Methods for Solving Inverse Laplace Transform | 拉普拉斯逆变换的求解方法

📚 Methods for Solving Inverse Laplace Transform | 拉普拉斯逆变换的求解方法

The inverse Laplace transform is a fundamental tool in applied mathematics, control theory, and differential equations. Given a function \( F(s) \) in the complex frequency domain, we seek a time-domain function \( f(t) \) whose Laplace transform equals \( F(s) \).

拉普拉斯逆变换是应用数学、控制理论和微分方程中的基本工具。给定复频域中的函数 \( F(s) \)，我们寻找一个时域函数 \( f(t) \)，使得它的拉普拉斯变换等于 \( F(s) \)。

---

1. Definition and Uniqueness | 定义与唯一性

The inverse Laplace transform is formally defined by the Bromwich integral:
\[
f(t) = \frac{1}{2\pi i} \int_{\sigma – i\infty}^{\sigma + i\infty} F(s) e^{st} \, ds,
\]
where the integration path is a vertical line to the right of all singularities of \( F(s) \).

拉普拉斯逆变换的正式定义是布罗米奇积分：
\[
f(t) = \frac{1}{2\pi i} \int_{\sigma – i\infty}^{\sigma + i\infty} F(s) e^{st} \, ds,
\]
其中积分路径是位于 \( F(s) \) 所有奇点右侧的一条竖直线。

In practice, we rarely evaluate this integral directly. Instead, we use tables, algebraic manipulation, and theorems that guarantee uniqueness under mild conditions.

在实践中，我们很少直接计算这个积分。相反，我们使用变换表、代数运算和在一定温和条件下保证唯一性的定理。

---

2. Partial Fraction Decomposition | 部分分式分解法

The most common method for rational functions \( F(s) = P(s)/Q(s) \) is to decompose them into simpler fractions whose inverse transforms are known.

对于有理函数 \( F(s) = P(s)/Q(s) \)，最常用的方法是将它们分解为逆变换已知的更简单分式。

For distinct linear factors \( (s – a_1)(s – a_2)\cdots(s – a_n) \), write:

对于互异的线性因子 \( (s – a_1)(s – a_2)\cdots(s – a_n) \)，写成：

F(s) = A₁/(s – a₁) + A₂/(s – a₂) + … + Aₙ/(s – aₙ)

Then each term transforms as \( A_k/(s – a_k) \to A_k e^{a_k t} \).

然后每一项的逆变换为 \( A_k/(s – a_k) \to A_k e^{a_k t} \)。

For repeated factors such as \( (s – a)^r \), include terms \( B_1/(s-a) + B_2/(s-a)^2 + \dots + B_r/(s-a)^r \). Use the transform pair \( 1/(s-a)^r \to \frac{t^{r-1}}{(r-1)!} e^{a t} \).

对于重因子如 \( (s – a)^r \)，需要包含项 \( B_1/(s-a) + B_2/(s-a)^2 + \dots + B_r/(s-a)^r \)。使用变换对 \( 1/(s-a)^r \to \frac{t^{r-1}}{(r-1)!} e^{a t} \)。

The coefficients \( A_k, B_k \) can be found by the cover-up method or by equating coefficients.

系数 \( A_k, B_k \) 可以通过遮盖法或比较系数法求得。

---

3. Using Standard Transform Tables | 使用标准变换表

Familiarity with basic Laplace transform pairs is essential. Here is a short table for reference:

熟悉基本的拉普拉斯变换对至关重要。下表是常用参考对：

F(s)	f(t)
1/s	1
1/s²	t
1/(s+a)	e⁻ᵃᵗ
ω/(s²+ω²)	sin(ωt)
s/(s²+ω²)	cos(ωt)
ω/((s+a)²+ω²)	e⁻ᵃᵗ sin(ωt)
(s+a)/((s+a)²+ω²)	e⁻ᵃᵗ cos(ωt)

When \( F(s) \) is not in the table, algebraic manipulation — completing the square, shifting, or splitting into simpler parts — often reduces it to tabulated forms.

