📚 The Unit Step Function and the Second Shifting Theorem of Laplace Transforms | 单位阶跃函数与拉普拉斯变换的延迟法则
In engineering and physics, many systems are not switched on at time \( t = 0 \). Instead, inputs such as voltages, forces, or signals may start at a later time \( t = a \), or may be turned off after a certain interval. The unit step function—also called the Heaviside function—provides a concise way to model such “switching-on” behaviour. When combined with the Laplace transform, it gives rise to a powerful rule known as the second shifting theorem (or the delay theorem), which allows us to handle time-delayed functions with ease.
在工程与物理中，许多系统并非在 \( t = 0 \) 时刻启动。输入信号（如电压、力或数据流）可能是在稍后的时刻 \( t = a \) 才被施加，也可能在一段时间后被关闭。单位阶跃函数（又称海维赛德函数）为这种“开关”行为提供了一种简洁的建模工具。当它与拉普拉斯变换结合时，便引出了强大的“第二平移定理”（即延迟法则），让我们能够轻松处理时间延迟函数。
1. Definition of the Unit Step Function | 单位阶跃函数的定义
The unit step function \( u(t) \) is defined as a piecewise constant function that is zero before a certain instant and one from that instant onward. In many textbooks, the function is written as \( u(t-a) \) to indicate that the “step” occurs at time \( t = a \). Formally, we write
单位阶跃函数 \( u(t) \) 是一个分段常数函数：在某个时刻之前取值为 0，从该时刻起取值为 1。许多教材使用 \( u(t-a) \) 的记号，以表明“跳跃”发生在 \( t = a \) 处。其正式定义为
u(t − a) = 0 for t < a; u(t − a) = 1 for t ≥ a
When \( a = 0 \), the function is simply \( u(t) \), which is 0 for \( t < 0 \) and 1 for \( t \ge 0 \). The value at the jump point is usually taken to be 1 for consistency with the Laplace transform theory, although the exact value at a single point does not affect the integral.
当 \( a = 0 \) 时，函数简化为 \( u(t) \)，它在 \( t < 0 \) 时为 0，在 \( t \ge 0 \) 时为 1。在跳跃点处的取值通常规定为 1，这是为了与拉普拉斯变换理论保持一致；事实上，单点处的取值并不影响积分结果。
2. Why Do We Need the Unit Step Function? | 为什么需要单位阶跃函数？
Many real-world inputs are not smooth or continuous. For example, a switch is flipped at \( t = 5 \) seconds, or a periodic signal is turned on for a limited time window. Writing such functions using traditional piecewise notation can be cumbersome, especially when manipulating them algebraically or transforming them. The unit step function allows us to express piecewise functions in a single closed-form expression, which we can then transform using standard Laplace rules.
许多实际输入并非光滑或连续的。例如，开关在 \( t = 5 \) 秒时被拨动，或周期性信号只在有限时间窗口内起作用。如果使用传统分段记号，这些函数写起来十分冗长，特别是当需要代数操作或变换时。单位阶跃函数允许我们用单个闭式表达式来表示分段函数，进而利用标准的拉普拉斯变换规则进行处理。
For instance, a function \( f(t) \) that is 0 for \( t < 3 \) and equals \( \sin(t) \) for \( t \ge 3 \) can be written as \( f(t) = \sin(t) \cdot u(t-3) \). This compact notation is the key to applying the delay theorem.
例如，一个函数 \( f(t) \)：当 \( t < 3 \) 时为 0，当 \( t \ge 3 \) 时为 \( \sin(t) \)，可以写成 \( f(t) = \sin(t) \cdot u(t-3) \)。这种紧凑记法是运用延迟法则的关键。
3. The Laplace Transform of the Unit Step Function | 单位阶跃函数的拉普拉斯变换
To appreciate the delay theorem, we first need the Laplace transform of the unit step function itself. Using the definition of the Laplace transform, we compute
要理解延迟法则，首先需要掌握单位阶跃函数自身的拉普拉斯变换。根据拉普拉斯变换的定义，
ℒ{u(t − a)} = ∫₀ᵃ 0·e⁻ˢᵗ dt + ∫ₐᵝ 1·e⁻ˢᵗ dt = e⁻ᵃˢ / s, s > 0
Here, \( e^{-as} \) is called the delay factor. It encodes the information that the step occurs at time \( a \). When \( a = 0 \), the transform reduces to \( 1/s \), which we already know for \( u(t) \).
