The Impulse Function and Its Laplace Transform | 冲激函数及其拉普拉斯变换

📚 The Impulse Function and Its Laplace Transform | 冲激函数及其拉普拉斯变换

The impulse function, also called the Dirac delta function δ(t), is a mathematical idealization used to model instantaneous phenomena such as a sudden force, a spike in voltage, or a point mass. Although it is not a function in the ordinary sense, it plays a central role in calculus, differential equations, and the theory of Laplace transforms. In this article, we explore its definition, key properties, and how to work with its Laplace transform for IB Mathematics Higher Level and further studies.

冲激函数又称狄拉克δ函数 δ(t)，是一种数学理想化模型，用于描述瞬时现象，例如突然的力、电压尖峰或点质量。尽管它不是普通意义下的函数，但在微积分、微分方程以及拉普拉斯变换理论中占据核心地位。本文将探讨其定义、关键性质，以及如何在IB数学高级水平及后续学习中运用它的拉普拉斯变换。

1. What Is an Impulse? | 什么是冲激？

In physics, an impulse is the product of a force and the short time during which it acts. If a force F(t) is applied over a very small interval, the total impulse is the area under the force-time graph. For example, a cricket bat hitting a ball exerts a huge force for only a few milliseconds, yet the total change in momentum is finite and measurable.

在物理学中，冲激是力与其作用时间的乘积。如果力 F(t) 在极短的时间间隔内作用，则总冲激就是力—时间图像下的面积。例如，板球拍击球时，在几毫秒内施加巨大但短暂的力，然而动量的总变化是有限且可测量的。

The idea behind the delta function is to compress this finite area into an infinitesimally short time interval. Instead of describing the exact shape of the force, we simply record its total effect: the impulse value.

δ函数背后的思想正是将这一有限面积压缩到无穷短的时间间隔内。我们不再描述力的具体形状，而只记录其总效果：冲激值。

2. Defining the Dirac Delta Function | 狄拉克δ函数的定义

Heuristically, the Dirac delta function δ(t) is described by two conditions:

直观上，狄拉克δ函数 δ(t) 由两个条件描述：

δ(t) = 0 for all t ≠ 0

对所有 t ≠ 0，δ(t) = 0
The integral of δ(t) over the whole real line equals 1

δ(t) 在整个实数轴上的积分为 1

More rigorously, we can construct δ(t) as the limit of a narrow rectangular pulse. Let δ_ε(t) be a rectangular pulse of width ε and height 1/ε, centered at the origin. Then as ε approaches 0, the pulse becomes infinitely tall and infinitesimally narrow, while its area remains equal to 1.

更严格地说，我们可以把 δ(t) 看作一个窄矩形脉冲的极限。设 δ_ε(t) 是宽度为 ε、高度为 1/ε、以原点为中心的矩形脉冲。当 ε 趋近于 0 时，脉冲变得无限高且无限窄，但其面积始终保持为 1。

δ(t) = lim (ε→0) δ_ε(t), where δ_ε(t) = 1/ε for −ε/2 ≤ t ≤ ε/2, and 0 otherwise

Because δ(t) cannot be defined pointwise like an ordinary function, it is technically known as a generalized function or distribution. For IB Mathematics and engineering applications, the intuitive limit interpretation is usually sufficient.

由于 δ(t) 不能像普通函数那样逐点定义，它在严格意义下被称为广义函数或分布。对于IB数学和工程应用而言，极限的直观解释通常已经足够。

3. Key Properties of the Delta Function | δ函数的关键性质

The following properties are essential for manipulating expressions that contain δ(t).

以下性质是处理含 δ(t) 表达式时的关键工具。

Zero away from the origin: δ(t) = 0 for t ≠ 0

在原点以外为零：当 t ≠ 0 时，δ(t) = 0
Total area equals 1: ∫ from −∞ to ∞ of δ(t) dt = 1

总面积为 1：∫ 从 −∞ 到 ∞ 的 δ(t) dt = 1
Evenness: δ(−t) = δ(t)

偶函数性质：δ(−t) = δ(t)
Scaling: δ(at) = (1/|a|) δ(t), for a ≠ 0

伸缩性质：δ(at) = (1/|a|) δ(t)，其中 a ≠ 0

The scaling property is particularly useful. For instance, δ(2t) = (1/2)δ(t), because the factor 2 compresses the “concentration” by half, so the area is preserved only after dividing by 2.

伸缩性质尤其常用。例如，δ(2t) = (1/2)δ(t)，因为因子 2 将“集中程度”压缩一半，只有除以 2 后才能保持面积不变。

4. The Sifting Property | 筛选性质

The most important operational property of the delta function is the sifting (or sampling) property. It states that for any continuous function f(t),

δ函数最重要的运算性质是筛选（或采样）性质。它指出，对任意连续函数 f(t)，

∫ from −∞ to ∞ of f(t) δ(t − a) dt = f(a)

In words, multiplying f(t) by δ(t − a) and integrating across all time selects exactly the value of f at the point t = a. This is why the delta function is often called a “pick-and-choose” device.

