Steady Forced Oscillations: Phasors, Impedance, Transfer Functions | 稳态受迫振动：相量、阻抗与传递函数

📚 Steady Forced Oscillations: Phasors, Impedance, Transfer Functions | 稳态受迫振动：相量、阻抗与传递函数

When a damped oscillating system is subjected to a periodic external force, the long-term behaviour settles into a steady forced oscillation at the frequency of the driving force. The mathematics of this phenomenon draws on complex numbers, differential equations, and algebraic techniques that simplify the analysis enormously. In this article we explore the concepts of phasors, mechanical and electrical impedance, and transfer functions, showing how they transform differential equations into simple algebraic relations.

当一个有阻尼的振动系统受到周期性的外力作用时，其长时间行为最终会稳定在与驱动力同频率的稳态受迫振动。这一现象的数学分析利用了复数、微分方程以及代数方法，能够极大地简化计算。本文探讨相量、机械阻抗与电阻抗以及传递函数的概念，展示如何将微分方程转化为简单的代数关系。

---

1. Introduction to Forced Oscillations | 受迫振动简介

Many physical systems, from a child on a swing to an electrical circuit containing a resistor, inductor, and capacitor, can be modelled as damped harmonic oscillators. When such a system is driven by a sinusoidal external source, the homogeneous solution (the transient part) decays exponentially due to damping, leaving only the particular integral – the steady-state response. This steady response oscillates at the same angular frequency ω as the drive, but with a shift in amplitude and phase relative to the input.

许多物理系统，从荡秋千的孩子到含有电阻、电感和电容的电路，都可以建模为有阻尼的简谐振子。当这样的系统受到正弦外部激励时，齐次解（暂态部分）因阻尼而指数衰减，最后只剩下特解——稳态响应。这个稳态响应以与驱动相同的角频率 ω 振动，但振幅和相位相对于输入发生了改变。

---

2. Governing Differential Equation | 控制微分方程

A generic second-order linear system with viscous damping and a sinusoidal forcing term can be written as

m d²x/dt² + c dx/dt + kx = F₀ cos(ωt)

where m is the mass (or inductance), c the damping coefficient (or resistance), k the stiffness (or reciprocal capacitance), and F₀ the amplitude of the applied force (or voltage). For mathematical convenience we often replace the cosine driving term with the complex exponential F₀ e^iωt, taking the real part at the end, because phasor algebra handles exponentials far more easily than trigonometric functions.

一个具有黏性阻尼和正弦强迫项的通用二阶线性系统可以写成

m d²x/dt² + c dx/dt + kx = F₀ cos(ωt)

其中 m 是质量（或电感），c 是阻尼系数（或电阻），k 是刚度（或电容的倒数），F₀ 是外加力（或电压）的幅值。为数学处理方便，我们常把余弦驱动项替换为复指数 F₀ e^iωt，最后取实部，因为相量代数处理指数远比三角函数容易。

---

3. Steady-State Solution and Undetermined Coefficients | 稳态解与待定系数法

The traditional approach seeks a particular solution of the form x_p(t) = A cos(ωt) + B sin(ωt). Substituting into the differential equation and equating coefficients yields two linear equations for A and B. Solving them gives expressions for the amplitude and phase. Though systematic, this method becomes algebraically heavy for anything beyond the simplest cases.

传统方法寻求形如 x_p(t) = A cos(ωt) + B sin(ωt) 的特解。代入微分方程并比较系数，可得到关于 A 和 B 的两个线性方程。解出它们即可得到振幅和相位的表达式。尽管系统化，但对于稍微复杂的系统，这种方法在代数上会变得很繁琐。

---

4. Complex Exponential Forcing and the Birth of Phasors | 复指数强迫与相量的产生

Instead of working with sine and cosine directly, we write the forcing as F(t) = F₀ e^iωt and guess a solution x(t) = X e^iωt, where X is a complex constant (the phasor) that contains both the amplitude and the phase information of the steady-state displacement. The real physical displacement is then Re{X e^iωt}.

我们不再直接处理正弦和余弦，而是将强迫项写成 F(t) = F₀ e^iωt，并假设解的形式为 x(t) = X e^iωt，其中 X 是一个复常数（相量），集中了稳态位移的振幅和相位信息。真实的物理位移就是 Re{X e^iωt}。

---

5. Phasor Representation of Oscillations | 振荡的相量表示

A phasor is a complex number that represents a sinusoidally varying quantity at a fixed frequency. For a displacement x(t) = X₀ cos(ωt + φ), the corresponding phasor is X = X₀ e^iφ. The benefit is profound: differentiation with respect to time becomes multiplication by iω: if x = X e^iωt, then dx/dt = iω X e^iωt and d²x/dt² = (iω)² X e^iωt = –ω² X e^iωt. Thus time-domain calculus turns into frequency-domain algebra.

