Laplace Transforms of Common Functions: tⁿ, e±ᵗ, sin t, cos t | 常见函数的拉普拉斯变换：tⁿ、e±ᵗ、sin t、cos t

📚 Laplace Transforms of Common Functions: tⁿ, e±ᵗ, sin t, cos t | 常见函数的拉普拉斯变换：tⁿ、e±ᵗ、sin t、cos t

In this article, we will derive and understand the Laplace transforms of the most common functions encountered in IB Mathematics HL and engineering mathematics: powers tⁿ, exponentials e±ᵗ, and trigonometric functions sin t and cos t. We will focus on the definitions, standard results, and worked examples.

本文将推导并理解 IB 数学 HL 与工程数学中最常见函数的拉普拉斯变换：幂函数 tⁿ、指数函数 e±ᵗ、三角函数 sin t 和 cos t。我们将重点讲解定义、标准结果以及典型例题。

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1. Definition of the Laplace Transform | 拉普拉斯变换的定义

The Laplace transform of a function f(t), defined for t ≥ 0, is given by the integral

L{f(t)} = F(s) = ∫₀^∞ e^(−st) f(t) dt

where s is a complex variable, and the integral converges for values of s in a certain region. The original function f(t) is called the inverse Laplace transform of F(s), written as f(t) = L⁻¹{F(s)}.

拉普拉斯变换将时间域中的函数 f(t) 转换为复频率域中的函数 F(s)。对于 t ≥ 0 的函数 f(t)，其拉普拉斯变换定义为上述积分，其中 s 为复变量，积分在 s 的某个收敛区域内存在。原函数 f(t) 称为 F(s) 的逆拉普拉斯变换。

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2. Transform of the Constant Function f(t) = 1 | 常数函数 f(t) = 1 的变换

Although not explicitly listed in the title, the transform of a constant is essential. Set f(t) = 1, then

L{1} = ∫₀^∞ e^(−st) dt = [−e^(−st)/s]₀^∞ = 1/s, for s > 0

This result is used repeatedly when finding inverse transforms and solving differential equations.

虽然标题没有明确列出常数函数，但常数的变换是基础。设 f(t)=1，则计算结果如上。当 s>0 时积分收敛，结果为 1/s。这个结果在求逆变换和解微分方程时会反复使用。

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3. Transform of f(t) = e^(at) | 指数函数 e^(at) 的变换

Consider the exponential function f(t) = e^(at), where a is a real constant. By definition,

L{e^(at)} = ∫₀^∞ e^(−st) e^(at) dt = ∫₀^∞ e^(−(s−a)t) dt

This integral converges provided s > a, giving

L{e^(at)} = 1/(s − a)

For a = 1, we get L{e^t} = 1/(s − 1); for a = −1, we get L{e^(−t)} = 1/(s + 1).

考虑指数函数 f(t)=e^(at)，其中 a 为实常数。根据定义，积分简化为 ∫₀^∞ e^(−(s−a)t) dt。当且仅当 s>a 时积分收敛，结果为 1/(s−a)。特别地，当 a=1 时 L{e^t}=1/(s−1)；当 a=−1 时 L{e^(−t)}=1/(s+1)。

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4. Transform of sin t and cos t | 正弦函数 sin t 与余弦函数 cos t 的变换

Using Euler’s formula, e^(it) = cos t + i sin t, we can derive the transforms of sine and cosine elegantly. First,

L{e^(it)} = 1/(s − i) = (s + i)/(s² + 1)

Separating real and imaginary parts:

L{cos t} = s/(s² + 1), L{sin t} = 1/(s² + 1)

Alternatively, one may integrate by parts twice. These results are valid for s > 0.

利用欧拉公式 e^(it)=cos t+i sin t，可以优雅地推导出正弦和余弦的变换。首先计算 L{e^(it)}=1/(s−i)，有理化分母后得到 (s+i)/(s²+1)。分离实部和虚部，即可得到 L{cos t}=s/(s²+1) 和 L{sin t}=1/(s²+1)。换一种方式，也可以通过两次分部积分得到相同的结果。这些结果在 s>0 时成立。

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5. Transform of f(t) = t | 幂函数 t 的变换

The simplest power function is f(t) = t. Using integration by parts,

L{t} = ∫₀^∞ t e^(−st) dt = [−t e^(−st)/s]₀^∞ + (1/s)∫₀^∞ e^(−st) dt

The boundary term vanishes at infinity for s > 0, and at t = 0 it is zero. Therefore,

L{t} = 1/s × 1/s = 1/s²

So L{t} = 1/s² for s > 0.

最简单的幂函数是 f(t)=t。使用分部积分：边界项在无穷远处为 0（当 s>0 时），在 t=0 处也为 0。剩下的积分恰好是 L{1}=1/s，因此 L{t}=1/s²。这个结果很重要，因为它是推导 tⁿ 变换的基础。

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6. General Formula for L{tⁿ} | tⁿ 的通用公式

Applying integration by parts recursively to f(t) = tⁿ, the structure emerges:

L{tⁿ} = n! / s^(n+1), s > 0

Set n = 1 gives 1!/s² = 1/s². Set n = 2: L{t²} = 2!/s³ = 2/s³. Set n = 3: L{t³} = 3!/s⁴ = 6/s⁴. Notice the factorial pattern.

