Laplace Transforms, Inverses, and Rules | 拉普拉斯变换、逆变换及运算法则

📚 Laplace Transforms, Inverses, and Rules | 拉普拉斯变换、逆变换及运算法则

The Laplace transform is a powerful integral transform used extensively in solving differential equations, system analysis, and control theory. It converts functions of a real variable t (often time) into functions of a complex variable s (frequency domain), turning differentiation and integration into algebraic operations.

拉普拉斯变换是一种重要的积分变换，广泛用于求解微分方程、系统分析和控制理论。它将实变量 t（通常为时间）的函数变换为复变量 s（频域）的函数，使微分和积分运算转化为代数运算。

1. Introduction to Laplace Transforms | 拉普拉斯变换简介

The Laplace transform named after Pierre-Simon Laplace transforms a time‑domain function f(t) into a complex frequency‑domain function F(s). This transformation simplifies linear differential equations into algebraic equations, which are much easier to solve.

以皮埃尔‑西蒙·拉普拉斯命名的拉普拉斯变换，将时域函数 f(t) 变换为复频域函数 F(s)。这种变换将线性微分方程简化为代数方程，求解起来更为简便。

A key advantage is that initial conditions are automatically incorporated into the algebraic equation, and the inverse transform readily gives the time‑domain solution. The method is widely applied in engineering, physics, and applied mathematics.

其一大优势是初始条件自动融入代数方程，并且逆变换可方便地得到时域解。该方法在工程、物理和应用数学领域应用广泛。

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2. Definition and Notation | 定义与符号

aplace transform of a function f(t), defined for t ≥ 0, is given by the improper integral:

F(s) = L{f(t)} = ∫₀^∞ e⁻ˢᵗ f(t) dt

provided the integral converges for some value of s.

对于定义在 t ≥ 0 的函数 f(t)，其拉普拉斯变换由下述反常积分定义：

F(s) = L{f(t)} = ∫₀^∞ e⁻ˢᵗ f(t) dt

只要该积分对于某些 s 值收敛。

transform exists if f(t) is piecewise continuous on [0, ∞) and of exponential order, i.e. |f(t)| ≤ Meᵅᵗ for some constants M, a, and all sufficiently large t.

这里 s = σ + iω 是复变量。如果 f(t) 在 [0, ∞) 上分段连续且具有指数阶，即存在常数 M、a 使得对所有足够大的 t 有 |f(t)| ≤ Meᵅᵗ，则变换存在。

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3. Laplace Transforms of Basic Functions | 基本函数的拉普拉斯变换

al. The simplest is the unit step function f(t) = 1, whose transform is 1/s for Re(s) > 0.

熟悉基本函数的变换至关重要。最简单的是单位阶跃函数 f(t) = 1，其变换为 1/s（Re(s) > 0）。

(s − a), a result that underpins many shifting theorems. The power function tⁿ (n = 0,1,2,…) gives n! / sⁿ⁺¹, while sin(ωt) and cos(ωt) yield ω/(s²+ω²) and s/(s²+ω²) respectively.

对于指数函数 f(t) = eᵅᵗ，积分得到 1/(s − a)，这一结果是许多平移定理的基础。幂函数 tⁿ (n = 0,1,2,…) 给出 n! / sⁿ⁺¹，而 sin(ωt) 和 cos(ωt) 分别给出 ω/(s²+ω²) 和 s/(s²+ω²)。

• L{1} = 1/s
• L{tⁿ} = n! / sⁿ⁺¹
• L{eᵅᵗ} = 1/(s − a)
• L{sin(ωt)} = ω/(s² + ω²)
• L{cos(ωt)} = s/(s² + ω²)

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4. Linearity Property | 线性性质

is property allows us to decompose complex functions into simpler components and transform term by term.

拉普拉斯变换是线性算子：对于任意常数 a 和 b，有 L{ a f(t) + b g(t) } = a L{f(t)} + b L{g(t)}。这一性质使我们能将复杂函数分解为更简单的部分，并逐项变换。

Linearity is crucial for handling sums of functions, such as polynomial expressions combined with trigonometric terms, because each term's transform can be found separately from standard tables.

线性性质对于处理函数之和至关重要，比如多项式与三角项的组合，因为每一项的变换可从标准表格中单独查找。

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5. First Shifting Theorem (s-shift) | 第一平移定理（s-域平移）

L{f(t)} = F(s), then multiplying f(t) by an exponential eᵅᵗ shifts the transform in the s‑domain:

L{eᵅᵗ f(t)} = F(s − a)

provided the transform exists for s − a.

