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拉普拉斯变换的性质

Reflect314数学公式

拉普拉斯变换的性质

为了叙述方便, 在下面的性质中, 均假设所涉及的拉氏变换存在, 且满足拉普拉斯变换的定义。

  1. 线性性质：设 α,β\alpha, \beta 为常数，且有 L[f(t)]=F(s)\mathcal{L}[f (t)] = F (s) ， L[g(t)]=G(s)\mathcal{L}[g (t)] = G (s) ，则有
    • L[αf(t)+βg(t)]=αF(s)+βG(s)\mathcal {L} [ \alpha f(t)+ \beta g(t)] = \alpha F (s) + \beta G (s)
    • L−1[αF(s)+βG(s)]=αf(t)+βg(t)\mathcal {L}^{- 1} \left[ \alpha F (s) + \beta G (s) \right] = \alpha f(t)+ \beta g (t)
  2. 相似性质：设 L[f(t)]=F(s)\mathcal{L}[f (t)] = F (s) ，则对任一常数 a>0a > 0 有
    • L[f(at)]=1aF(sa)\mathcal {L} [ f (a t) ] = \frac {1}{a} F \left (\frac {s}{a}\right)
  3. 微分性质：
    1. 导数的像函数：设 L[f(t)]=F(s)\mathcal{L}[f (t)] = F (s) ，则有
      • L[f′(t)]=sF(s)−f(0)\mathcal {L} \left[ f^{\prime}(t)\right] = s F (s) - f (0)
      • L[f(n)(t)]=snF(s)−sn−1f(0)−sn−2f′(0)−⋯−f(n−1)(0)\mathcal {L} \left[ f^{(n)}(t)\right] = s^{n} F (s) - s^{n - 1} f (0) - s^{n - 2} f^{\prime} (0) - \dots - f^{(n - 1)} (0)
        • 其中， f(k)(0)f^{(k)}(0) 应理解为 lim⁡t→0+f(k)(t)\lim_{t\to 0^{+}}f^{(k)}(t)
    2. 像函数的导数：设 L[f(t)]=F(s)\mathcal{L}[f (t)] = F (s) ，则有
      • F′(s)=−L[tf(t)]F^{\prime} (s) = - \mathscr {L} [ t f(t)]
      • F(n)(s)=(−1)nL[tnf(t)]F^{(n)} (s) = (- 1)^{n} \mathscr {L} [ t^{n} f(t)]
  4. 积分性质：
    1. 积分的像函数：设 L[f(t)]=F(s)\mathcal{L}[f (t)] = F (s) ，则有
      • L[∫0tf(τ)dτ]=1sF(s)\mathcal {L} \left[ \int_{0}^{t} f (\tau) \mathrm{d} \tau \right] = \frac {1}{s} F (s)
      • L[∫0tdt∫0tdt⋯∫0t⏟n次f(t)dt]=1snF(s)\mathcal {L} \left[ \underbrace {\int_{0}^{t} \mathrm{d} t \int_{0}^{t} \mathrm{d} t \cdots \int_{0}^{t}}_{n \text {次}} f(t) \mathrm{d} t \right] = \frac {1}{s^{n}} F (s)
