📚 Inverting a Laplace Transform | 拉普拉斯逆变换
The inverse Laplace transform is a fundamental tool in applied mathematics, especially for solving differential equations and analysing linear systems. Given a function F(s) in the complex frequency domain, the inverse transform recovers the original time-domain function f(t). In the IB Mathematics HL curriculum (particularly within the calculus option), students learn how to invert Laplace transforms using tables, algebraic manipulation, and key theorems. This article provides a comprehensive guide to inverting a Laplace transform, covering essential techniques and common pitfalls.
逆拉普拉斯变换是应用数学中的一个基本工具，尤其在求解微分方程和分析线性系统时发挥着重要作用。给定复频域中的函数 F(s)，逆变换可以恢复原始的时域函数 f(t)。在 IB 数学高等级课程（尤其是微积分选修部分）中，学生需要学习如何利用变换表、代数操作和关键定理来求拉普拉斯逆变换。本文将全面介绍拉普拉斯逆变换的各种方法，涵盖基本技巧和常见错误。
1. Definition and Notation | 定义与符号
ally, it is given by the Bromwich integral, but in IB, we rely on the property that the Laplace transform is one-to-one for piecewise continuous functions of exponential order. Therefore, we can recover f(t) from a table of known transforms or by algebraic manipulation.
函数 F(s) 的拉普拉斯逆变换定义为使 L{f(t)} = F(s) 的函数 f(t)，其中 L 表示拉普拉斯变换算子。我们记作 f(t) = L⁻¹{F(s)}。形式上，逆变换由 Bromwich 积分给出，但在 IB 课程中，我们利用拉普拉斯变换对分段连续且具有指数阶的函数是单射的性质。因此，可以通过已知的变换表或代数操作来还原 f(t)。
is property is used repeatedly to break down complicated expressions into manageable pieces. Always verify that the function F(s) is a proper rational function before attempting inversion; if not, perform polynomial division first.
必须牢记逆拉普拉斯变换是线性的：L⁻¹{a F(s) + b G(s)} = a f(t) + b g(t)。这个性质被反复用来将复杂表达式拆分为可管理的部分。在尝试求逆之前，务必检查 F(s) 是否为有理真分式；若不是，请先进行多项式除法。
2. Linearity Property in Action | 线性性质的应用
os(2t). The constant coefficients are preserved, and the result is f(t) = 3eᵗ − 2 cos(2t).
逆拉普拉斯变换的线性性质允许我们逐项求逆。例如，若 F(s) = 3/(s−1) − 2s/(s²+4)，我们可以分别将 3/(s−1) 逆变换为 3eᵗ，将 2s/(s²+4) 逆变换为 2 cos(2t)。常数系数保持不变，得到 f(t) = 3eᵗ − 2 cos(2t)。
This property becomes even more powerful when combined with partial fraction decomposition. A rational function that looks daunting can be split into a sum of simple fractions, each matching a standard entry in the table. Always write the expression as a sum of terms before looking up the inverse.
当与部分分式分解结合时，这一性质更为强大。看起来很棘手的理性函数可以被拆分为简单分式之和，每一项都与表中的标准条目匹配。在查表求逆之前，务必先将表达式写为各项之和。
3. Using the Table of Transforms | 使用变换表
A crucial skill is to recognise standard forms and match them with entries in the Laplace transform table. The table below lists some fundamental transform pairs that appear frequently in IB problems.
识别标准形式并与拉普拉斯变换表中的条目进行匹配是一项关键技能。下表列出了一些在 IB 问题中经常出现的基本变换对。
| f(t) | F(s) = L{f(t)} |
|---|---|
| 1 | 1/s |
| tⁿ (n = 0,1,2,...) | n! / sⁿ⁺¹ |
| eᵃᵗ | 1/(s − a) |
| sin(kt) | k/(s² + k²) |
| cos(kt) | s/(s² + k²) |
| sinh(kt) | k/(s² − k²) |
| cosh(kt) | s/(s² − k²) |
| tⁿ eᵃᵗ | n!/(s − a)ⁿ⁺¹ |
not yet in the exact form k/(s²+k²); rewrite it as (5/2) × 2/(s²+2²) to get (5/2) sin(2t).
当你遇到像 1/(s−2) 这样的有理函数时，可以直接读出其逆变换为 e²ᵗ。对于如 3s/(s²+9) 的项，可看出它是 cos(3t) 变换的 3 倍，从而得到 3 cos(3t)。类似地，5/(s²+4) 并非直接的 k/(s²+k²) 形式；将其改写为 (5/2) × 2/(s²+2²)，即得 (5/2) sin(2t)。
4. Partial Fraction Decomposition | 部分分式分解
When F(s) is a rational function whose denominator can be factorised, the method of partial fractions is the primary algebraic tool. The goal is to express F(s) as a sum of simpler fractions, each of which corresponds to a known inverse (often exponential or sine/cosine). The general procedure is: (1) ensure the degree of the numerator is less than the denominator; if not, perform polynomial long division. (2) Completely factor the denominator into linear and irreducible quadratic factors. (3) Write the decomposition with unknown constants. (4) Multiply through by the common denominator and equate coefficients to solve for the constants. (5) Invert term by term using the table.
