📚 An Overview of Laplace Transforms for IB Mathematics | IB数学：拉普拉斯变换综述
The Laplace transform is a powerful integral transform that converts functions of time into functions of a complex variable. It allows us to turn differential equations into algebraic equations, making many problems in physics and engineering much easier to solve. For IB Mathematics students, this topic is often encountered in higher-level applications, particularly when studying differential equations and linear systems.
拉普拉斯变换是一种强大的积分变换，它将时间函数转换为复变量函数。它使我们能够将微分方程转化为代数方程，从而使物理和工程中的许多问题更容易求解。对于IB数学学生而言，这一主题通常在高阶应用中出现，尤其是在研究微分方程和线性系统时。
1. Definition and Notation | 定义与记号
Laplace transform of a function f(t), defined for t ≥ 0, is given by the improper integral
(单边)拉普拉斯变换定义为：对于定义在 t ≥ 0 上的函数 f(t)，其变换式为下面的广义积分
F(s) = L{f(t)} = ∫₀∞ e−st f(t) dt
on f(t) is called the time-domain function, while F(s) is said to be in the frequency domain or s-domain. The symbol L is the Laplace operator.
其中 s 是复变量，通常写成 s = σ + iω。原函数 f(t) 称为时域函数，而 F(s) 则被称为频域或 s 域函数。符号 L 表示拉普拉斯算子。
A common alternative notation is L{f} or simply F(s), where the lowercase letter corresponds to the original function and the uppercase letter denotes its transform.
另一种常见记法是 L{f} 或直接写 F(s)，其中小写字母对应原函数，大写字母表示其变换。
2. Existence and Convergence | 存在性与收敛性
interval [0, a] and is of exponential order, meaning that there exist constants M, c, T such that |f(t)| ≤ M ect for all t ≥ T.
并不是每个函数都有拉普拉斯变换。积分在 t → ∞ 时必须收敛。一个充分条件是：f(t) 在每个区间 [0, a] 上分段连续，并且是指数阶的，即存在常数 M、c、T，使得对所有 t ≥ T，有 |f(t)| ≤ M ect。
When the integral converges, it does so for all s with Re(s) > c. The region of convergence is therefore a half-plane in the complex s-plane. In most IB problems, the functions are simple enough that convergence is automatically guaranteed.
当积分收敛时，它对于所有满足 Re(s) > c 的 s 都收敛。因此，收敛区域是复平面上 s 的一个半平面。在大多数IB题目中，函数足够简单，收敛性通常自动满足。
3. Linearity | 线性性质
The Laplace transform is a linear operator. If a and b are constants and f(t) and g(t) have Laplace transforms F(s) and G(s), then
拉普拉斯变换是线性算子。如果 a 和 b 是常数，f(t) 和 g(t) 的拉普拉斯变换分别为 F(s) 和 G(s)，则有
L{af(t) + bg(t)} = aF(s) + bG(s)
This property is extremely useful when computing transforms of linear combinations of known functions. It follows directly from the linearity of the integral.
这个性质在计算已知函数线性组合的变换时非常有用。它直接由积分的线性性推出。
4. Transforms of Common Functions | 常见函数的拉普拉斯变换
Using the definition, we can derive the Laplace transforms of elementary functions. The following table lists the most frequently used pairs in IB Mathematics.
利用定义，我们可以推导出初等函数的拉普拉斯变换。下表列出了IB数学中最常用的变换对。
| f(t) | F(s) = L{f(t)} |
|---|---|
| 1 | 1/s, s > 0 |
| t (n ∈ Z⁺) | n! / sn+1, s > 0 |
| eat | 1 / (s − a), s > a |
| sin(at) | a / (s² + a²) |
| cos(at) | s / (s² + a²) |
| tn eat | n! / (s − a)n+1 |
For example, L{1} = ∫₀∞ e⁻ˢᵗ dt = 1/s. Similarly, L{eat} = ∫₀∞ e−(s−a)t ). These fundamental pairs are often used together with linearity to build new transforms.