换言之，将 f(t) 乘以 δ(t − a) 并对所有时间积分，恰好选出 f 在 t = a 处的值。因此，δ函数常被称为“取数器”或“筛选器”。

If the interval of integration does not include the point a, the result is 0. For example, ∫ from 2 to 5 of f(t) δ(t − 7) dt = 0, since t = 7 lies outside the interval.

如果积分区间不包含点 a，则结果为零。例如，∫ 从 2 到 5 的 f(t) δ(t − 7) dt = 0，因为 t = 7 在区间之外。

5. Relation to the Heaviside Step Function | 与赫维赛德阶跃函数的关系

The Heaviside step function u(t) is defined as u(t) = 1 for t > 0, and u(t) = 0 for t < 0. It is used to "switch on" a function at t = 0.

赫维赛德阶跃函数 u(t) 定义为：当 t > 0 时 u(t) = 1，当 t < 0 时 u(t) = 0。它用于在 t = 0 时“开启”某个函数。

A beautiful connection between the step function and the delta function is that differentiation of u(t) yields δ(t):

阶跃函数与δ函数之间有一个漂亮的联系：对 u(t) 求导即得 δ(t)：

u'(t) = δ(t)

Conversely, the integral of δ(t) from −∞ to t gives the step function: ∫ from −∞ to t of δ(τ) dτ = u(t).

反过来，δ(t) 从 −∞ 到 t 的积分给出阶跃函数：∫ 从 −∞ 到 t 的 δ(τ) dτ = u(t)。

This relationship is why the Laplace transform of the step function, ℒ{u(t)} = 1/s, is naturally connected to the Laplace transform of the delta function, ℒ{δ(t)} = 1.

正是这种关系，使得阶跃函数的拉普拉斯变换 ℒ{u(t)} = 1/s 与δ函数的拉普拉斯变换 ℒ{δ(t)} = 1 自然地联系在一起。

6. Laplace Transform of the Impulse | 冲激函数的拉普拉斯变换

By definition, the one-sided Laplace transform of a function f(t) is

根据定义，函数 f(t) 的单边拉普拉斯变换为

ℒ{f(t)} = ∫ from 0 to ∞ of e^(−st) f(t) dt

For f(t) = δ(t), the sifting property gives

对于 f(t) = δ(t)，利用筛选性质可得

ℒ{δ(t)} = ∫ from 0 to ∞ of e^(−st) δ(t) dt = e^(−s·0) = 1

Therefore, the Laplace transform of the unit impulse is the constant function 1.

因此，单位冲激的拉普拉斯变换是常数函数 1。

If the impulse occurs at a later time t = a, with a > 0, then

如果冲激出现在较晚的时刻 t = a（a > 0），则

ℒ{δ(t − a)} = ∫ from 0 to ∞ of e^(−st) δ(t − a) dt = e^(−as)

This result exactly parallels the shift property for ordinary functions: a time delay in the time domain corresponds to multiplication by e^(−as) in the s-domain.

这一结果与普通函数的时移性质完全平行：时域中的时间延迟对应于 s 域中乘以 e^(−as)。

f(t)	ℒ{f(t)} = F(s)
δ(t)	1
δ(t − a)	e^(−as)
u(t)	1/s
u(t − a)	e^(−as)/s

7. Laplace Transform of Derivatives Involving the Impulse | 含冲激导数的拉普拉斯变换

Sometimes we need the Laplace transform of δ'(t), the derivative of the delta function. This function is a “doublet” or dipole, and it arises in certain circuit and vibration problems.

有时我们需要 δ'(t)（δ函数的导数）的拉普拉斯变换。这个函数称为“偶极子”或“双极”，出现在某些电路和振动问题中。

Using the definition and integration by parts, we obtain

利用定义和分部积分，我们得到

ℒ{δ'(t)} = s

More generally, the Laplace transform of the n-th derivative of the delta function is

更一般地，δ函数的 n 阶导数的拉普拉斯变换为

ℒ{δ⁽ⁿ⁾(t)} = sⁿ

This follows because differentiation in the time domain corresponds to multiplication by s in the s-domain, after applying the usual initial condition assumptions at t = 0⁻.

这个结果之所以成立，是因为在时域中求导对应于在 s 域中乘以 s（在采用 t = 0⁻ 处通常的初值假设之后）。

8. Solving Differential Equations with Impulse Forcing | 求解含冲激输入的微分方程

The Laplace transform is exceptionally effective for solving differential equations with impulse inputs. Consider the initial value problem

拉普拉斯变换在求解含冲激输入的微分方程时极为有效。考虑初值问题

y” + 3y’ + 2y = δ(t − 1), y(0) = 0, y'(0) = 0

Step 1: Take the Laplace transform of both sides.

第一步：对方程两边同时取拉普拉斯变换。

(s²Y(s) − s·0 − 0) + 3(sY(s) − 0) + 2Y(s) = e^(−s)

Since the initial conditions are zero, this simplifies to

由于初值为零，上式简化为

(s² + 3s + 2)Y(s) = e^(−s)

Step 2: Solve for Y(s).

