📚 Laplace Transform Theorems: Scaling, Shifting, and Common Factors | 拉普拉斯变换的尺度法则、移位法则与常见因子
The Laplace transform is one of the most powerful tools in engineering mathematics, particularly for solving linear differential equations and analyzing control systems. In this revision guide, we focus on three essential properties: the scaling theorem, the shifting theorems, and the treatment of common multiplicative factors such as powers of t and exponential functions.
拉普拉斯变换是工程数学中最强大的工具之一，特别适用于求解线性微分方程和分析控制系统。在本复习指南中，我们聚焦于三个核心性质：尺度法则、移位法则，以及幂函数、指数函数等常见因子的处理方法。
1. Definition and Notation | 定义与记号
For a function f(t) defined for t ≥ 0, the Laplace transform is defined by the improper integral:
F(s) = L{f(t)} = ∫₀⁺∞ e^(−st) f(t) dt
The transform converts a function of time t into a function of the complex variable s. The inverse Laplace transform, denoted L⁻¹{F(s)}, recovers the original function f(t). We assume all functions are piecewise continuous and of exponential order, ensuring convergence of the integral for Re(s) sufficiently large.
对于定义在 t ≥ 0 上的函数 f(t)，其拉普拉斯变换定义为广义积分：
F(s) = L{f(t)} = ∫₀⁺∞ e^(−st) f(t) dt
该变换将时间变量 t 的函数转换为复变量 s 的函数。逆拉普拉斯变换记作 L⁻¹{F(s)}，可恢复原函数 f(t)。我们假设所有函数都是分段连续的且具有指数阶，以确保积分在 Re(s) 足够大时收敛。
2. The Linearity Property | 线性性质
Before discussing scaling and shifting, we must recall the linearity of the Laplace transform. For any constants a and b, and any functions f(t) and g(t):
L{a f(t) + b g(t)} = a L{f(t)} + b L{g(t)} = a F(s) + b G(s)
This property allows us to decompose complicated functions into simpler ones, transform each term separately, and then combine the results. It is the backbone of solving linear differential equations using Laplace transforms.
在讨论尺度与移位之前，我们必须回顾拉普拉斯变换的线性性质。对于任意常数 a 和 b，以及任意函数 f(t) 和 g(t)：
L{a f(t) + b g(t)} = a L{f(t)} + b L{g(t)} = a F(s) + b G(s)
该性质允许我们将复杂函数分解为简单函数的组合，分别变换后再合并结果。这是用拉普拉斯变换求解线性微分方程的基石。
3. First Shifting Theorem: Frequency Shift | 第一移位定理：频域移位
The first shifting theorem, also called the s-shifting theorem, deals with multiplication of f(t) by an exponential factor e^(at). If L{f(t)} = F(s), then:
L{e^(at) f(t)} = F(s − a)
This means that multiplying a function by e^(at) in the time domain corresponds to shifting the transform F(s) by a units to the right in the s-domain. The theorem holds for any real or complex constant a.
第一移位定理，也称为 s 域移位定理，处理 f(t) 乘以指数因子 e^(at) 的情形。若 L{f(t)} = F(s)，则：
L{e^(at) f(t)} = F(s − a)
这意味着在时域中乘以 e^(at)，相当于在 s 域中将变换 F(s) 向右平移 a 个单位。该定理对任意实数或复数常数 a 均成立。
4. Applications of the First Shift | 第一移位定理的应用
Consider the basic transform L{1} = 1/s for s > 0. Applying the first shift theorem with a = −2 gives:
L{e^(−2t)} = 1/(s + 2)
Similarly, since L{sin(ωt)} = ω/(s² + ω²), we obtain:
L{e^(at) sin(ωt)} = ω/[(s − a)² + ω²]
This is particularly useful when solving differential equations with exponential forcing terms. For example, the transform of e^(3t) t² can be found from L{t²} = 2/s³ by shifting: L{e^(3t) t²} = 2/(s − 3)³.
考虑基本变换 L{1} = 1/s（s > 0）。将第一移位定理应用于 a = −2，得到：
L{e^(−2t)} = 1/(s + 2)
类似地，由于 L{sin(ωt)} = ω/(s² + ω²)，我们得到：
L{e^(at) sin(ωt)} = ω/[(s − a)² + ω²]
这在求解含指数强迫项的微分方程时尤为有用。例如，e^(3t) t² 的变换可由 L{t²} = 2/s³ 移位得到：L{e^(3t) t²} = 2/(s − 3)³。
5. Second Shifting Theorem: Time Delay | 第二移位定理：时域延迟
The second shifting theorem, or t-shifting theorem, addresses functions that are delayed in time. If L{f(t)} = F(s) and u(t − a) is the unit step (Heaviside) function, then:
L{u(t − a) f(t − a)} = e^(−as) F(s)
Equivalently, multiplying the transform F(s) by e^(−as) corresponds to shifting the time function f(t) forward by a units and setting it to zero for t < a.