若 L{f(t)} = F(s)，则 f(t) 乘以指数函数 eᵅᵗ 会导致 s 域的平移：

L{eᵅᵗ f(t)} = F(s − a)

只要变换对 s − a 存在。

example, since L{sin(ωt)} = ω/(s²+ω²), it follows that L{eᵅᵗ sin(ωt)} = ω/((s − a)² + ω²).

该定理特别适用于处理诸如 eᵅᵗ sin(ωt) 或 eᵅᵗ tⁿ 等函数。例如，由 L{sin(ωt)} = ω/(s²+ω²) 可得 L{eᵅᵗ sin(ωt)} = ω/((s − a)² + ω²)。

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6. Second Shifting Theorem (t-shift) | 第二平移定理（t-域平移）

When a function is delayed by a constant a > 0 and multiplied by a unit step function u(t − a), the transform gains an exponential factor:

L{ u(t − a) f(t − a) } = e⁻ᵃˢ F(s)

where u(t − a) is the Heaviside step function that is 0 for t < a and 1 for t ≥ a.

当函数被延迟常数 a > 0 并乘以单位阶跃函数 u(t − a) 时，其变换中包含一个指数因子：

L{ u(t − a) f(t − a) } = e⁻ᵃˢ F(s)

其中 u(t − a) 是海维赛德阶跃函数，当 t < a 时为 0，t ≥ a 时为 1。

This theorem is essential for piecewise‑ t) delayed without the shift in the argument, extra care is needed: L{u(t − a) f(t)} = e⁻ᵃˢ L{f(t + a)}.

该定理对分段定义函数和电路中的开关输入建模至关重要。若函数仅是被延迟而未平移自变量，则需特别注意：L{u(t − a) f(t)} = e⁻ᵃˢ L{f(t + a)}。

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7. Transform of Derivatives | 导数的拉普拉斯变换

The transform of the first derivative f′(t) is given by:

L{f′(t)} = s F(s) − f(0)

This key formula shows how differentiation in the time domain becomes multiplication by s in the s‑domain, minus the initial condition.

一阶导数 f′(t) 的变换为：

L{f′(t)} = s F(s) − f(0)

该关键公式表明时域内的微分运算如何变成 s 域中的乘法，并减去初始条件。

second derivative: L{f″(t)} = s² F(s) − s f(0) − f′(0). In general, the transform of the n‑ ves sⁿ F(s) minus a polynomial in s whose coefficients are the initial values f(0), f′(0), …, f⁽ⁿ⁻¹⁾(0).

对于二阶导数：L{f″(t)} = s² F(s) − s f(0) − f′(0)。一般地，n 阶导数的变换包含 sⁿ F(s) 减去一个 s 的多项式，其系数为初始值 f(0), f′(0), …, f⁽ⁿ⁻¹⁾(0)。

This property is the main reason the Laplace transform is so effective for initial value problems: it automatically incorporates the initial conditions into the algebraic equation.

该性质是拉普拉斯变换在初值问题中如此有效的主要原因：它将初始条件自动纳入代数方程。

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8. Transform of Integrals | 积分的拉普拉斯变换

The transform of the integral of a function is closely related to division by s:

L{ ∫₀ᵗ f(τ) dτ } = F(s) / s

This mirrors the fact that integration in the time domain corresponds to multiplication by 1/s in the s‑domain.

函数积分的变换与除以 s 密切相关：

L{ ∫₀ᵗ f(τ) dτ } = F(s) / s

这反映出时域中的积分对应于 s 域中的乘以 1/s。

For definite integrals where the lower limit is not zero, the formula is slightly modified. This property is often used in conjunction with the derivative rule to solve integro‑differential equations.

对于下限不为零的定积分，公式会有所调整。该性质常与导数法则联用，用于求解积分-微分方程。

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9. Inverse Laplace Transforms and Partial Fractions | 逆拉普拉斯变换与部分分式

valuate the complex inversion integral; instead, we manipulate F(s) into a sum of terms whose inverses are known from tables.

逆拉普拉斯变换从 F(s) 恢复 f(t)，记为 f(t) = L⁻¹{F(s)}。实际中我们很少计算复反演积分，而是将 F(s) 化为若干已知逆变换项的和。

F(s) = (2s+1)/((s−1)(s+2)), we split it into A/(s−1) + B/(s+2) and use L⁻¹{1/(s−a)} = eᵅᵗ.

对于有理函数 F(s) = P(s)/Q(s)，常用部分分式分解的方法。若 F(s) 有不同的线性因子，例如 F(s) = (2s+1)/((s−1)(s+2))，我们将其拆分为 A/(s−1) + B/(s+2)，然后利用 L⁻¹{1/(s−a)} = eᵅᵗ 求逆。

For repeated factors (s − a)ⁿ, the inverse involves tⁿ⁻¹ eᵅᵗ/(n−1)!. For irreducible quadratic factors, completing the square and using combinations of s and constants yield sines and cosines multiplied by exponentials.

