Convolution Theorem for Laplace Transforms | 拉普拉斯变换的卷积定理

📚 Convolution Theorem for Laplace Transforms | 拉普拉斯变换的卷积定理

The convolution theorem is one of the most powerful tools in the theory of Laplace transforms. It allows us to find the inverse Laplace transform of a product of two transforms by computing a single integral, and it plays a central role in solving differential equations, integral equations, and control systems analysis. In this article, we will explore the statement, proof, and applications of this theorem with carefully matched examples.

卷积定理是拉普拉斯变换理论中最强大的工具之一。它使我们能够通过计算一个积分来求得两个变换之积的逆拉普拉斯变换，并且在求解微分方程、积分方程和控制系统分析中发挥着核心作用。本文将系统地介绍这一定理的表述、证明及其应用，并配以精心匹配的例题。

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1. Definition of Convolution | 卷积的定义

Let \(f(t)\) and \(g(t)\) be two piecewise continuous functions defined for \(t \ge 0\). Their convolution, denoted by \((f * g)(t)\), is defined by the integral

设 \(f(t)\) 和 \(g(t)\) 是定义在 \(t \ge 0\) 上的两个分段连续函数。它们的卷积记为 \((f * g)(t)\)，定义为如下积分

(f * g)(t) = ∫₀ᵗ f(τ) g(t − τ) dτ

Notice that the integral runs from 0 to \(t\), which reflects the causal nature of the functions—values of \(g\) after time \(t\) do not contribute. The convolution is commutative: \((f * g)(t) = (g * f)(t)\).

注意积分限从 0 到 \(t\)，这反映了函数的因果性——即 \(g\) 在时间 \(t\) 之后的值不会产生影响。卷积满足交换律：\((f * g)(t) = (g * f)(t)\)。

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2. The Laplace Transform: A Quick Review | 拉普拉斯变换：简要回顾

The one-sided Laplace transform of a function \(f(t)\) is defined as

函数 \(f(t)\) 的单边拉普拉斯变换定义为

F(s) = ℒ{f(t)} = ∫₀^∞ e⁻ˢᵗ f(t) dt

for values of \(s\) where the integral converges. The inverse Laplace transform is denoted by \(f(t) = ℒ⁻¹{F(s)}\). A standard property is that \(\mathcal{L}\{f'(t)\} = sF(s) – f(0)\), which converts differentiation into algebra.

其中 \(s\) 取使积分收敛的值。逆拉普拉斯变换记为 \(f(t) = ℒ⁻¹{F(s)}\)。一个标准性质是 \(\mathcal{L}\{f'(t)\} = sF(s) – f(0)\)，它把微分运算转化为代数运算。

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3. Statement of the Convolution Theorem | 卷积定理的表述

The convolution theorem states that if \(F(s) = \mathcal{L}\{f(t)\}\) and \(G(s) = \mathcal{L}\{g(t)\}\), then

卷积定理指出：若 \(F(s) = \mathcal{L}\{f(t)\}\) 且 \(G(s) = \mathcal{L}\{g(t)\}\)，则

ℒ{(f * g)(t)} = F(s) · G(s)

Equivalently, for inverse transforms we have

等价地，对于逆变换有

ℒ⁻¹{F(s) G(s)} = ∫₀ᵗ f(τ) g(t − τ) dτ

This theorem turns a product in the \(s\)-domain into a convolution in the time domain, which is often much easier to evaluate.

这一定理将 \(s\) 域中的乘积转化为时域中的卷积，而时域卷积往往更容易计算。

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4. Proof of the Convolution Theorem | 卷积定理的证明

To prove the theorem, we write the product \(F(s)G(s)\) as a double integral:

为了证明该定理，我们将乘积 \(F(s)G(s)\) 写成二重积分：

F(s)G(s) = ∫₀^∞ ∫₀^∞ e⁻ˢ⁽ᵘ⁺ᵛ⁾ f(u) g(v) du dv

Introduce the change of variables \(t = u + v\) and \(\tau = u\). The region \(u, v \ge 0\) maps to \(0 \le \tau \le t < \infty\), with \(v = t - \tau\). The Jacobian is 1, hence

引入变量替换 \(t = u + v\)、\(\tau = u\)。区域 \(u, v \ge 0\) 映射到 \(0 \le \tau \le t < \infty\)，且 \(v = t − \tau\)。雅可比行列式为 1，因此

F(s)G(s) = ∫₀^∞ e⁻ˢᵗ [ ∫₀ᵗ f(τ) g(t − τ) dτ ] dt = ℒ{(f * g)(t)}

The interchange of integration order is justified by absolute convergence under typical assumptions of exponential order.