相量是一个复数，表示在固定频率下作正弦变化的量。对于位移 x(t) = X₀ cos(ωt + φ)，对应的相量是 X = X₀ e^iφ。其好处极大：对时间的求导变成了乘以 iω：若 x = X e^iωt，则 dx/dt = iω X e^iωt，d²x/dt² = (iω)² X e^iωt = –ω² X e^iωt。于是时域微积分转变为频域代数。

---

6. Transforming the Differential Equation to an Algebraic One | 将微分方程转化为代数方程

Substituting the phasor forms into the equation of motion m d²x/dt² + c dx/dt + kx = F₀ e^iωt gives

(–mω² + i cω + k) X = F₀

which is immediately solvable for the phasor X. The whole fourth-order differential equation becomes a simple complex algebraic expression. This is the power of the phasor method – it eliminates differential operators entirely.

将相量形式代入运动方程 m d²x/dt² + c dx/dt + kx = F₀ e^iωt，得到

(–mω² + i cω + k) X = F₀

从中可立即解出相量 X。原本四阶的微分方程（这里实际是二阶，上面写了四阶，笔误，应表述为“二阶微分方程”）转化为了一个简单的复数代数表达式。这就是相量方法的威力——它完全消去了微分算子。

---

7. Definition of Impedance | 阻抗的定义

In direct analogy with electrical engineering, we define the impedance Z of a mechanical system as the complex ratio of the driving force phasor to the velocity phasor. Since v = dx/dt = iω X, we have

Z = F / v = F / (iω X) = (k – mω² + i cω) / (iω) = c + i(mω – k/ω)

For an electrical RLC series circuit, the impedance takes the form Z = R + i(ωL – 1/(ωC)). Impedance encodes both the amplitude reduction and the phase shift introduced by the system at a given frequency.

与电气工程完全类似，我们将机械系统的阻抗 Z 定义为驱动力相量与速度相量的复数比值。由于 v = dx/dt = iω X，我们有

Z = F / v = F / (iω X) = (k – mω² + i cω) / (iω) = c + i(mω – k/ω)

对于 RLC 串联电路，阻抗的形式为 Z = R + i(ωL – 1/(ωC))。阻抗同时编码了系统在给定频率下引入的振幅衰减和相位移。

---

8. Transfer Function: Output-over-Input Ratio | 传递函数：输出与输入之比

A more general concept is the transfer function H(iω), defined as the ratio of the output phasor to the input phasor. For the displacement response it is

H(iω) = X / F₀ = 1 / (k – mω² + i cω)

If we take velocity as the output, the transfer function becomes H_v(iω) = iω / (k – mω² + i cω) = 1/Z. Transfer functions are complex-valued and depend only on the system parameters and frequency; they fully characterise the steady-state behaviour.

更一般的概念是传递函数 H(iω)，定义为输出相量与输入相量之比。对于位移响应而言

H(iω) = X / F₀ = 1 / (k – mω² + i cω)

若取速度作为输出，则传递函数变为 H_v(iω) = iω / (k – mω² + i cω) = 1/Z。传递函数是复值的，仅依赖于系统参数和频率；它们完全刻画了稳态行为。

---

9. Amplitude and Phase Response | 幅频与相频响应

From the transfer function we extract a real amplitude gain |H(iω)| and a phase angle ∠H(iω). The steady-state output amplitude is the product of the input amplitude and the gain, while the output phase is the sum of the input phase and the phase shift introduced by the system. Explicitly, for displacement:

|X| = F₀ / √((k – mω²)² + (cω)²), φ = –arctan( cω/(k – mω²) )

These formulas reveal how the response grows near resonance and how the phase lag approaches –π/2 as the driving frequency passes through the natural frequency.

从传递函数我们可以提取实数的幅值增益 |H(iω)| 和相位角 ∠H(iω)。稳态输出振幅等于输入振幅与增益的乘积，而输出相位是输入相位与系统引入的相移之和。对于位移，具体有：

|X| = F₀ / √((k – mω²)² + (cω)²), φ = –arctan( cω/(k – mω²) )

这些公式揭示了响应在共振附近如何增大，以及当驱动频率经过固有频率时相位滞后如何趋近 –π/2。

---

10. Resonance and Quality Factor | 共振与品质因数

Resonance occurs when the driving frequency ω maximises the amplitude of the steady-state response. For the displacement amplitude, the resonant frequency is ω_r = √(k/m – c²/(2m²)) = √(ω₀² – (c²/(2m²))), slightly less than the natural frequency ω₀ = √(k/m). The sharpness of the resonance is measured by the quality factor Q = ω₀ m / c. A high Q implies a tall, narrow resonance peak.