对 f(t)=tⁿ 反复使用分部积分，可以发现规律。通用公式为 L{tⁿ}=n!/s^(n+1)，其中 s>0。验证一下：n=1 时得到 1!/s²=1/s²；n=2 时得到 2!/s³=2/s³；n=3 时得到 6/s⁴。可见阶乘模式非常清晰。

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7. Linearity of the Laplace Transform | 拉普拉斯变换的线性性质

The Laplace transform is a linear operator. If L{f(t)} = F(s) and L{g(t)} = G(s), then for any constants a and b,

L{a f(t) + b g(t)} = a F(s) + b G(s)

This property allows us to combine the transforms derived above without redoing integration.

拉普拉斯变换是线性算子。若 L{f(t)}=F(s)，L{g(t)}=G(s)，则对于任意常数 a、b，有 L{a f(t)+b g(t)}=aF(s)+bG(s)。这个性质使我们无需重新积分即可组合已经推导出的变换结果。

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8. Worked Example 1: Linear Combination | 例 1：线性组合

Find L{3t² − 2e^(−t) + 4 sin t}.

L{3t²} = 3 × (2!/s³) = 6/s³

L{−2e^(−t)} = −2 × (1/(s+1))

L{4 sin t} = 4 × (1/(s²+1))

Combining the three results, the final expression is

F(s) = 6/s³ − 2/(s+1) + 4/(s²+1)

求 L{3t²−2e^(−t)+4 sin t}。分别计算每一项：L{3t²}=6/s³，L{−2e^(−t)}=−2/(s+1)，L{4 sin t}=4/(s²+1)。根据线性性质相加，得到最终结果如上。

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

Solve y” − y = e^(−t), with y(0) = 0 and y'(0) = 1.

Let L{y(t)} = Y(s). Then L{y’} = sY(s) − y(0) = sY(s), and L{y”} = s²Y(s) − s y(0) − y'(0) = s²Y(s) − 1. Also L{e^(−t)} = 1/(s+1).

Substitute into the equation:

(s²Y − 1) − Y = 1/(s+1) ⇒ (s² − 1)Y = 1 + 1/(s+1)

Recognize s²−1 = (s−1)(s+1), and 1 + 1/(s+1) = (s+2)/(s+1). Thus

Y(s) = (s+2)/[(s−1)(s+1)²]

Using partial fractions, one can find the inverse transform, but the main point here is to demonstrate how the standard Laplace transforms of e^(−t), t, and constants are applied.

设 L{y(t)}=Y(s)。利用微分性质，L{y’}=sY(s)−y(0)=sY(s)，L{y”}=s²Y(s)−s y(0)−y'(0)=s²Y(s)−1。代入微分方程，整理得到代数方程。因式分解并化简，得到 Y(s) 的表达式。通过部分分式可以进一步求逆变换，但这里的主要目的是展示如何使用 e^(−t) 等标准变换来求解微分方程。

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10. Summary Table | 常用拉普拉斯变换表

f(t)	F(s) = L{f(t)}
1	1/s
t	1/s²
tⁿ (n positive integer)	n! / s^(n+1)
e^(at)	1/(s−a)
e^(−t)	1/(s+1)
sin t	1/(s²+1)
cos t	s/(s²+1)

The table above summarizes all standard transforms discussed in this article. They form the building blocks for solving linear differential equations and for analyzing control systems.

上表总结了本文讨论的所有标准变换。它们是求解线性微分方程和分析控制系统的基础构件。

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11. Common Mistakes and Tips | 常见错误与学习建议

Students often make the following mistakes:

学生常犯以下错误：

• Forgetting the initial conditions when applying L{y’} = sY(s) − y(0). Always write them explicitly.

  应用 L{y’}=sY(s)−y(0) 时忘记初值条件。务必显式写出 y(0) 和 y'(0)。

• Using L{e^(−t)} = 1/(s−1) with wrong sign. Remember that e^(−t) gives 1/(s+1).

  误将 L{e^(−t)} 写作 1/(s−1)。请记住 e^(−t) 对应的分母为 s+1。

• Confusing the transforms of sin t and cos t. The sine transform has numerator 1, while the cosine transform has numerator s.

  混淆 sin t 与 cos t 的变换。正弦变换的分子为 1，而余弦变换的分子为 s。

• Forgetting the domain condition s > a. The Laplace transform is only defined when the integral converges.

  忘记收敛条件 s>a。只有当积分收敛时拉普拉斯变换才有意义。

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12. Conclusion | 总结

This article has derived and summarized the Laplace transforms of the most common functions: tⁿ, e^(at), e^(−t), sin t and cos t. Together with the linearity property, these transforms provide a powerful tool for solving differential equations and for engineering analysis.

本文推导并总结了最常见函数的拉普拉斯变换：tⁿ、e^(at)、e^(−t)、sin t 和 cos t。结合线性性质，这些变换为求解微分方程和工程分析提供了强有力的工具。

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