其中 \( e^{-as} \) 称为延迟因子，它表征了阶跃发生在时刻 \( a \) 这一信息。当 \( a = 0 \) 时，变换退化为 \( 1/s \)，这正是 \( u(t) \) 的拉普拉斯变换。
4. The Second Shifting Theorem (Delay Theorem) | 第二平移定理（延迟法则）
The second shifting theorem answers the following question: if we know the Laplace transform \( F(s) \) of a function \( f(t) \), what is the Laplace transform of the delayed and then “switched-on” function \( f(t-a) \cdot u(t-a) \)? The result is surprisingly elegant:
第二平移定理回答了如下问题：如果我们已知函数 \( f(t) \) 的拉普拉斯变换 \( F(s) \)，那么延迟后“开启”的函数 \( f(t-a) \cdot u(t-a) \) 的拉普拉斯变换是什么？结果非常优美：
ℒ{ f(t − a) · u(t − a) } = e⁻ᵃˢ F(s), a ≥ 0
Equivalently, multiplying the transform by \( e^{-as} \) corresponds to shifting the original time function to the right by \( a \) units, with the step function ensuring the function is zero before \( t = a \). This theorem is also known as the time-shifting property.
换句话说，对变换乘以 \( e^{-as} \)，等价于把原始时间函数向右平移 \( a \) 个单位；单位阶跃函数确保了函数在 \( t < a \) 时为零。这一定理也称为时移性质。
5. Derivation of the Delay Theorem | 延迟法则的推导
Let us derive this result directly from the definition. We start with the Laplace integral of the delayed function:
下面从定义出发直接推导。设延迟函数的拉普拉斯积分为：
ℒ{ f(t − a) · u(t − a) } = ∫₀ᵃ 0·e⁻ˢᵗ dt + ∫ₐᵝ f(t − a) · e⁻ˢᵗ dt
We then change variables. Let \( τ = t − a \), so \( t = τ + a \) and \( dt = dτ \). The lower limit \( t = a \) becomes \( τ = 0 \), and the upper limit \( t → ∞ \) becomes \( τ → ∞ \). Therefore
接着进行变量代换。令 \( τ = t − a \)，则 \( t = τ + a \)，\( dt = dτ \)。积分下限 \( t = a \) 变为 \( τ = 0 \)，上限 \( t → ∞ \) 变为 \( τ → ∞ \)。因此
∫ₐᵝ f(t − a) e⁻ˢᵗ dt = ∫₀ᵝ f(τ) e⁻ˢ⁽ᵗ⁺ᵃ⁾ dτ = e⁻ᵃˢ ∫₀ᵝ f(τ) e⁻ˢᵗ dτ
The remaining integral is exactly \( F(s) \), the Laplace transform of \( f(t) \). This completes the proof.
剩余积分恰好为 \( F(s) \)，即 \( f(t) \) 的拉普拉斯变换。证明完毕。
6. Key Identities and Notation | 关键恒等式与记号
Below are the most useful identities that frequently appear in IB exam questions. Note that the function before the step must be written in the form \( f(t-a) \), not \( f(t) \), before applying the theorem.