在通常的指数阶条件下，积分次序的交换可由绝对收敛性保证。

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5. Key Properties of Convolution | 卷积的主要性质

• Commutativity: \(f * g = g * f\), so we may choose whichever function is easier to integrate.

  交换律： \(f * g = g * f\)，因此我们可以选择更容易积分的那个函数。

• Associativity: \((f * g) * h = f * (g * h)\), allowing grouping without changing the result.

  结合律： \((f * g) * h = f * (g * h)\)，组合方式不影响结果。

• Distributivity: \(f * (g + h) = f * g + f * h\), which is useful in decomposing complex inputs.

  分配律： \(f * (g + h) = f * g + f * h\)，这有助于分解复杂输入。

• Identity element: \(f * \delta(t) = f(t)\), where \(\delta(t)\) is the Dirac delta function.

  单位元： \(f * \delta(t) = f(t)\)，其中 \(\delta(t)\) 是狄拉克 δ 函数。

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6. Example 1: Finding an Inverse Transform | 例 1：求逆变换

Find \(\mathcal{L}^{-1}\left\{\frac{1}{(s^2 + 1)^2}\right\}\). Notice that \(\frac{1}{(s^2 + 1)^2} = F(s) G(s)\) with \(F(s) = G(s) = \frac{1}{s^2 + 1}\). We know \(\mathcal{L}^{-1}\{\frac{1}{s^2 + 1}\} = \sin t\). Thus

求 \(\mathcal{L}^{-1}\left\{\frac{1}{(s^2 + 1)^2}\right\}\)。注意到 \(\frac{1}{(s^2 + 1)^2} = F(s) G(s)\)，其中 \(F(s) = G(s) = \frac{1}{s^2 + 1}\)。已知 \(\mathcal{L}^{-1}\{\frac{1}{s^2 + 1}\} = \sin t\)，因此

ℒ⁻¹{1/(s² + 1)²} = ∫₀ᵗ sin τ · sin(t − τ) dτ

Using the trigonometric identity \(\sin A \sin B = \frac{1}{2}[\cos(A-B) – \cos(A+B)]\), we get

利用三角恒等式 \(\sin A \sin B = \frac{1}{2}[\cos(A−B) − \cos(A+B)]\)，得到

∫₀ᵗ sin τ sin(t−τ) dτ = ½∫₀ᵗ [cos(2τ−t) − cos t] dτ = ½( sin t − t cos t )

Therefore \(\mathcal{L}^{-1}\left\{\frac{1}{(s^2 + 1)^2}\right\} = \frac{1}{2}(\sin t – t \cos t)\).

因此 \(\mathcal{L}^{-1}\left\{\frac{1}{(s^2 + 1)^2}\right\} = \frac{1}{2}(\sin t − t \cos t)\)。

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7. Example 2: Solving a Differential Equation | 例 2：求解微分方程

Solve the initial value problem \(y” + y = f(t)\), with \(y(0) = 0\), \(y'(0) = 0\). Taking the Laplace transform gives

求解初值问题 \(y” + y = f(t)\)，其中 \(y(0) = 0\)、\(y'(0) = 0\)。对两边取拉普拉斯变换得到

(s² + 1) Y(s) = F(s) ⇒ Y(s) = F(s) · 1/(s² + 1)

Since \(\mathcal{L}^{-1}\{1/(s^2+1)\} = \sin t\), the convolution theorem gives

由于 \(\mathcal{L}^{-1}\{1/(s^2+1)\} = \sin t\)，卷积定理给出

y(t) = ∫₀ᵗ f(τ) sin(t − τ) dτ

This formula expresses the response of an undamped harmonic oscillator to any forcing function \(f(t)\). It is called the Duhamel integral.

这个公式表达了无阻尼谐振子对任意强迫函数 \(f(t)\) 的响应，称为杜哈梅积分。

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8. Convolution with the Dirac Delta | 与狄拉克 δ 函数的卷积

The Dirac delta function \(\delta(t-a)\) satisfies \(\mathcal{L}\{\delta(t-a)\} = e^{-as}\) for \(a \ge 0\). Therefore, by the convolution theorem,

狄拉克 δ 函数 \(\delta(t−a)\) 满足 \(\mathcal{L}\{\delta(t−a)\} = e^{−as}\)（\(a \ge 0\)）。因此，由卷积定理，

ℒ⁻¹{e⁻ᵃˢ F(s)} = ∫₀ᵗ δ(τ − a) f(t − τ) dτ = f(t − a) u(t − a)

This is exactly the second shift theorem. It shows that multiplying by \(e^{-as}\) in the \(s\)-domain corresponds to a time delay of \(a\) in the time domain.