    2. 像函数的积分：设 L[f(t)]=F(s)\mathcal{L}[f (t)] = F (s) ，则有
      • ∫s∞F(s)ds=L[f(t)t]\int_{s}^{\infty} F (s) \mathrm{d} s = \mathcal {L} \left[ \frac {f (t)}{t} \right]
      • ∫s∞ds∫s∞ds⋯∫s∞⏟n次F(s)ds=L[f(t)tn]\underbrace {\int_{s}^{\infty} \mathrm{d} s \int_{s}^{\infty} \mathrm{d} s \cdots \int_{s}^{\infty}}_{n \text {次}} F (s) \mathrm{d} s = \mathcal {L} \left[ \frac {f (t)}{t^{n}} \right]
  5. 延迟性质：设 L[f(t)]=F(s)\mathcal{L}[f (t)]=F (s) ，当 t<0t<0 时， f(t)=0f (t)=0 ，则对任一非负实数 τ\tau 有
    • L[f(t−τ)]=e−sτF(s)\mathcal {L} [ f (t - \tau) ] = \mathrm{e}^{- s \tau} F (s)
    • 必须注意的是本性质对 f(t)f (t) 的要求，即当 t<0t<0 时 f(t)=0f (t)=0 。此时 f(t−τ)f (t-\tau) 在 t<τt<\tau 时为零，故 f(t−τ)f (t-\tau) 应理解为 f(t−τ)u(t−τ)f (t-\tau) u (t-\tau) ，而不是 f(t−τ)u(t)f (t-\tau) u (t) 。因此完整的写法应为
    • L[f(t−τ)u(t−τ)]=e−sτF(s)\mathcal {L} [ f (t - \tau) u (t - \tau) ] = \mathrm{e}^{- s \tau} F (s)
    • 相应地就有 L−1[e−sτF(s)]=f(t−τ)u(t−τ).\mathcal{L}^{-1}\left[\mathrm{e}^{-s\tau}F (s)\right] = f (t - \tau) u (t - \tau).
  6. 位移性质：设 L[f(t)]=F(s)\mathcal{L}[f (t)] = F (s) ，则有
    • L[eatf(t)]=F(s−a)(a为一复常数).\mathcal {L} \left[ \mathrm{e}^{a t} f(t)\right] = F (s - a) \quad (a \text {为一复常数}).
  7. 周期函数的像函数：设 f(t)f (t) 是 [0,+∞)[0, +\infty) 内以 TT 为周期的函数，且 f(t)f (t) 在一个周期内逐段光滑，则
    • L[f(t)]=11−e−sT∫0Tf(t)e−stdt\mathcal {L} [ f(t)] = \frac {1}{1 - \mathrm{e}^{- s T}} \int_{0}^{T} f(t)\mathrm{e}^{- s t} \mathrm{d} t
  8. 卷积定理：设 L[f(t)]=F(s)\mathcal{L}[f (t)] = F(s) ， L[g(t)]=G(s)\mathcal{L}[g(t)] = G(s) ，则有
    • 时域卷积定理：L[f(t)∗g(t)]=F(s)⋅G(s)\mathcal {L} \left[ f(t)* g(t)\right] = F (s) \cdot G(s)
    • 频域卷积定理： L[f(t)⋅g(t)]=12πjF(s)∗G(s)\mathcal {L} \left[ f (t) \cdot g (t) \right] = \frac {1}{2 \pi j} F (s) * G (s)