对于重因子 (s − a)ⁿ，其逆变换涉及 tⁿ⁻¹ eᵅᵗ/(n−1)!。对于不可约的二次因子，通过配方并利用 s 和常数的组合可得到指数函数乘正余弦的形式。

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10. Convolution Theorem | 卷积定理

The convolution of two functions f and g, defined as (f ∗ g)(t) = ∫₀ᵗ f(τ) g(t − τ) dτ, has a particularly simple Laplace transform:

L{f ∗ g} = F(s) G(s)

This theorem is invaluable for finding the inverse transform of a product of two transforms.

两个函数 f 和 g 的卷积定义为 (f ∗ g)(t) = ∫₀ᵗ f(τ) g(t − τ) dτ，其拉普拉斯变换特别简洁：

L{f ∗ g} = F(s) G(s)

该定理对于求两个变换乘积的逆变换极有价值。

convolution gives ∫₀ᵗ 1·eᵃ⁽ᵗ⁻ᵗ⁾ dτ = (eᵅᵗ − 1)/a, which matches the inverse of 1/(s(s−a)).

卷积使我们能够在已知每个因子的逆变换时，避免繁琐的部分分式展开。例如，若 F(s) = 1/s、G(s) = 1/(s−a)，卷积得出 ∫₀ᵗ 1·eᵃ⁽ᵗ⁻ᵗ⁾ dτ = (eᵅᵗ − 1)/a，这与 1/(s(s−a)) 的逆变换一致。

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11. Table of Common Laplace Transforms | 常用拉普拉斯变换表

A concise table of transforms is indispensable for quick reference. Below is a summary of the most frequently used pairs.

一个简明的变换表是快速查阅所必需的。以下是常用变换对的汇总。

f(t)	F(s) = L{f(t)}
1	1/s
tⁿ (n = 0,1,2,…)	n! / sⁿ⁺¹
eᵅᵗ	1/(s − a)
sin(ωt)	ω/(s² + ω²)
cos(ωt)	s/(s² + ω²)
tⁿ eᵅᵗ	n!/(s − a)ⁿ⁺¹
eᵅᵗ sin(ωt)	ω/((s − a)² + ω²)
eᵅᵗ cos(ωt)	(s − a)/((s − a)² + ω²)
u(t − a) (unit step)	e⁻ᵃˢ / s
δ(t − a) (Dirac delta)	e⁻ᵃˢ

Note that the transform is linear, so combinations of these basic forms can be handled directly.

注意变换是线性的，因此这些基本形式的组合可直接处理。

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12. Applications and Examples | 应用与实例

quations with given initial conditions. Consider y″ + 3y′ + 2y = e⁻ᵗ, y(0) = 1, y′(0) = 0.

拉普拉斯变换法擅长求解带给定初始条件的线性常微分方程。考虑 y″ + 3y′ + 2y = e⁻ᵗ，y(0) = 1，y′(0) = 0。

of both sides, applying the derivative rules: (s²Y − s·1 − 0) + 3(sY − 1) + 2Y = 1/(s+1).

步骤1：对等式两边取拉普拉斯变换，应用导数法则：(s²Y − s·1 − 0) + 3(sY − 1) + 2Y = 1/(s+1)。

Y and use partial fractions: Y(s) = (s²+4s+4) / ((s+1)(s²+3s+2)) = (s+2)² / ((s+1)²(s+2)).

步骤2：合并 Y(s) 的项：(s² + 3s + 2)Y = s + 3 + 1/(s+1)。解出 Y 并使用部分分式：Y(s) = (s²+4s+4) / ((s+1)(s²+3s+2)) = (s+2)² / ((s+1)²(s+2))。

ain terms like 1/(s+1)² and 1/(s+2). Taking inverse transforms yields y(t) = t e⁻ᵗ + e⁻²ᵗ.

经化简和部分分式分解后，可得诸如 1/(s+1)² 和 1/(s+2) 的项。取逆变换得 y(t) = t e⁻ᵗ + e⁻²ᵗ。

Similarly, in circuit analysis, the Laplace transform converts integro‑differential equations of RLC circuits into algebraic relations involving impedance, making it straightforward to find the response to various inputs.

类似地，在电路分析中，拉普拉斯变换将 RLC 电路的积分-微分方程转化为包含阻抗的代数关系，使求解对各类输入的响应变得直接。

Mastery of these rules and the art of inverting transforms through table lookup, partial fractions, and convolution is a cornerstone of applied mathematics.

掌握这些规则以及通过查表、部分分式和卷积进行逆变换的技巧，是应用数学的基石之一。

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