第二移位定理，即 t 域移位定理，处理时间延迟的函数。若 L{f(t)} = F(s)，且 u(t − a) 为单位阶跃（海维赛德）函数，则：
L{u(t − a) f(t − a)} = e^(−as) F(s)
等价地，将变换 F(s) 乘以 e^(−as)，相当于将时间函数 f(t) 向前平移 a 个单位，并在 t < a 时将其置为零。
6. Working with Delayed Functions | 延迟函数的运算
To apply the second shift theorem, we must rewrite a delayed function in the form u(t − a) f(t − a). For instance, suppose g(t) = t² for t ≥ 2 and g(t) = 0 otherwise. We write g(t) = u(t − 2) (t − 2 + 2)² = u(t − 2) [(t − 2)² + 4(t − 2) + 4].
Taking transforms term by term using the second shift:
L{g(t)} = e^(−2s) (2/s³ + 4/s² + 4/s)
This technique is invaluable for piecewise-defined functions and for modeling switching phenomena in electrical circuits.
要应用第二移位定理，我们必须将延迟函数改写为 u(t − a) f(t − a) 的形式。例如，设 g(t) = t²（t ≥ 2），否则 g(t) = 0。我们将 g(t) 改写为 g(t) = u(t − 2) (t − 2 + 2)² = u(t − 2) [(t − 2)² + 4(t − 2) + 4]。
利用第二移位定理逐项求变换：
L{g(t)} = e^(−2s) (2/s³ + 4/s² + 4/s)
这一技巧在分段函数建模以及电路开关现象分析中至关重要。
7. The Scaling Theorem | 尺度法则
The scaling theorem describes the effect of stretching or compressing time. If L{f(t)} = F(s) and a is a positive constant, then:
L{f(at)} = (1/a) F(s/a)
Intuitively, compressing time by a factor of a (i.e., making the function evolve faster) causes the transform to be scaled in both amplitude and frequency. This theorem is fundamental in control theory, where time constants are adjusted to meet performance specifications.
尺度法则描述了时间轴拉伸或压缩的效果。若 L{f(t)} = F(s)，且 a 为正数，则：
L{f(at)} = (1/a) F(s/a)
直觉上，将时间压缩 a 倍（即让函数演化更快）会导致变换在幅度和频率上都发生缩放。该定理在控制理论中具有基础性地位，因为时间常数常常需要调整以满足性能指标。
8. Verification of the Scaling Theorem | 尺度法则的验证
Let us verify the scaling theorem for f(t) = sin(t). We know L{sin(t)} = 1/(s² + 1). For a = 3, we compute L{sin(3t)} directly:
L{sin(3t)} = 3/(s² + 9)
Using the scaling theorem with a = 3: (1/3) × 1/((s/3)² + 1) = (1/3) × 1/(s²/9 + 1) = (1/3) × 9/(s² + 9) = 3/(s² + 9). The two results agree perfectly.
让我们验证 f(t) = sin(t) 的尺度法则。已知 L{sin(t)} = 1/(s² + 1)。对于 a = 3，我们直接计算 L{sin(3t)}：
L{sin(3t)} = 3/(s² + 9)
利用尺度法则（a = 3）：(1/3) × 1/((s/3)² + 1) = (1/3) × 1/(s²/9 + 1) = (1/3) × 9/(s² + 9) = 3/(s² + 9)。两种方法得到完全一致的结果。
9. Common Factors: Powers of t | 常见因子：t 的幂
Powers of t in the time domain correspond to derivatives in the s-domain. The general formula for n = 1, 2, 3, … is:
L{tⁿ f(t)} = (−1)ⁿ dⁿF/dsⁿ
In particular, taking f(t) = 1 with F(s) = 1/s gives:
- L{t} = 1/s²
- L{t²} = 2/s³
- L{tⁿ} = n!/sⁿ⁺¹
These are indispensable when solving differential equations with polynomial coefficients or when dealing with ramp inputs in engineering systems.