例题

例题：求 cos⁡ωt\cos \omega t 的拉氏变换.

解：

由 cos⁡ωt=12(ejωt+e−jωt)\cos \omega t = \frac{1}{2} (\mathrm{e}^{\mathrm{j}\omega t} + \mathrm{e}^{-\mathrm{j}\omega t}) 及 L[ejωt]=1s−jω\mathcal{L}[\mathrm{e}^{\mathrm{j}\omega t}] = \frac{1}{s - \mathrm{j}\omega} ，有$$\mathcal {L} [ \cos \omega t ] = \frac {1}{2} (\mathcal {L} [ \mathrm{e}^{\mathrm{j} \omega t} ] + \mathcal {L} [ \mathrm{e}^{- \mathrm{j} \omega t} ]) = \frac {1}{2} \left (\frac {1}{s - j \omega} + \frac {1}{s + j \omega}\right) = \frac {s}{s^{2} + \omega^ {2}}.$$

同样可得 L[sin⁡ωt]=ωs2+ω2.\mathcal{L}\left[\sin\omega t\right]=\frac{\omega}{s^{2}+\omega^{2}}.

例题：已知 F(s)=5s−1(s+1)(s−2)F (s) = \frac{5s - 1}{(s + 1)(s - 2)} ，求 L−1[F(s)]\mathcal{L}^{-1}[F (s)]

解：

由 F(s)=5s−1(s+1)(s−2)=21s+1+31s−2F (s) = \frac{5s - 1}{(s + 1)(s - 2)} = 2\frac{1}{s + 1} +3\frac{1}{s - 2} 及 L[eat]=1s−a\mathcal{L}[e^{at}] = \frac{1}{s - a} ，有$$\mathcal {L}^{- 1} [ F (s) ] = 2 \mathcal {L}^{- 1} \left[ \frac {1}{s + 1} \right] + 3 \mathcal {L}^{- 1} \left[ \frac {1}{s - 2} \right] = 2 \mathrm{e}^{- t} + 3 \mathrm{e}^{2 t}.$$

例题：求解常微分方程$$y^{\prime \prime}(t)+ \omega^ {2} y(t)= 0, \quad y (0) = 0, \quad y^{\prime} (0) = \omega .$$

解：

对方程两边取拉氏变换, 并利用线性性质及 L[f(n)(t)]=snF(s)−sn−1f(0)−sn−2f′(0)−⋯−f(n−1)(0)\mathcal {L} \left[ f^{(n)}(t)\right] = s^{n} F (s) - s^{n - 1} f (0) - s^{n - 2} f^{\prime} (0) - \dots - f^{(n - 1)} (0) 有$$s^{2} Y (s) - s y (0) - y^{\prime} (0) + \omega^ {2} Y (s) = 0,$$

其中 Y(s)=L[y(t)]Y (s) = \mathcal{L}[y (t)] ，代入初值即得$$Y (s) = \frac {\omega}{s^{2} + \omega^ {2}}.$$

根据Laplace 变换#^90d59的结果, 有 y(t)=L−1[Y(s)]=sin⁡ωt.y (t)=\mathcal{L}^{-1}[Y (s)]=\sin\omega t.

例题：求 f(t)=tmf (t)=t^{m} 的拉氏变换 ( m⩾1m \geqslant 1 且为正整数).

解法一：直接利用定义求解.

L[tm]=∫0+∞tme−stdt=−1s∫0+∞tmd(e−st)=−1stme−st∣0+∞+1s∫0+∞e−stmtm−1dt,\begin{array}{r l} \mathcal {L} [ t^{m} ] & = \int_{0}^{+ \infty} t^{m} \mathrm{e}^{- s t} \mathrm{d} t = - \frac {1}{s} \int_{0}^{+ \infty} t^{m} \mathrm{d} (\mathrm{e}^{- s t}) \\ & = - \frac {1}{s} t^{m} \mathrm{e}^{- s t} \Bigg |_{0}^{+ \infty} + \frac {1}{s} \int_{0}^{+ \infty} \mathrm{e}^{- s t} m t^{m - 1} \mathrm{d} t, \end{array}

可得递推关系 L[tm]=msL[tm−1]\mathcal{L}[t^m] = \frac{m}{s}\mathcal{L}[t^{m - 1}] ，又由 L[1]=1s\mathcal{L}[1] = \frac{1}{s} 有$$\mathcal {L} \left[ t^{m} \right] = \frac {m !}{s^{m + 1}}.$$

解法二：利用导数的像函数性质求解.

设 f(t)=tmf (t)=t^{m} ，则 f(m)(t)=m!f^{(m)}(t)=m! 且 f(0)=f′(0)=⋯=f(m−1)(0)=0f (0)=f' (0)=\cdots=f^{(m-1)}(0)=0 ，由微分性质有 L[f(m)(t)]=smL[f(t)]\mathcal{L}[f^{(m)}(t)]=s^{m}\mathcal{L}[f (t)] ，即 $$\mathcal {L} \left[ t^{m} \right] = \frac {1}{s^{m}} \mathcal {L} \left[ m! \right] = \frac {m !}{s^{m + 1}}.$$

例题：求函数 f(t)=tsin⁡ωtf (t)=t\sin\omega t 的拉氏变换.

解：

由于已知 L[sin⁡ωt]=ωs2+ω2\mathcal{L}[\sin \omega t] = \frac{\omega}{s^2 + \omega^2} ，根据 F′(s)=−L[tf(t)]F^{\prime} (s) = - \mathscr {L} [ t f(t)] 有$$\mathcal {L} \left[ t \sin \omega t \right] = - \frac {\mathrm{d}}{\mathrm{d} s} \left (\frac {\omega}{s^{2} + \omega^ {2}}\right) = \frac {2 \omega s}{\left (s^{2} + \omega^ {2}\right)^{2}}.$$

例题：求函数 f(t)=t2cos⁡2tf (t)=t^{2}\cos^{2}t 的拉氏变换.