这正是第二位移定理。它表明在 \(s\) 域中乘以 \(e^{−as}\) 对应于时域中的时间延迟 \(a\)。

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9. Application to Integral Equations | 在积分方程中的应用

Consider the Volterra integral equation \(y(t) = t + \int_0^t y(\tau) \sin(t-\tau)\,d\tau\). We can rewrite the integral as \(y * \sin t\). Taking the Laplace transform,

考虑 Volterra 积分方程 \(y(t) = t + \int_0^t y(\tau) \sin(t−\tau)\,d\tau\)。可将积分改写为 \(y * \sin t\)。取拉普拉斯变换得

Y(s) = 1/s² + Y(s) · 1/(s² + 1)

Solving for \(Y(s)\) yields

解出 \(Y(s)\) 得

Y(s) = (s² + 1)/s⁴ = 1/s² + 1/s⁴

The inverse transform gives \(y(t) = t + \frac{t^3}{6}\). This demonstrates how convolution converts an integral equation into an algebraic equation.

逆变换得到 \(y(t) = t + \frac{t^3}{6}\)。这展示了卷积如何将积分方程转化为代数方程。

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10. Connection with Transfer Functions | 与传递函数的联系

In control engineering, a linear time-invariant system with impulse response \(h(t)\) has transfer function \(H(s) = \mathcal{L}\{h(t)\}\). The output \(y(t)\) for an input \(x(t)\) is given by the convolution \(y(t) = x * h(t)\). Taking Laplace transforms, \(Y(s) = X(s) H(s)\).

在控制工程中，具有冲激响应 \(h(t)\) 的线性时不变系统的传递函数为 \(H(s) = \mathcal{L}\{h(t)\}\)。系统对输入 \(x(t)\) 的输出为卷积 \(y(t) = x * h(t)\)。取拉普拉斯变换得 \(Y(s) = X(s) H(s)\)。

This simple algebraic relationship allows engineers to compute system responses using block diagrams and frequency-domain methods, avoiding the direct evaluation of convolution integrals.

这种简单的代数关系使工程师能够利用方块图和频域方法计算系统响应，避免了直接计算卷积积分。

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11. A Table Comparing Time and s-Domain Operations | 时域与 s 域运算对照表

Time domain (t)	s-domain (s)
\(f(t)\)	\(F(s)\)
\(\int_0^t f(\tau)g(t-\tau)\,d\tau\)	\(F(s)G(s)\)
\(f(t-a)u(t-a)\)	\(e^{-as}F(s)\)
\(f'(t)\)	\(sF(s)-f(0)\)
\(\int_0^t f(\tau)\,d\tau\)	\(\frac{1}{s}F(s)\)

The first row in the table is the essence of the convolution theorem. The second row is its special case for a delay, which follows from convolving with a delta function.

表中的第一行正是卷积定理的本质。第二行是延迟情形的特例，它由与 δ 函数卷积得到。

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12. Conclusion and Examination Tips | 总结与考试要点

The convolution theorem provides a bridge between multiplication in the \(s\)-domain and integration in the time domain. It is indispensable for inverting products of transforms, solving differential and integral equations, and modeling system responses.

卷积定理连接了 \(s\) 域中的乘法与时域中的积分。它在求变换乘积的逆变换、求解微分方程和积分方程以及建模系统响应方面不可或缺。

• Always check that both \(f\) and \(g\) are causal (zero for \(t < 0\)).

  始终检查 \(f\) 和 \(g\) 是否都是因果的（即 \(t < 0\) 时取零）。

• When computing convolutions, choose the simpler function to integrate first.

  计算卷积时，选择较简单的函数先积分。

• Recognize common transform pairs, such as \(\mathcal{L}\{e^{at}\} = \frac{1}{s-a}\), to speed up inverse calculations.

  熟记常见变换对，如 \(\mathcal{L}\{e^{at}\} = \frac{1}{s−a}\)，以加快逆变换计算。

• In IB exams, state the theorem explicitly before applying it, and always show the convolution integral for partial credit.

  在 IB 考试中，应用定理前要明确写出定理内容，并且写出卷积积分以获得步骤分。

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