当驱动频率 ω 使稳态响应的振幅达到最大时，便发生共振。对于位移振幅，共振频率为 ω_r = √(k/m – c²/(2m²)) = √(ω₀² – (c²/(2m²)))，略小于固有频率 ω₀ = √(k/m)。共振的尖锐程度由品质因数 Q = ω₀ m / c 衡量。高 Q 值意味着又高又窄的共振峰。

---

11. Worked Example 1: Mass–Spring–Damper System | 实例一：质量–弹簧–阻尼系统

A machine of mass m = 2 kg is mounted on a spring of stiffness k = 200 N/m and a damper with coefficient c = 8 N·s/m. It is driven by a force F(t) = 10 cos(ωt) N. Determine the steady-state displacement phasor and the amplitude when ω = 8 rad/s.

Substituting into the phasor equation: (–2·8² + i·8·8 + 200) X = 10, i.e. (–128 + i64 + 200) X = 10, so (72 + i64) X = 10. Thus X = 10 / (72 + i64). The denominator magnitude is √(72² + 64²) = √(5184 + 4096) = √9280 ≈ 96.33. Hence |X| ≈ 10/96.33 ≈ 0.1038 m. The phase is φ = –arctan(64/72) ≈ –0.726 rad. At 8 rad/s the oscillation amplitude is about 10.4 cm.

一机器质量 m = 2 kg，安装在刚度 k = 200 N/m 的弹簧和阻尼系数为 c = 8 N·s/m 的阻尼器上，受驱动力 F(t) = 10 cos(ωt) N 作用。求当 ω = 8 rad/s 时的稳态位移相量和振幅。

代入相量方程：(–2·8² + i·8·8 + 200) X = 10，即 (–128 + i64 + 200) X = 10，所以 (72 + i64) X = 10。于是 X = 10 / (72 + i64)。分母的模为 √(72² + 64²) = √(5184 + 4096) = √9280 ≈ 96.33。因此 |X| ≈ 10/96.33 ≈ 0.1038 m。相位为 φ = –arctan(64/72) ≈ –0.726 rad。在 8 rad/s 时振动幅度约为 10.4 cm。

---

12. Worked Example 2: Electrical RLC Circuit | 实例二：RLC电路

Consider a series RLC circuit driven by an AC voltage v_in(t) = V₀ cos(ωt). The differential equation for the charge q on the capacitor is L d²q/dt² + R dq/dt + (1/C) q = V₀ cos(ωt). Using phasors with v_in = V₀ e^iωt and q = Q e^iωt, we obtain (–Lω² + i Rω + 1/C) Q = V₀. The transfer function for the charge is Q/V₀ = 1/(1/C – Lω² + i Rω). The voltage across the capacitor is V_C = Q/C, giving |V_C| = V₀ / √((1 – LC ω²)² + (RC ω)²). This expression exhibits a resonance peak when ω ≈ 1/√(LC), the electrical natural frequency.

考虑一个由交流电压 v_in(t) = V₀ cos(ωt) 驱动的串联 RLC 电路。电容器上电荷 q 的微分方程为 L d²q/dt² + R dq/dt + (1/C) q = V₀ cos(ωt)。采用相量 v_in = V₀ e^iωt、q = Q e^iωt，我们得到 (–Lω² + i Rω + 1/C) Q = V₀。电荷的传递函数为 Q/V₀ = 1/(1/C – Lω² + i Rω)。电容两端的电压是 V_C = Q/C，故 |V_C| = V₀ / √((1 – LC ω²)² + (RC ω)²)。当 ω ≈ 1/√(LC)（电学固有频率）时，该表达式呈现共振峰。

---

13. Comparison and Summary | 比较与总结

The beauty of the phasor–impedance–transfer-function framework is its universality. Whether the system is a mechanical spring–mass–damper, an acoustic resonator, or an electrical circuit, the mathematics is structurally identical. A table summarises the analogies:

Quantity	Mechanical	Electrical
Inertia	m	L
Damping	c	R
Stiffness/elastance	k	1/C
Driving	F	V
Response	x or v	q or i
Impedance Z	c + i(mω – k/ω)	R + i(ωL – 1/(ωC))

By mastering phasors, a learner can sidestep cumbersome differential-equation solving and design systems with desired frequency responses. The concepts of impedance and transfer functions are fundamental not only in physics and engineering but also in any IB Mathematics Higher Level exploration involving mathematical modelling of real-world oscillatory phenomena.

相量–阻抗–传递函数框架的美妙之处在于其普遍性。无论系统是机械弹簧–质量–阻尼器、声学谐振器还是电路，其数学结构完全相同。下表总结了这些类比。

Published by TutorHao | IB Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

物理量	机械系统	电学系统
惯性	m	L
阻尼	c	R
刚度/倒电容	k	1/C
驱动量	F	V
响应	x 或 v