📚 Laplace Transforms of tⁿ, e±t, sin t, cos t | tⁿ, e±t, sin t, cos t 的拉普拉斯变换
The Laplace transform is a powerful integral transform used extensively in mathematics, physics, and engineering to convert differential equations into algebraic equations. It transforms a function of time f(t) into a function of complex frequency s, making it easier to analyse linear time-invariant systems such as electrical circuits, mechanical systems, and control processes. In the International Baccalaureate (IB) Mathematics curriculum, particularly within the Analysis and Approaches HL option or Further Mathematics topics, understanding the Laplace transforms of elementary functions like tⁿ, e±t, sin t, and cos t is fundamental for solving initial value problems and understanding system dynamics.
拉普拉斯变换是一种强大的积分变换，广泛应用于数学、物理和工程中，可将微分方程转化为代数方程。它将时间函数 f(t) 变换为复频率 s 的函数，从而简化对线性时不变系统（如电路、机械系统和控制过程）的分析。在国际文凭（IB）数学课程中，特别是在分析与方法 HL 的选修部分或进阶数学专题里，掌握 tⁿ、e±t、sin t 和 cos t 等基本函数的拉普拉斯变换，是求解初值问题和理解系统动力学的基础。
1. Definition and Notation | 定义与符号
The Laplace transform of a function f(t), defined for t ≥ 0, is given by the integral ℒ{f(t)} = F(s) = ∫₀∞ e⁻ˢ l converges. Here, s is a complex variable, often written as s = σ + iω. We use the script ℒ to denote the transform operator. The inverse transform is denoted by ℒ = f(t). In this article, we will compute the forward transforms for several key functions.
对于定义在 t ≥ 0 上的函数 f(t)，拉普拉斯变换由积分 ℒ{f(t)} = F(s) = ∫₀∞ e⁻ˢᵗ f(t) dt 给出，条件是积分收敛。这里 s 是一个复变量，通常写为 s = σ + iω。我们用花体 ℒ 表示变换算子。逆变换记为 ℒ⁻¹{F(s)} = f(t)。在本文中，我们将计算几个关键函数的正向变换。
2. Existence Conditions and the Heaviside Step | 存在条件与赫维赛德阶跃函数
must be of exponential order as t → ∞, meaning there exist constants M, c, and T such that |f(t)| ≤ M ec t for all t ≥ T. The integral converges for Re(s) > c. All functions studied here – tⁿ, e±t, sin t, cos t – fine causal signals, we assume the functions are defined for t ≥ 0 and are zero for t < 0.
拉普拉斯变换存在的条件是，当 t → ∞ 时 f(t) 必须为指数阶，即存在常数 M、c 和 T，使得对所有 t ≥ T 有 |f(t)| ≤ M ec t。积分在 Re(s) > c 时收敛。本文研究的全部函数——tⁿ、e±t、sin t、cos t——均满足这一条件。尽管通常会用赫维赛德阶跃函数 u(t) 明确定义因果信号，但我们默认这些函数在 t ≥ 0 上有定义，且 t < 0 时为零。
3. Laplace Transform of tⁿ (n a Non‑Negative Integer) | tⁿ (n 为非负整数) 的拉普拉斯变换
The Laplace transform of tⁿ, where n is a non‑negative integer, is ℒ{tⁿ} = n! d integration by parts or by recognising the connection with the Gamma function Γ(n+1) = n!. For n = 0, we have ℒ{1} = 1/s. For n = 1, ℒ{t} = 1/s². For n = 2, ℒ{t²} = 2! / s³ = 2/s³, and this pattern continues.
对于 n 为非负整数的 tⁿ，其拉普拉斯变换为 ℒ{tⁿ} = n! / sⁿ⁺¹，Re(s) > 0。这个简洁的结果可以通过重复分部积分获得，或借助伽马函数 Γ(n+1) = n! 的联系得到。当 n = 0 时，ℒ{1} = 1/s；n = 1 时，ℒ{t} = 1/s²；n = 2 时，ℒ{t²} = 2! / s³ = 2/s³，以此类推。
4. Derivation and Worked Examples for tⁿ | tⁿ 的推导与示例
To verify the formula, start with n = 0: ∫₀∞ e⁻ˢᵗ dt = 1/s. For n = 1, use integration by parts: ∫₀∞ t e⁻ˢᵗ dt = [−t e⁻ˢᵗ / s]₀∞ + (1/s) ∫₀∞ e⁻ˢᵗ dt = 1/s². For a general n, define Iₙ = ∫₀∞ tⁿ e⁻ˢᵗ dt. Integration by parts yields Iₙ = (n/s) Iₙ₋₁. With I₀ = 1/s, induction gives Iₙ = n! / sⁿ⁺¹.
要验证这个公式，从 n = 0 开始：∫₀∞ e⁻ˢᵗ dt = 1/s。n = 1 时利用分部积分：∫₀∞ t e⁻ˢᵗ dt = [−t e⁻ˢᵗ / s]₀∞ + (1/s) ∫₀∞ e⁻ˢᵗ dt = 1/s²。对于一般的 n，令 Iₙ = ∫₀∞ tⁿ e⁻ˢᵗ dt，分部积分得 Iₙ = (n/s) Iₙ₋₁。结合 I₀ = 1/s，用归纳法即得 Iₙ = n! / sⁿ⁺¹。
Example: Find ℒ{t³}. Solution: Here n = 3, so ℒ{t³} = 3! / s⁴ = 6 / s⁴.
示例：求 ℒ{t³}。解：此处 n = 3，因此 ℒ{t³} = 3! / s⁴ = 6 / s⁴。
5. Laplace Transform of eat – General Exponential | eat 的一般指数变换
For any constant a, the Laplace transform of the exponential function is ℒ{eat} = ∫₀∞ ea−s e(s) > Re(a). This formula is one of the most frequently used in engineering mathematics. It shows that multiplying by an exponential in the time domain corresponds to a simple shift in the s‑domain.
对于任意常数 a，指数函数的拉普拉斯变换为 ℒ{eat} = ∫₀∞ ea−sᵗ dt = 1 / (s − a)，当 Re(s) > Re(a) 时收敛。该公式是工程数学中最常用的变换之一。它表明，时域中乘以指数函数相当于 s 域中的简单平移。
6. Special Cases: et and e−t | 特殊情况：et 与 e−t
Substituting a = 1 gives ℒ{et} = 1 / (s − 1), with region of convergence Re(s) > 1. Substituting a = −1 gives ℒ{e−t s + 1), valid for Re(s) > −1. These two transforms appear constantly in solutions of first‑ fferential equations and are fundamental building blocks for more complicated expressions.
代入 a = 1 得 ℒ{et} = 1 / (s − 1)，收敛域为 Re(s) > 1。代入 a = −1 得 ℒ{e−t} = 1 / (s + 1)，对 Re(s) > −1 成立。这两个变换在求解一阶线性微分方程时频繁出现，是构造更复杂表达式的基本积木。
7. Laplace Transforms of sin(kt) and cos(kt) | sin(kt) 与 cos(kt) 的拉普拉斯变换
Using Euler's formula, sin(kt) = (ei k t − e−i k t) / (2i) and cos(kt) = (ei k t + e−i k t) / 2. Applying linearity and the exponential transform, we obtain ℒ{sin(kt)} = k / (s² + k²) and ℒ{cos(kt)}
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