)+βg(t))e−stdt=α∫0+∞​f(t)e−stdt+β∫0+∞​g(t)e−stdt=αF(s)+βG(s)​ 证毕

设 L [ f ( t ) ] = F ( s ) \mathscr{L}\left[f(t)\right]=F(s) L[f(t)]=F(s)，则对任一常数 a > 0 a>0 a>0有 L [ f ( a t ) ] = ∫ 0 + ∞ f ( a t ) e − s t d t = 令 x = a t 1 a ∫ 0 + ∞ f ( x ) e − ( s a ) x d x = 1 a F ( s a ) \begin{aligned} \mathscr{L}[f(at)]&=\int_0^{+\infty}f(at)e^{-st}dt\\& \xlongequal{令x=at}\frac{1}{a}\int_0^{+\infty}f(x)e^{-(\frac{s}{a})x}dx\\&=\frac{1}{a}F\left({\frac{s}{a}}\right) \end{aligned} L[f(at)]​=∫0+∞​f(at)e−stdt令x=at a1​∫0+∞​f(x)e−(as​)xdx=a1​F(as​)​ 证毕

设 L [ f ( t ) ] = F ( s ) \mathscr{L}\left[f(t)\right]=F(s) L[f(t)]=F(s)，则有 L [ f ′ ( t ) ] = ∫ 0 + ∞ f ′ ( t ) e − s t d t = 分部积分 f ( t ) e − s t ∣ 0 + ∞ + s ∫ 0 + ∞ f ( t ) e − s t d t = s F ( s ) − f ( 0 ) \begin{aligned} \mathscr{L}[f'(t)]&=\int_0^{+\infty}f'(t)e^{-st}dt\\ &\xlongequal{分部积分}f(t)e^{-st}\big|_0^{+\infty}+s\int_0^{+\infty}f(t)e^{-st}dt\\ &=sF(s)-f(0) \end{aligned} L[f′(t)]​=∫0+∞​f′(t)e−stdt分部积分 f(t)e−st ​0+∞​+s∫0+∞​f(t)e−stdt=sF(s)−f(0)​ 经过数学归纳法可得： L [ f ( n ) ( t ) ] = s n F ( s ) − s n − 1 f ( 0 ) − s n − 2 f ′ ( 0 ) − ⋯ − f ( n − 1 ) ( 0 ) \mathscr{L}[{f}^{(n)}(t)]=s^nF(s)-s^{n-1}f(0)-s^{n-2}f'(0)-\cdots-{f}^{(n-1)}(0) L[f(n)(t)]=snF(s)−sn−1f(0)−sn−2f′(0)−⋯−f(n−1)(0) 证毕

设 L [ f ( t ) ] = F ( s ) \mathscr{L}\left[f(t)\right]=F(s) L[f(t)]=F(s)，则有 F ′ ( s ) = d d s ∫ 0 + ∞ f ( t ) e − s t d t = ∫ 0 + ∞ ∂ ∂ s [ f ( t ) e − s t ] d t = − ∫ 0 + ∞ t f ( t ) e − s t d t = − L [ t f ( t ) ] \begin{aligned} F'(s)&=\frac{d}{ds}\int_0^{+\infty}f(t)e^{-st}dt\\&=\int_0^{+\infty}\frac{\partial}{\partial s}\left[f(t)e^{-st}\right]dt\\ &=-\int_0^{+\infty}tf(t)e^{-st}dt\\ &=-\mathscr{L}[tf(t)] \end{aligned} F′(s)​=dsd​∫0+∞​f(t)e−stdt=∫0+∞​∂s∂​[f(t)e−st]dt=−∫0+∞​tf(t)e−stdt=−L[tf(t)]​ 对 F ( s ) F(s) F(s)施行同样步骤，反复进行可得： F ( n ) ( s ) = ( − 1 ) n L [ t n f ( t ) ] {F}^{(n)}(s)=(-1)^n\mathscr{L}[t^nf(t)] F(n)(s)=(−1)nL[tnf(t)] 证毕