解：

L[t2cos⁡2t]=12L[t2(1+cos⁡2t)]=12d2ds2(1s+ss2+4)=2(s6+24s2+32)s3(s2+4)3.\begin{array}{r l} \mathcal {L} [ t^{2} \cos^ {2} t ] & = \frac {1}{2} \mathcal {L} [ t^{2} (1 + \cos 2 t) ] = \frac {1}{2} \frac {\mathrm{d}^{2}}{\mathrm{d} s^{2}} \left (\frac {1}{s} + \frac {s}{s^{2} + 4}\right) \\ & = \frac {2 (s^{6} + 2 4 s^{2} + 3 2)}{s^{3} (s^{2} + 4)^{3}}. \end{array}

例题：求函数 f(t)=sin⁡ttf (t)=\frac{\sin t}{t} 的拉氏变换.

解：

由 L[sin⁡t]=11+s2\mathcal{L}[\sin t] = \frac{1}{1 + s^2} 及 ∫s∞F(s)ds=L[f(t)t]\int_{s}^{\infty} F (s) \mathrm{d} s = \mathcal {L} \left[ \frac {f (t)}{t} \right] 有$$\mathcal {L} \left[ \frac {\sin t}{t} \right] = \int_{s}^{\infty} \frac {1}{1 + s^{2}} \mathrm{d} s = \operatorname{arccot} s$$

即 $$\int_{0}^{+ \infty} \frac {\sin t}{t} \mathrm{e}^{- s t} \mathrm{d} t = \operatorname{arccot} s$$

在上式中, 如果令 s=0s=0 有 $$\int_{0}^{+ \infty} \frac {\sin t}{t} \mathrm{d} t = \frac {\pi}{2}$$

例题：计算积分 (1) ∫0+∞e−3tcos⁡2tdt;\int_0^{+\infty}\mathrm{e}^{-3t}\cos 2t\mathrm{d}t; (2) ∫0+∞1−cos⁡tte−tdt.\int_0^{+\infty}\frac{1 - \cos t}{t}\mathrm{e}^{-t}\mathrm{d}t.

解：

(1)由 L[cos⁡2t]=ss2+4\mathcal{L}[\cos 2t] = \frac{s}{s^2 + 4} ，有$$\int_{0}^{+ \infty} \mathrm{e}^{- 3 t} \cos 2 t \mathrm{d} t = \frac {s}{s^{2} + 4} \Big |_{s = 3} = \frac {3}{1 3}$$

(2) 由∫s∞F(s)ds=L[f(t)t]\int_{s}^{\infty} F (s) \mathrm{d} s = \mathcal {L} \left[ \frac {f (t)}{t} \right]有$$\begin{array}{r l} \mathcal {L} \left[ \frac {1 - \cos t}{t} \right] & = \int_{s}^{\infty} \mathcal {L} [ 1 - \cos t ] \mathrm{d} s \ & = \int_{s}^{\infty} \frac {1}{s (s^{2} + 1)} \mathrm{d} s = \frac {1}{2} \ln \frac {s{2}}{s + 1} \Bigg |_{s}^{\infty} = \frac {1}{2} \ln \frac {s^{2} + 1}{s^{2}}, \end{array}$$

即$$\int_{0}^{+ \infty} \frac {1 - \cos t}{t} \mathrm{e}^{- s t} \mathrm{d} t = \frac {1}{2} \ln \frac {s^{2} + 1}{s^{2}},$$

令 s=1s = 1 得$$\int_{0}^{+ \infty} \frac {1 - \cos t}{t} \mathrm{e}^{- t} \mathrm{d} t = \frac {1}{2} \ln 2.$$

例题：设 f(t)=sin⁡tf(t)= \sin t ，求 L[f(t−π2)]\mathcal{L}\left[f\left (t - \frac{\pi}{2}\right)\right]