当 \( F(s) \) 不在表中时，通过代数运算——配方、平移或拆分成更简单的部分——通常可以化为表中的形式。

---

4. Linearity and Shifting Properties | 线性与平移性质

The inverse transform is linear: if \( c_1 \) and \( c_2 \) are constants, then
\[
\mathcal{L}^{-1}\{c_1 F_1(s) + c_2 F_2(s)\} = c_1 f_1(t) + c_2 f_2(t).
\]

逆变换是线性的：如果 \( c_1 \) 和 \( c_2 \) 是常数，那么
\[
\mathcal{L}^{-1}\{c_1 F_1(s) + c_2 F_2(s)\} = c_1 f_1(t) + c_2 f_2(t)。
\]

The first shifting theorem states that \( \mathcal{L}^{-1}\{F(s-a)\} = e^{a t} f(t) \). This is extremely useful when \( F(s) \) contains terms like \( (s+a)^2 \) that suggest completing the square.

第一平移定理说明 \( \mathcal{L}^{-1}\{F(s-a)\} = e^{a t} f(t) \)。当 \( F(s) \) 包含像 \( (s+a)^2 \) 这样的项时，这个定理非常有用，提示我们进行配方。

For example, \( \frac{s+2}{(s+1)^2+4} \) can be written as \( \frac{(s+1)+1}{(s+1)^2+4} \), leading to \( e^{-t}[\cos(2t) + \tfrac{1}{2}\sin(2t)] \).

例如，\( \frac{s+2}{(s+1)^2+4} \) 可以写成 \( \frac{(s+1)+1}{(s+1)^2+4} \)，结果是 \( e^{-t}[\cos(2t) + \tfrac{1}{2}\sin(2t)] \)。

---

5. Differentiation and Integration Properties | 微分与积分性质

The derivative property in the \( s \)-domain gives \( \mathcal{L}\{t^n f(t)\} = (-1)^n F^{(n)}(s) \). Consequently, if \( F(s) = G'(s) \), then \( f(t) = -t\, g(t) \). This trick often simplifies inverse transforms of functions involving derivatives of known transforms.

\( s \) 域中的微分性质给出 \( \mathcal{L}\{t^n f(t)\} = (-1)^n F^{(n)}(s) \)。因此，如果 \( F(s) = G'(s) \)，那么 \( f(t) = -t\, g(t) \)。这个技巧常可简化涉及已知变换导数的函数的逆变换。

The integration property states that if \( f(t) \) is piecewise continuous, then
\[
\mathcal{L}^{-1}\left\{\frac{F(s)}{s}\right\} = \int_0^t f(u)\, du.
\]

积分性质说明，如果 \( f(t) \) 分段连续，那么
\[
\mathcal{L}^{-1}\left\{\frac{F(s)}{s}\right\} = \int_0^t f(u)\, du。
\]

Thus dividing by \( s \) corresponds to integrating from 0 to \( t \), a useful step for functions with \( 1/s \) factors.

因此除以 \( s \) 对应从 0 到 \( t \) 的积分，这对于含有 \( 1/s \) 因子的函数很有用。

---

6. Convolution Theorem | 卷积定理

If \( F(s) = \mathcal{L}\{f(t)\} \) and \( G(s) = \mathcal{L}\{g(t)\} \), then
\[
\mathcal{L}^{-1}\{F(s)G(s)\} = (f * g)(t) = \int_0^t f(\tau) g(t – \tau)\, d\tau.
\]

如果 \( F(s) = \mathcal{L}\{f(t)\} \) 且 \( G(s) = \mathcal{L}\{g(t)\} \)，那么
\[
\mathcal{L}^{-1}\{F(s)G(s)\} = (f * g)(t) = \int_0^t f(\tau) g(t – \tau)\, d\tau。
\]

This is especially helpful when \( F(s) \) is a product of two functions whose individual inverse transforms are known but the product does not appear directly in tables.

当 \( F(s) \) 是两个已知逆变换的函数的乘积，但乘积并未直接出现在表中时，这个定理特别有用。

Example: \( \frac{1}{s^2(s^2+1)} \) can be treated as \( \frac{1}{s^2} \cdot \frac{1}{s^2+1} \), whose convolution gives \( t – \sin t \).