设 L [ f ( t ) ] = F ( s ) , g ( t ) = ∫ 0 t f ( t ) d t \mathscr{L}\left[f(t)\right]=F(s),g(t)=\int_0^tf(t)dt L[f(t)]=F(s),g(t)=∫0t​f(t)dt，则 g ′ ( t ) = f ( t ) g'(t)=f(t) g′(t)=f(t)，且 g ( 0 ) = 0 g(0)=0 g(0)=0，利用微分之导数的像函数可得： L [ g ′ ( t ) ] = s L [ g ( t ) ] − g ( 0 ) = s L [ g ( t ) ] = s L [ ∫ 0 t f ( t ) d t ] \mathscr{L}[g'(t)]=s\mathscr{L}[g(t)]-g(0)=s\mathscr{L}[g(t)]=s\mathscr{L}\left[\int_0^tf(t)dt\right] L[g′(t)]=sL[g(t)]−g(0)=sL[g(t)]=sL[∫0t​f(t)dt] 即有 L [ ∫ 0 t f ( t ) d t ] = 1 s F ( s ) \mathscr{L}\left[\int_0^tf(t)dt\right]=\frac{1}{s}F(s) L[∫0t​f(t)dt]=s1​F(s)，反复利用上式可得： L [ ∫ 0 t d t ∫ 0 t d t ⋯ ∫ 0 t ⏟ n 个 f ( t ) d t ] = 1 s n F ( s ) \mathscr{L}[\underbrace{\int_0^tdt\int_0^tdt\cdots\int_0^t}_{n个}f(t)dt]=\frac{1}{s^n}F(s) L[n个 ∫0t​dt∫0t​dt⋯∫0t​​​f(t)dt]=sn1​F(s) 证毕

设 L [ f ( t ) ] = F ( s ) \mathscr{L}\left[f(t)\right]=F(s) L[f(t)]=F(s)，则有 ∫ s + ∞ F ( s ) d s = ∫ s + ∞ [ ∫ 0 + ∞ f ( t ) e − s t d t ] d s = ∫ 0 + ∞ f ( t ) [ ∫ s + ∞ e − s t d s ] d t = ∫ 0 + ∞ f ( t ) ⋅ [ − 1 t e − s t ] ∣ s + ∞ d t = ∫ 0 + ∞ f ( t ) t e − s t = L [ f ( t ) t ] \begin{aligned} \int_s^{+\infty}F(s)ds&=\int_s^{+\infty}\left[\int_0^{+\infty}f(t)e^{-st}dt\right]ds\\ &=\int_0^{+\infty}f(t)\left[\int_s^{+\infty}e^{-st}ds\right]dt\\ &=\int_0^{+\infty}f(t)\cdot\left[-\frac{1}{t}e^{-st}\right]\big|_s^{+\infty}dt\\ &=\int_0^{+\infty}\frac{f(t)}{t}e^{-st}\\ &=\mathscr{L}\left[\frac{f(t)}{t}\right] \end{aligned} ∫s+∞​F(s)ds​=∫s+∞​[∫0+∞​f(t)e−stdt]ds=∫0+∞​f(t)[∫s+∞​e−stds]dt=∫0+∞​f(t)⋅[−t1​e−st] ​s+∞​dt=∫0+∞​tf(t)​e−st=L[tf(t)​]​ 反复利用上式可得： ∫ s ∞ d s ∫ s ∞ d s ⋯ ∫ s ∞ ⏟ n 个 F ( s ) d s = L [ f ( t ) t n ] \underbrace{\int_s^\infty ds\int_s^\infty ds\cdots\int_s^\infty}_{n个} F(s)ds=\mathscr{L}[\frac{f(t)}{t^n}] n个 ∫s∞​ds∫s∞​ds⋯∫s∞​​​F(s)ds=L[tnf(t)​] 证毕

设 L [ f ( t ) ] = F ( s ) \mathscr{L}\left[f(t)\right]=F(s) L[f(t)]=F(s)，当 t < 0 t