解：

由于 L[sin⁡t]=1s2+1\mathcal{L}[\sin t] = \frac{1}{s^2 + 1} , 根据 L[f(t−τ)]=e−sτF(s)\mathcal {L} [ f (t - \tau) ] = \mathrm{e}^{- s \tau} F (s) 有 $$\mathcal {L} \left[ f \left (t - \frac {\pi}{2}\right) \right] = \mathcal {L} \left[ \sin \left (t - \frac {\pi}{2}\right) \right] = \mathrm{e}^{- \frac {\pi}{2} s} \mathcal {L} [ \sin t ] = \frac {1}{s^{2} + 1} \mathrm{e}^{- \frac {\pi}{2} s}.$$

按照前面的解释, 则应有$$\begin{array}{r l} \mathcal {L}^{- 1} \left[ \frac {1}{s^{2} + 1} \mathrm{e}^{- \frac {\pi}{2} s} \right] & = \sin \left (t - \frac {\pi}{2}\right) u \left (t - \frac {\pi}{2}\right) \ & = \left{ \begin{array}{l l} - \cos t, & t > \frac {\pi}{2}, \ 0, & t < \frac {\pi}{2}. \end{array} \right. \end{array}$$

例题：求 L−1[1s−1e−s]L^{-1}\left[\frac{1}{s-1}e^{-s}\right] .

解：

由 L−1[1s−1]=etu(t)\mathcal{L}^{-1}\left[\frac{1}{s - 1}\right] = \mathrm{e}^t u (t) ，有$$\mathcal {L}^{- 1} \left[ \frac {1}{s - 1} \mathrm{e}^{- s} \right] = \mathrm{e}^{t - 1} u (t - 1) = \left{ \begin{array}{l l} \mathrm{e}^{t - 1}, & t > 1, \ 0, & t < 1. \end{array} \right.$$

例题：求全波整流后的正弦波 f(t)=∣sin⁡ωt∣f (t)=\left|\sin\omega t\right| 的像函数.

解：

f(t)f (t) 的周期为 T=πωT = \frac{\pi}{\omega} , 故有$$\begin{array}{r l} \mathcal {L} [ f(t)] & = \frac {1}{1 - \mathrm{e}^{- s T}} \int_{0}^{T} \mathrm{e}^{- s t} \sin \omega t \mathrm{d} t \ & = \frac {1}{1 - \mathrm{e}^{- s T}} \cdot \frac {\mathrm{e}^{- s t} (- s \sin \omega t - \omega \cos \omega t)}{s^{2} + \omega^ {2}} \Bigg |_{0}^{T} \ & = \frac {\omega}{s^{2} + \omega^ {2}} \cdot \frac {1 + \mathrm{e}^{- s T}}{1 - \mathrm{e}^{- s T}} = \frac {\omega}{s^{2} + \omega^ {2}} \mathrm{cth} \frac {s \pi}{2 \omega}. \end{array}$$

例题：已知 F(s)=s2(s2+1)2F (s) = \frac{s^2}{(s^2 + 1)^2} , 求 f(t)=L−1[F(s)]f(t)= \mathcal{L}^{-1}[F (s)]

解：

由于 F(s)=ss2+1⋅ss2+1,L−1[ss2+1]=cos⁡tF (s) = \frac{s}{s^2 + 1} \cdot \frac{s}{s^2 + 1}, \mathcal{L}^{-1}\left[\frac{s}{s^2 + 1}\right] = \cos t ，故有

f(t)=L−1[F(s)]=cos⁡t∗cos⁡t=∫0tcos⁡τcos⁡(t−τ)dτ=12∫0t[cos⁡t+cos⁡(2τ−t)]dτ=12(tcos⁡t+sin⁡t).\begin{array}{r l} f(t)& = \mathcal {L}^{- 1} [ F (s) ] = \cos t * \cos t = \int_{0}^{t} \cos \tau \cos (t - \tau) \mathrm{d} \tau \\ & = \frac {1}{2} \int_{0}^{t} [ \cos t + \cos (2 \tau - t) ] \mathrm{d} \tau = \frac {1}{2} (t \cos t + \sin t). \end{array}

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