以下是 IB 考试中最常用的几个恒等式。需要注意：在应用第二平移定理之前，阶跃函数前面的函数必须写成 \( f(t-a) \) 的形式，而不能是 \( f(t) \)。
| Function (时域) | Laplace Transform (频域) |
| u(t−a) | e⁻ᵃˢ / s |
| f(t−a)·u(t−a) | e⁻ᵃˢ F(s) |
| u(t−a) (constant 1) | e⁻ᵃˢ / s |
| (t−a)·u(t−a) | e⁻ᵃˢ / s² |
| e^(b(t−a))·u(t−a) | e⁻ᵃˢ / (s−b) |
7. Worked Example 1: A Simple Shift | 例题 1：简单平移
Let us compute the Laplace transform of \( g(t) = t^2 \cdot u(t – 4) \). We need to rewrite the coefficient \( t^2 \) in terms of \( (t-4) \). Let \( w = t – 4 \), so \( t = w + 4 \). Then \( t^2 = (w+4)^2 = w^2 + 8w + 16 \). Thus,
我们来计算 \( g(t) = t^2 \cdot u(t – 4) \) 的拉普拉斯变换。需要把系数 \( t^2 \) 改写为 \( (t-4) \) 的表达式。令 \( w = t – 4 \)，则 \( t = w + 4 \)，于是 \( t^2 = (w+4)^2 = w^2 + 8w + 16 \)。因此
g(t) = [ (t−4)² + 8(t−4) + 16 ] · u(t−4)
Now we apply the delay theorem term by term:
接着逐项应用延迟定理：
ℒ{g(t)} = e⁻⁴ˢ( 2/s³ + 8/s² + 16/s )
This example illustrates the crucial algebraic step: express the function as \( f(t-a) \) before “shifting”.
该例子展示了关键代数步骤：先化为 \( f(t-a) \) 的形式，再进行平移。
8. Worked Example 2: A Piecewise Function | 例题 2：分段函数
Consider the function \( h(t) \) defined as: \( h(t) = 1 \) for \( 0 \le t < 2 \), and \( h(t) = e^{-t} \) for \( t \ge 2 \). We want its Laplace transform. First, write \( h(t) \) using step functions:
设函数 \( h(t) \) 定义为：当 \( 0 \le t < 2 \) 时 \( h(t) = 1 \)，当 \( t \ge 2 \) 时 \( h(t) = e^{-t} \)。我们来求其拉普拉斯变换。首先用阶跃函数表示 \( h(t) \)：
h(t) = 1 · [1 − u(t−2)] + e⁻ᵗ · u(t−2)
To apply the delay theorem to the term \( e^{-t} \cdot u(t-2) \), we write \( e^{-t} = e^{-(t-2)} \cdot e^{-2} \). Therefore,
为了对 \( e^{-t} \cdot u(t-2) \) 应用延迟定理，我们改写 \( e^{-t} = e^{-(t-2)} \cdot e^{-2} \)。因此，
ℒ{h(t)} = 1/s − e⁻²ˢ/s + e⁻²ˢ · 1/(s+1)
We can combine the last two terms if desired: \( \mathcal{L}\{h\} = 1/s + e^{-2s}[1/(s+1) − 1/s] \).
如有需要可合并后两项：\( \mathcal{L}\{h\} = 1/s + e^{-2s}[1/(s+1) − 1/s] \)。
9. Inverse Transform Using the Delay Theorem | 利用延迟定理求逆变换
The delay theorem is equally useful in the reverse direction. If a Laplace transform has the form \( e^{-as} F(s) \), then its inverse is \( f(t-a)\cdot u(t-a) \). For example, consider
延迟定理在求逆变换时同样非常有用。如果某个拉普拉斯变换具有 \( e^{-as} F(s) \) 的形式，那么其逆变换就是 \( f(t-a)\cdot u(t-a) \)。例如，考虑
H(s) = ( 3e⁻²ˢ ) / ( s² + 9 )
We recognize \( 3/(s^2 + 9) \) as the transform of \( \sin(3t) \). Since \( a = 2 \), we immediately obtain
我们识别出 \( 3/(s^2 + 9) \) 是 \( \sin(3t) \) 的拉普拉斯变换。由于 \( a = 2 \)，立即得到
h(t) = sin[3(t−2)] · u(t−2)
Notice how the inverse transform must also be expressed with a shifted argument \( t-a \).
