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) 。

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1. Definition and Uniqueness | 定义与唯一性

The inverse Laplace transform is formally defined by the Bromwich integral: f(t) = 12π i ∫_{σ - i∞}^{σ + i∞} 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) = 12π i ∫_{σ - i∞}^{σ + i∞} 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。

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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 - a₁)(s - a₂)⋯(s - a_n) , write:

对于互异的线性因子 (s - a₁)(s - a₂)⋯(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) → A_k e^{a_k t} .

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

For repeated factors such as (s - a)^r , include terms B₁/(s-a) + B₂/(s-a)² + … + B_r/(s-a)^r . Use the transform pair 1/(s-a)^r → ^tr-1⁄_(r-1)! e^{a t} .

对于重因子如 (s - a)^r ，需要包含项 B₁/(s-a) + B₂/(s-a)² + … + B_r/(s-a)^r 。使用变换对 1/(s-a)^r → ^tr-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。

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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。

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

The inverse transform is linear: if c₁ and c₂ are constants, then L^-1{c₁ F₁(s) + c₂ F₂(s)} = c₁ f₁(t) + c₂ f₂(t).

逆变换是线性的：如果 c₁ 和 c₂ 是常数，那么 L^-1{c₁ F₁(s) + c₂ F₂(s)} = c₁ f₁(t) + c₂ f₂(t)。

The first shifting theorem states that 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 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, ^s+2⁄_(s+1)²+4 can be written as ^(s+1)+1⁄_(s+1)²+4 , leading to e^-t[cos(2t) + 12sin(2t)] .

例如， ^s+2⁄_(s+1)²+4 可以写成 ^(s+1)+1⁄_(s+1)²+4 ，结果是 e^-t[cos(2t) + 12sin(2t)] 。

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

The derivative property in the s -domain gives L{tⁿ 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 L{tⁿ 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 L^-1{F(s)s} = ∫₀^t f(u) du.

积分性质说明，如果 f(t) 分段连续，那么 L^-1{F(s)s} = ∫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。

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6. Convolution Theorem | 卷积定理

If F(s) = L{f(t)} and G(s) = L{g(t)} , then L^-1{F(s)G(s)} = (f * g)(t) = ∫₀^t f(τ) g(t - τ) dτ.

如果 F(s) = L{f(t)} 且 G(s) = L{g(t)} ，那么 L^-1{F(s)G(s)} = (f * g)(t) = ∫₀^t f(τ) g(t - τ) dτ。

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: ¹⁄_s²(s²+1) can be treated as ¹⁄_s² · ¹⁄_s²+1 , whose convolution gives t - sin t .

例如： ¹⁄_s²(s²+1) 可以看作 ¹⁄_s² · ¹⁄_s²+1 ，其卷积给出 t - sin t 。

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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) = ∑ 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) = ∑ 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→ sk} (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→ sk} (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√s , where partial fractions fail.

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

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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 δ(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 δ(t-a) 。它们的变换分别是 e^{-a s}/s 和 e^{-a s} 。

Using the second shifting theorem: 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： 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 δ(t-a) , representing an instantaneous impulse.

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

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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) = ^{∫₀T 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) = ^{∫₀T 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: ¹⁄_{1 - e^{-T s}} = ∑_n=0^∞ 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： ¹⁄_{1 - e^{-T s}} = ∑_n=0^∞ 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。

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10. Worked Examples | 实例计算

Example 1: Find L^-1{2s+5(s+1)(s+2)} .

例 1：求 L^-1{2s+5(s+1)(s+2)} 。

Partial fractions: 2s+5(s+1)(s+2) = 3s+1 - 1s+2 . Hence f(t) = 3e^-t - e^-2t .

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

Example 2: Compute L^-1{¹⁄_s²+4s+13} .

例 2：计算 L^-1{¹⁄_s²+4s+13} 。

Complete the square: s²+4s+13 = (s+2)²+9 . Then L^-1{¹⁄_(s+2)²+3²} = 13 e^-2t sin(3t).

配方： s²+4s+13 = (s+2)²+9 。于是 L^-1{¹⁄_(s+2)²+3²} = 13 e^-2t sin(3t)。

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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 L^-1{¹⁄_(s-a)^m} = ^tm-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 L^-1{¹⁄_(s-a)^m} = ^tm-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。

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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。

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