第二步：解出 Y(s)。

Y(s) = e^(−s) / ((s + 1)(s + 2))

Step 3: Decompose into partial fractions.

第三步：化为部分分式。

1 / ((s + 1)(s + 2)) = 1/(s + 1) − 1/(s + 2)

Step 4: Apply the inverse Laplace transform, using the shift e^(−s).

第四步：利用因子 e^(−s) 应用拉普拉斯逆变换。

y(t) = u(t − 1)(e^(−(t − 1)) − e^(−2(t − 1)))

The step function u(t − 1) ensures that the response is zero before the impulse arrives at t = 1. After t = 1, the system responds with a decaying exponential combination.

阶跃函数 u(t − 1) 确保在冲激于 t = 1 到达之前响应为零。在 t = 1 之后，系统以衰减指数组合的形式作出响应。

9. Applications in Physics and Engineering | 在物理与工程中的应用

The impulse function is not merely a theoretical curiosity; it appears in many real-world contexts.

冲激函数不仅仅是理论上的珍品，它在许多现实场景中都会出现。

Mechanics: an impulsive force, such as a hammer blow, is modeled by a large force acting over a short time. The delta function gives the total momentum change directly.

力学：锤击等冲激力可以用短时间内的巨大作用力来建模。δ函数可以直接给出总动量变化。
Electrical circuits: a sudden voltage spike injected into an RC or RLC circuit is represented by δ(t). The impulse response of a circuit is the fundamental building block for analysing its behaviour under any input.

电路：注入RC或RLC电路的瞬时电压尖峰可以用 δ(t) 表示。电路的冲激响应是分析其在任意输入下行为的基本构件。
Signal processing: the impulse response of a linear time-invariant system completely characterizes the system. Any input signal can be decomposed into impulses via convolution.

信号处理：线性时不变系统的冲激响应完全刻画了系统的特性。任何输入信号都可以通过卷积分解为一系列冲激。
Probability and statistics: although not the focus here, the delta function is also used to represent discrete probability densities in continuous frameworks.

概率与统计：虽然不是本文的重点，δ函数也可用于在连续框架中表示离散概率密度。

10. Common Pitfalls and Exam Tips | 常见误区与考试技巧

Students often lose marks because of a few recurring mistakes. Here is what to watch out for.

学生常常因为一些反复出现的错误而失分。以下是你需要留意的要点。

Treating δ(t) as an ordinary function: δ(0) is not a finite number; only its integral has meaning. Never plug t = 0 into δ(t) and try to evaluate it directly.

把 δ(t) 当作普通函数：δ(0) 不是有限数值；只有它的积分才有意义。切勿直接将 t = 0 代入 δ(t) 并试图求值。
Forgetting the shift: ℒ{δ(t − a)} = e^(−as), not e^(as) or 1. The sign of the exponent depends on the direction of the shift.

忘记时移：ℒ{δ(t − a)} = e^(−as)，不是 e^(as) 或 1。指数中的正负号取决于时移的方向。
Using the wrong lower limit: in the one-sided Laplace transform, the integral starts at 0. If an impulse occurs at t = 0, the sifting property still gives 1, but careful textbook conventions use 0⁻ to include the impulse fully.

下限用错：单边拉普拉斯变换的积分从 0 开始。如果冲激发生在 t = 0，筛选性质仍然给出 1，但严格教材通常使用 0⁻ 以完整包含该冲激。
Partial fraction errors: when solving differential equations with impulse forcing, always verify your partial fractions before applying the inverse transform.

部分分式错误：在求解含冲激输入的微分方程时，应用逆变换之前一定要验证部分分式分解是否正确。

Exam tip: when you see an impulse forcing term, immediately write down its Laplace transform from the standard table. This saves time and reduces algebraic errors.

考试技巧：当你看到冲激输入项时，立即从标准表中写出其拉普拉斯变换。这样可以节省时间并减少代数错误。

11. Practice Problems | 练习题

Try these problems to consolidate your understanding.

尝试以下题目以巩固你的理解。

Problem 1: Compute ℒ{5δ(t) + 3δ(t − 4)}.

题目1：计算 ℒ{5δ(t) + 3δ(t − 4)}。

Answer: 5 + 3e^(−4s)

Problem 2: Solve y” + 4y = 2δ(t − π), y(0) = 0, y'(0) = 0.

题目2：求解 y” + 4y = 2δ(t − π)，y(0) = 0，y'(0) = 0。

Answer: y(t) = u(t − π) sin(2(t − π))

Problem 3: Use the sifting property to evaluate ∫ from −∞ to ∞ of (t² + 1) δ(t − 3) dt.

题目3：利用筛选性质计算 ∫ 从 −∞ 到 ∞ 的 (t² + 1) δ(t − 3) dt。

Answer: 10

Work through each problem systematically: take the transform, solve algebraically, then transform back. This is the core skill tested in IB Mathematics HL differential equations questions involving impulse functions.

请按系统步骤完成每道题：取变换、代数求解、再逆变换。这是IB数学高级水平中涉及冲激函数的微分方程题所考察的核心技能。

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The Impulse Function and Its Laplace Transform | 冲激函数及其拉普拉斯变换