例如：\( \frac{1}{s^2(s^2+1)} \) 可以看作 \( \frac{1}{s^2} \cdot \frac{1}{s^2+1} \)，其卷积给出 \( t – \sin t \)。

---

7. Complex Inversion via Residues | 利用留数进行复反演

The Bromwich integral can be evaluated by closing the contour to the left and applying the residue theorem. For a rational \( F(s) \), the inverse transform is
\[
f(t) = \sum \text{Res}\{F(s) e^{s t}, s_k\},
\]
where the sum is over all poles of \( F(s) \).

布罗米奇积分可以通过向左闭合围道并应用留数定理来计算。对于有理函数 \( F(s) \)，逆变换为
\[
f(t) = \sum \text{Res}\{F(s) e^{s t}, s_k\},
\]
其中求和遍历 \( F(s) \) 的所有极点。

If \( s_k \) is a simple pole, the residue is \( \lim_{s\to s_k} (s – s_k) F(s) e^{s t} \). For a pole of order \( m \), use the formula with derivatives.

如果 \( s_k \) 是单极点，留数为 \( \lim_{s\to s_k} (s – s_k) F(s) e^{s t} \)。对于 \( m \) 阶极点，使用含导数的公式。

This method is powerful for non-rational functions, such as those involving \( e^{-a\sqrt{s}} \), where partial fractions fail.

这个方法对于非有理函数（例如包含 \( e^{-a\sqrt{s}} \) 的函数）非常强大，此时部分分式法失效。

---

8. Step and Impulse Functions | 阶跃函数与冲激函数

In engineering applications, we often encounter Heaviside step \( u(t-a) \) and Dirac delta \( \delta(t-a) \). Their transforms are \( e^{-a s}/s \) and \( e^{-a s} \), respectively.

在工程应用中，我们经常遇到海维赛德阶跃函数 \( u(t-a) \) 和狄拉克冲激函数 \( \delta(t-a) \)。它们的变换分别是 \( e^{-a s}/s \) 和 \( e^{-a s} \)。

Using the second shifting theorem: \( \mathcal{L}^{-1}\{e^{-a s} F(s)\} = u(t-a) f(t-a) \). This allows us to handle piecewise-defined forcing functions.

使用第二平移定理：\( \mathcal{L}^{-1}\{e^{-a s} F(s)\} = u(t-a) f(t-a) \)。这使我们能够处理分段定义的激励函数。

When \( F(s) = e^{-a s} \) alone, the inverse is \( \delta(t-a) \), representing an instantaneous impulse.

当 \( F(s) = e^{-a s} \) 单独出现时，逆变换是 \( \delta(t-a) \)，表示瞬时冲激。

---

9. Periodic Functions | 周期函数的逆变换

For a periodic function \( f(t) \) with period \( T \), the Laplace transform has a known form:
\[
F(s) = \frac{\int_0^T e^{-s t} f(t)\, dt}{1 – e^{-T s}}.
\]

对于周期为 \( T \) 的周期函数 \( f(t) \)，其拉普拉斯变换具有已知形式：
\[
F(s) = \frac{\int_0^T e^{-s t} f(t)\, dt}{1 – e^{-T s}}。
\]

To invert such an \( F(s) \), expand \( 1/(1 – e^{-T s}) \) as a geometric series:
\[
\frac{1}{1 – e^{-T s}} = \sum_{n=0}^{\infty} e^{-n T s}.
\]
Then each term corresponds to a delayed copy of the base waveform.

要对此类 \( F(s) \) 求逆变换，将 \( 1/(1 – e^{-T s}) \) 展开为几何级数：
\[
\frac{1}{1 – e^{-T s}} = \sum_{n=0}^{\infty} e^{-n T s}。
\]
然后每一项对应基本波形的一个延迟副本。

Thus \( f(t) \) is constructed by repeating the first-period function every \( T \) seconds.