例如，L{1} = ∫₀∞ e⁻ˢᵗ dt = 1/s。类似地，L{eat} = ∫₀∞ e−(s−a)t dt = 1/(s − a)。这些基本的变换对常与线性性质结合，用于构造新的变换。
5. Differentiation and Integration Properties | 微分与积分性质
e valuable is how they handle derivatives. If F(s) = L{f(t)} and f′(t) is continuous, then
拉普拉斯变换之所以有价值，主要原因之一是它对导数的处理。如果 F(s) = L{f(t)} 且 f′(t) 连续，则有
L{f′(t)} = sF(s) − f(0)
More generally, for the n-th derivative,
更一般地，对于 n 阶导数，有
L{f⁽ⁿ⁾(t)} = sⁿ F(s) − sⁿ⁻¹ f(0) − sⁿ⁻² f′(0) − … − f⁽ⁿ⁻¹⁾(0)
For integration, the transform of the integral of f(t) is given by
对于积分，f(t) 的积分的变换为
L{∫₀ᵗ f(τ) dτ} = F(s) / s
These properties allow us to replace differential equations in t with algebraic equations in s.
这些性质使我们能够将关于 t 的微分方程替换为关于 s 的代数方程。
6. Shifting Theorems | 平移定理
The first shifting theorem, also called the s-shift, states that multiplying f(t) by eat shifts the transform by a:
第一平移定理，也称为 s 域平移，指出将 f(t) 乘以 eat 会使变换在 s 域中移动 a：
L{eat f(t)} = F(s − a)
For example, L{eat ² + b²). This theorem is extremely useful when working with exponential damping or growth.
例如，L{eat cos(bt)} = (s − a) / ((s − a)² + b²)。这个定理在研究指数衰减或增长时非常有用。
The second shifting theorem deals with time delays. If u(t − a) is the unit step function (Heaviside function), then
第二平移定理处理的是时间延迟。如果 u(t − a) 是单位阶跃函数(赫维赛德函数)，则
L{f(t − a) u(t − a)} = e−as F(s)
This result is used to model impulses and piecewise-defined functions, which commonly appear in electrical circuits and control theory.
该结果用于建模脉冲和分段定义函数，这些函数常见于电路和控制理论中。
7. Inverse Laplace Transform | 逆拉普拉斯变换
Recovering f(t) from F(s) is called finding the inverse Laplace transform and is denoted L⁻¹{F(s)}. In IB-level problems, this is usually done by rearranging F(s) into a form that matches known transform pairs, then reading off the time-domain function.
从 F(s) 恢复 f(t) 称为求逆拉普拉斯变换，记作 L⁻¹{F(s)}。在IB水平的问题中，通常通过将 F(s) 整理成已知变换对的形式，然后直接读出时域函数来完成。
Because the transform is linear, the inverse is also linear:
由于变换是线性的，逆变换也是线性的：
L⁻¹{aF(s) + bG(s)} = a f(t) + b g(t)
efficient is 2, we write 1/(s² + 4) = (1/2) · 2/(s² + 4), so the inverse is (1/2) sin(2t).
例如，要求 L⁻¹{1/(s² + 4)}，我们识别出 2/(s² + 4) 对应 sin(2t)。因为分母系数为2，所以写 1/(s² + 4) = (1/2) · 2/(s² + 4)，因此逆变换为 (1/2) sin(2t)。
8. Partial Fractions in Inverse Transforms | 部分分式在逆变换中的应用
Many rational functions F(s) need to be decomposed into simpler fractions before the inverse transform can be applied. Suppose
许多有理函数 F(s) 在应用逆变换前需要分解为更简单的分式。假设
F(s) = (s + 3) / ((s + 1)(s + 2))
We set
我们设
(s + 3) / ((s + 1)(s + 2)) = A/(s + 1) + B/(s + 2)
uting s = −2 gives B = −1. Thus F(s) = 2/(s + 1) − 1/(s + 2), and the inverse is f(t) = 2e−t − e−2t.
两边乘以分母得 s + 3 = A(s + 2) + B(s + 1)。代入 s = −1 得 A = 2；代入 s = −2 得 B = −1。因此 F(s) = 2/(s + 1) − 1/(s + 2)，逆变换为 f(t) = 2e−t − e−2t。
When the denominator contains repeated or irreducible quadratic factors, the decomposition becomes slightly more involved, but the idea remains the same.