时域中的 t 的幂对应于 s 域中的导数。对于 n = 1, 2, 3, … 的一般公式为：
L{tⁿ f(t)} = (−1)ⁿ dⁿF/dsⁿ
特别地，取 f(t) = 1 且 F(s) = 1/s，可得：
- L{t} = 1/s²
- L{t²} = 2/s³
- L{tⁿ} = n!/sⁿ⁺¹
这些公式在求解含多项式系数的微分方程或处理工程系统中的斜坡输入时不可或缺。
10. Combining Theorems: A Comprehensive Table | 定理综合：常用变换表
The true power of these properties emerges when they are combined. Below is a concise table of frequently used Laplace transform pairs, all derivable using the theorems discussed above:
这些性质的真正威力在于组合运用。下表汇总了常用拉普拉斯变换对，均可通过上述定理推导得到：
| f(t) | F(s) = L{f(t)} |
|---|---|
| 1 | 1/s |
| tⁿ (n = 0, 1, 2, …) | n!/sⁿ⁺¹ |
| e^(at) | 1/(s − a) |
| sin(ωt) | ω/(s² + ω²) |
| cos(ωt) | s/(s² + ω²) |
| e^(at) tⁿ | n!/(s − a)ⁿ⁺¹ |
| e^(at) sin(ωt) | ω/[(s − a)² + ω²] |
| e^(at) cos(ωt) | (s − a)/[(s − a)² + ω²] |
| u(t − a) f(t − a) | e^(−as) F(s) |
| f(at) | (1/a) F(s/a) |
These pairs are given in the IB Mathematics Analysis and Approaches HL formula booklet, but understanding their derivation from the basic theorems is crucial for exam success.
这些变换对在 IB 数学分析与方法（AA）HL 的公式手册中均有列出，但理解其从基本定理推导而来的过程对于考试成功至关重要。
11. Worked Example: Solving an ODE | 实例分析：求解常微分方程
We now illustrate the combined use of these theorems. Solve y” − 4y’ + 3y = e^(2t), with y(0) = 0 and y'(0) = 0.
Taking the Laplace transform of both sides and using L{y’} = sY(s) − y(0) and L{y”} = s²Y(s) − s y(0) − y'(0):
(s² − 4s + 3) Y(s) = 1/(s − 2)
Since s² − 4s + 3 = (s − 1)(s − 3), we decompose:
Y(s) = 1/[(s − 1)(s − 2)(s − 3)]
Using partial fractions: 1/[(s − 1)(s − 2)(s − 3)] = A/(s − 1) + B/(s − 2) + C/(s − 3). Solving gives A = 1/2, B = −1, C = 1/2. Hence:
y(t) = ½ e^t − e^(2t) + ½ e^(3t)
我们通过一个综合实例展示这些定理的组合应用。求解 y” − 4y’ + 3y = e^(2t)，初始条件 y(0) = 0，y'(0) = 0。
对两边取拉普拉斯变换，并利用 L{y’} = sY(s) − y(0) 和 L{y”} = s²Y(s) − s y(0) − y'(0)：
(s² − 4s + 3) Y(s) = 1/(s − 2)
由于 s² − 4s + 3 = (s − 1)(s − 3)，我们进行部分分式分解：
Y(s) = 1/[(s − 1)(s − 2)(s − 3)]
设 1/[(s − 1)(s − 2)(s − 3)] = A/(s − 1) + B/(s − 2) + C/(s − 3)。解得 A = 1/2，B = −1，C = 1/2。因此：
y(t) = ½ e^t − e^(2t) + ½ e^(3t)
12. Common Pitfalls and Exam Tips | 常见错误与应试技巧
Students frequently make the following errors when applying the scaling and shift theorems:
- Confusing the two shifting theorems: the first shift involves e^(at) f(t) with F(s − a), while the second shift involves u(t − a) f(t − a) with e^(−as) F(s). Remember: exponential in time ↔ shift in s; exponential in s ↔ shift in time.
- Forgetting the factor 1/a in the scaling theorem. The transform of f(at) is (1/a)F(s/a), not F(s/a).
- Not rewriting delayed functions into the form u(t − a) f(t − a) before applying the second shift theorem.
- When using L{tⁿ f(t)} = (−1)ⁿ F⁽ⁿ⁾(s), forgetting the alternating sign (−1)ⁿ for higher powers of t.
For IB exams, always write down the theorem being used and simplify your final answer to a single fraction with factored denominators. This not only earns method marks but also makes partial fraction decomposition easier.
学生在应用尺度法则和移位定理时常犯以下错误：
- 混淆两个移位定理：第一移位涉及 e^(at) f(t) 与 F(s − a)；第二移位涉及 u(t − a) f(t − a) 与 e^(−as) F(s)。记住：时域指数 ↔ s 域移位；s 域指数 ↔ 时域移位。
- 在尺度法则中忘记因子 1/a。f(at) 的变换是 (1/a)F(s/a)，而不是 F(s/a)。
- 在应用第二移位定理之前，未将延迟函数改写为 u(t − a) f(t − a) 的形式。
- 使用 L{tⁿ f(t)} = (−1)ⁿ F⁽ⁿ⁾(s) 时，忘记高次幂 t 对应的交替符号 (−1)ⁿ。
在 IB 考试中，务必写出所用定理，并将最终答案化为以因式分母为特征的单一分式。这不仅能获得方法分，还会让部分分式分解更加容易。
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