当 F(s) 是分母可分解的理性函数时，部分分式法是主要的代数工具。目的是将 F(s) 表示为简单分式之和，每一项都对应已知的逆变换（通常为指数函数或正弦/余弦）。一般步骤为：（1）确保分子的次数低于分母；若不是，先进行多项式长除法。（2）将分母完全分解为一次和不可约二次因式。（3）写出含待定常数的分解式。（4）两边同乘公分母并比较系数，解出常数。（5）利用变换表逐项求逆。
3, A = −1/3. Thus F(s) = −1/3·1/(s+2) + 10/3·1/(s−1). Inverting gives f(t) = −1/3 e⁻²ᵗ + 10/3 eᵗ.
考虑例子：求 L⁻¹{(3s+7)/(s²+s−2)}。首先分解分母：s²+s−2 = (s+2)(s−1)。设 (3s+7)/((s+2)(s−1)) = A/(s+2) + B/(s−1)。两边同乘 (s+2)(s−1) 得 3s+7 = A(s−1) + B(s+2)。比较系数：s 项，A + B = 3；常数项：−A + 2B = 7。解得 B = 10/3，A = −1/3。因此 F(s) = −1/3·1/(s+2) + 10/3·1/(s−1)，逆变换得 f(t) = −1/3 e⁻²ᵗ + 10/3 eᵗ。
quadratic factors like s²+2s+5 are handled by completing the square, which is discussed next.
当分母包含重一次因式（如 (s+1)²）时，部分分式必须包含 C/(s+1) + D/(s+1)² 的形式。D/(s+1)² 的逆变换为 D t e⁻ᵗ，利用 tⁿ eᵃᵗ 条目并取 n=1。对于像 s²+2s+5 这类不可约二次因式，则采用下一节介绍的配方法处理。
5. Completing the Square | 配方
When F(s) involves a quadratic denominator that cannot be factorised into real linear factors, we complete the square to rewrite it in the form (s − a)² + b² or (s − a)² − b². This allows us to use the shifting theorem alongside standard sine, cosine, sinh or cosh forms.
当 F(s) 含有无法分解为实一次因式的二次分母时，我们通过配方将其改写为 (s − a)² + b² 或 (s − a)² − b² 的形式。这样便可结合移位定理和标准的正弦、余弦、双曲正弦或双曲余弦形式进行逆变换。
ed to ½ · 2/((s+1)²+2²), yielding ½ e⁻ᵗ sin(2t). Hence f(t) = e⁻ᵗ cos(2t) + ½ e⁻ᵗ sin(2t).
例如，求 F(s) = (s+2)/(s²+2s+5) 的逆变换。分母 s²+2s+5 = (s+1)² + 2²。将分子写为 s+2 = (s+1) + 1。因此 F(s) = (s+1)/((s+1)²+4) + 1/((s+1)²+4)。第一项是 s/(s²+k²) 的形式并带有平移 a = −1，得到 e⁻ᵗ cos(2t)。第二项需调整为 ½ · 2/((s+1)²+2²)，即得 ½ e⁻ᵗ sin(2t)。故 f(t) = e⁻ᵗ cos(2t) + ½ e⁻ᵗ sin(2t)。
ommon source of error. For denominators of the form (s+α)² − β², you obtain hyperbolic functions because the transform table uses the difference of squares to produce sinh and cosh.
务必仔细核对平移参数：(s+1) 中的正号表示 a = −1。读错符号是常见的错误来源。对于形如 (s+α)² − β² 的分母，由于变换表利用平方差产生 sinh 和 cosh，将得到双曲函数。
6. First Shifting Theorem (s-shifting) | 第一移位定理（s-移位）
ing s by s − a in the Laplace domain corresponds to multiplying the time function by eᵃᵗ. This theorem justifies the patterns seen in entries like 1/(s−a)ⁿ and s/((s−a)² + k²).
第一移位定理指出：若 L⁻¹{F(s)} = f(t)，则 L⁻¹{F(s − a)} = eᵃᵗ f(t)。换言之，在拉普拉斯域中用 s − a 替换 s，相当于在时域中乘以 eᵃᵗ。该定理解释了诸如 1/(s−a)ⁿ 和 s/((s−a)² + k²) 等条目的规律。
s an illustration, find L⁻¹{2/(s−3)³}. We recognise that 2/(s)³ corresponds to t² (since 2! y, for any rational expression already in the form F(s−a), simply invert F(s) and then multiply the result by eᵃᵗ.
举例说明，求 L⁻¹{2/(s−3)³}。我们知道 2/s³ 对应 t²（因为 2!/s³ = 2/s³ 给出 t²）。平移 a = 3，得到 f(t) = e³ᵗ t²。更一般地，对于任何已形如 F(s−a) 的理性表达式，只需先对 F(s) 求逆，再将结果乘以 eᵃᵗ。
This theorem often works hand-in-hand with completing the square. After rewriting the denominator, you can identify the shift a immediately from the term (s − a)². Then proceed to invert the unshifted form, using sine or cosine as appropriate, and attach the exponential factor.
该定理常与配方法协同使用。在改写分母
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