当分母包含重根或不可约的二次因子时，分解会稍微复杂一些，但基本思路不变。
9. Application to Differential Equations | 在微分方程中的应用
Consider the initial value problem
考虑初值问题
y″ − 3y′ + 2y = 2et, y(0) = 1, y′(0) = 2
Taking the Laplace transform of both sides, using the derivative property, we obtain
对两边取拉普拉斯变换，利用导数性质，我们得到
(s²Y(s) − s y(0) − y′(0)) − 3(sY(s) − y(0)) + 2Y(s) = 2/(s − 1)
Substituting the initial conditions gives
代入初始条件得
(s² − 3s + 2)Y(s) − s − 2 + 3 = 2/(s − 1)
Therefore
因此
Y(s) = (2/(s − 1) + s − 1) / ((s − 1)(s − 2))
Using partial fractions and inverse transforms, the solution is found as y(t) = et + e2t + t et. This demonstrates how Laplace transforms turn a differential equation into an algebraic manipulation.
使用部分分式和逆变换，解为 y(t) = et + e2t + t et。这展示了拉普拉斯变换如何将微分方程转化为代数运算。
10. Convolution Theorem | 卷积定理
The convolution of two functions f and g is defined as
两个函数 f 和 g 的卷积定义为
(f * g)(t) = ∫₀ᵗ f(τ) g(t − τ) dτ
The Laplace transform converts convolution into ordinary multiplication:
拉普拉斯变换将卷积转化为普通乘法：
L{f * g} = F(s) G(s)
e of a product of two transforms. For example, if H(s) = F(s)G(s), then h(t) = (f * g)(t).
这一性质在求解积分方程或寻找两个变换乘积的逆变换时特别有用。例如，如果 H(s) = F(s)G(s)，则 h(t) = (f * g)(t)。
11. Connection to Fourier Transform | 与傅里叶变换的联系
gral becomes the Fourier transform. However, the Laplace transform includes the real part σ, allowing it to handle a wider class of functions, including those that do not decay to zero.
拉普拉斯变换可以看作是傅里叶变换的推广。如果令 s = iω，拉普拉斯积分就变成了傅里叶变换。然而，拉普拉斯变换包含了实部 σ，因此能够处理更广泛的函数，包括那些不衰减到零的函数。
In signal processing, the Laplace transform is used on continuous-time systems, while the Fourier transform is used to analyse frequency content of periodic or finite-energy signals. Both tools share many properties, such as linearity and convolution.
在信号处理中，拉普拉斯变换用于连续时间系统，而傅里叶变换用于分析周期或有限能量信号的频率成分。两者共享许多性质，如线性和卷积。
12. Exam Tips for IB Students | 给IB学生的考试建议
When solving exam problems, always check the initial conditions before applying the derivative property. A common mistake is to forget subtracting f(0) and its derivatives when computing L{f′} or L{f″}. Also remember to state the domain of convergence where required, even though it is often implicit.
在解答考试题目时，在应用导数性质前一定要检查初始条件。一个常见错误是计算 L{f′} 或 L{f″} 时忘记减去 f(0) 及其各阶导数。同时，在需要说明收敛域的地方一定要写出来，即使问题中常常隐含。
For inverse transforms, practise partial fractions thoroughly. Many IB questions include rational functions with simple linear or quadratic factors. Recognising standard pairs instantly will save you time. Also, review the first shifting theorem, as questions often combine eat with sine or cosine.
对于逆变换，请彻底练习部分分式。许多IB题目包含具有简单线性或二次因子的有理函数。迅速识别标准变换对能节省时间。此外，复习第一平移定理，因为题目常常将 eat 与正弦或余弦结合。
Finally, interpret the problem and check your solution. If the original differential equation has constant coefficients, the Laplace method gives a straightforward algebraic path. Verify that the solution satisfies the initial conditions and, if possible, substitute back into the original equation.
最后，要理解题意并检查你的解。如果原微分方程是常系数的，拉普拉斯方法会提供一条直接的代数路径。验证解是否满足初始条件，并在可能的情况下，将解代回原方程进行检验。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图