因此 \( f(t) \) 通过每隔 \( T \) 秒重复第一个周期的函数来构造。

---

10. Worked Examples | 实例计算

Example 1: Find \( \mathcal{L}^{-1}\left\{\frac{2s+5}{(s+1)(s+2)}\right\} \).

例 1：求 \( \mathcal{L}^{-1}\left\{\frac{2s+5}{(s+1)(s+2)}\right\} \)。

Partial fractions: \( \frac{2s+5}{(s+1)(s+2)} = \frac{3}{s+1} – \frac{1}{s+2} \). Hence \( f(t) = 3e^{-t} – e^{-2t} \).

部分分式：\( \frac{2s+5}{(s+1)(s+2)} = \frac{3}{s+1} – \frac{1}{s+2} \)。因此 \( f(t) = 3e^{-t} – e^{-2t} \)。

Example 2: Compute \( \mathcal{L}^{-1}\left\{\frac{1}{s^2+4s+13}\right\} \).

例 2：计算 \( \mathcal{L}^{-1}\left\{\frac{1}{s^2+4s+13}\right\} \)。

Complete the square: \( s^2+4s+13 = (s+2)^2+9 \). Then
\[
\mathcal{L}^{-1}\left\{\frac{1}{(s+2)^2+3^2}\right\} = \frac{1}{3} e^{-2t} \sin(3t).
\]

配方：\( s^2+4s+13 = (s+2)^2+9 \)。于是
\[
\mathcal{L}^{-1}\left\{\frac{1}{(s+2)^2+3^2}\right\} = \frac{1}{3} e^{-2t} \sin(3t)。
\]

---

11. Common Pitfalls and Tips | 常见错误与技巧

• Do not forget the Heaviside step function when the transform contains \( e^{-a s} \). Skipping it changes the domain of \( f(t) \).

  当变换中含有 \( e^{-a s} \) 时，不要忘记海维赛德阶跃函数。漏掉它会改变 \( f(t) \) 的定义域。

• When using partial fractions, check whether the degree of the numerator is less than that of the denominator. If not, perform polynomial division first.

  使用部分分式时，检查分子次数是否小于分母次数。若不是，先进行多项式除法。

• For repeated poles, use the formula \( \mathcal{L}^{-1}\left\{\frac{1}{(s-a)^m}\right\} = \frac{t^{m-1}}{(m-1)!} e^{a t} \) carefully; a common mistake is to use \( m! \) instead of \( (m-1)! \).

  对于重极点，小心使用公式 \( \mathcal{L}^{-1}\left\{\frac{1}{(s-a)^m}\right\} = \frac{t^{m-1}}{(m-1)!} e^{a t} \)；常见错误是使用 \( m! \) 而不是 \( (m-1)! \)。

• Always complete the square when facing quadratic denominators; then match with sine and cosine transforms involving a shift.

  遇到二次分母时，务必配方；然后与含有平移的正弦、余弦变换对应。

• If the inverse transform seems too complicated, try rewriting the expression using known properties first, such as multiplying by \( s/s \) or separating into even and odd parts.

  如果逆变换看起来过于复杂，尝试先利用已知性质重写表达式，例如乘以 \( s/s \) 或分成奇偶部分。

---

12. Summary | 总结

The inverse Laplace transform can be approached through several complementary strategies: partial fractions, table look-up, linearity and shifting, convolution, and residue theory. Mastery of these techniques is essential for solving linear differential equations and analyzing dynamic systems.

拉普拉斯逆变换可以通过几种互补的策略来处理：部分分式、查表、线性与平移、卷积以及留数理论。掌握这些技巧对于求解线性微分方程和分析动态系统至关重要。

Start by identifying the structure of \( F(s) \). If it is rational, try partial fractions. If it contains exponentials, use shifting theorems. If it is a product of known transforms, consider convolution. For advanced problems, residues provide a unified method.

首先识别 \( F(s) \) 的结构。如果是有理函数，尝试部分分式。如果包含指数项，使用平移定理。如果是已知变换的乘积，考虑卷积。对于更复杂的问题，留数法提供了统一的方法。

With consistent practice, selecting the most efficient method becomes intuitive.

通过持续练习，选择最有效的方法会变得直觉化。

---

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）