注意逆变换也必须包含平移后的自变量 \( t-a \)。
10. Solving Differential Equations with Delayed Inputs | 求解带延迟输入的微分方程
The delay theorem is especially valuable when solving initial value problems that involve forcing terms starting at a time other than \( t = 0 \). Consider the differential equation:
当微分方程的输入项并非从 \( t = 0 \) 开始作用时，延迟法则尤为实用。考虑以下微分方程：
y’ − 3y = u(t−2), y(0) = 0
Taking the Laplace transform of both sides, using the derivative rule \( \mathcal{L}\{y’\} = sY(s) − y(0) \), gives
对方程两边取拉普拉斯变换，利用导数法则 \( \mathcal{L}\{y’\} = sY(s) − y(0) \)，得到
sY(s) − 3Y(s) = e⁻²ˢ / s
Solving for \( Y(s) \):
解出 \( Y(s) \)：
Y(s) = e⁻²ˢ / [ s(s − 3) ]
We first find the partial fraction decomposition of the non-delayed part: \( 1/[s(s-3)] = -1/(3s) + 1/[3(s-3)] \). Therefore,
先对非延迟部分作部分分式分解：\( 1/[s(s-3)] = -1/(3s) + 1/[3(s-3)] \)。于是
Y(s) = e⁻²ˢ · ( −1/(3s) + 1/(3(s−3)) )
Inverse transforming using the delay theorem yields
利用延迟定理求逆变换，得到
y(t) = [ −1/3 + (1/3)e^{3(t−2)} ] · u(t−2)
This compact answer fully describes the solution: it is zero before \( t = 2 \) and follows the forced response afterward.
这个紧凑的答案完整描述了方程的解：在 \( t < 2 \) 时为零，在 \( t \ge 2 \) 之后遵循受迫响应。
11. Common Mistake and Pitfall | 常见错误与陷阱
The most frequent mistake in applying the second shifting theorem is to compute \( \mathcal{L}\{f(t)\cdot u(t-a)\} \) as \( e^{-as}F(s) \). This is incorrect unless \( f(t) \) is already of the form \( f(t-a) \). For instance, \( \mathcal{L}\{e^{-t} u(t-2)\} \) is not \( e^{-2s}/(s+1) \). The correct procedure is to rewrite \( e^{-t} = e^{-2}\cdot e^{-(t-2)} \), then apply the theorem to get \( e^{-2s} \cdot e^{-2}/(s+1) \).
应用第二平移定理时最常见的错误是把 \( \mathcal{L}\{f(t)\cdot u(t-a)\} \) 直接写成 \( e^{-as}F(s) \)。除非 \( f(t) \) 已经具有 \( f(t-a) \) 的形式，否则这是错误的。例如，\( \mathcal{L}\{e^{-t} u(t-2)\} \) 并不等于 \( e^{-2s}/(s+1) \)。正确做法是先改写 \( e^{-t} = e^{-2}\cdot e^{-(t-2)} \)，然后应用定理得到 \( e^{-2s} \cdot e^{-2}/(s+1) \)。
Another common pitfall is forgetting the step function in the inverse transform. The result must include \( u(t-a) \) to indicate that the solution is inactive before the delay time.
另一个常见陷阱是在逆变换结果中忘记阶跃函数。结果必须包含 \( u(t-a) \)，以表明解在延迟时刻之前不生效。
12. Summary and exam tips | 总结与考试提示
The second shifting theorem is a cornerstone of Laplace transform applications. To use it effectively: (1) rewrite any function multiplied by \( u(t-a) \) in terms of \( t-a \); (2) apply \( \mathcal{L}\{f(t-a)u(t-a)\} = e^{-as}F(s) \); (3) for inverse transforms, don’t forget the \( u(t-a) \) factor; (4) combine with partial fractions when dealing with rational functions.
第二平移定理是拉普拉斯变换应用的基石。有效使用它的要点是：(1) 将任何与 \( u(t-a) \) 相乘的函数改写为 \( t-a \) 的表达式；(2) 应用 \( \mathcal{L}\{f(t-a)u(t-a)\} = e^{-as}F(s) \)；(3) 求逆变换时，不要忘记 \( u(t-a) \) 因子；(4) 处理有理函数时，结合部分分式分解。
In the IB exam, this theorem often appears in paper 2 or paper 3 questions involving piecewise forcing functions or systems with signal delays. Practice rewriting functions into the correct shifted form—it is the single most important algebraic skill for this topic.
在 IB 考试中，该定理经常出现在 Paper 2 或 Paper 3 中涉及分段输入函数或系统延迟的题目中。反复练习将函数化为正确的平移形式——这是本主题中最重要的代数技巧。
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