Case Study in Cambridge Further Mathematics Year 12: Practical Application Walkthrough | Year 12 剑桥进阶数学：案例分析实战演练

📚 Case Study in Cambridge Further Mathematics Year 12: Practical Application Walkthrough | Year 12 剑桥进阶数学：案例分析实战演练

Case studies bridge the gap between abstract mathematical theory and real-world engineering problems. In AS-Level Further Mathematics, topics such as complex numbers, matrices, and differential equations can be woven into a single, integrated investigation. This walkthrough uses an AC circuit network to demonstrate phasor analysis with complex impedances, the solution of linear systems via matrices, and the differential equation approach to transient behaviour. By working through the entire process, you will strengthen your ability to tackle multi-step, cross-topic exam questions confidently.

案例研究在抽象数学理论与真实工程问题之间架起了桥梁。在 AS-Level 进阶数学中，复数、矩阵和微分方程等主题可以整合到一个完整的探究中。本次实战演练通过一个交流电路网络，展示如何利用复数阻抗进行相量分析、用矩阵求解线性方程组，以及借助微分方程描述瞬态行为。从头到尾走完这个过程，你将提升应对多步骤、跨主题考试题目的信心。

1. Understanding the Case: An RLC Network | 案例理解：一个 RLC 网络

We consider a two-mesh AC circuit. The voltage source is v_s(t) = 10 cos(2t) V clockwise direction. The shared resistor R₃ couples the two loops. Our objectives are to determine the sinusoidal steady-state currents and the capacitor voltage using phasor methods, then to analyse the natural response after the source is suddenly removed at t = 0.

我们考虑一个双网孔交流电路。电压源为 v_s(t) = 10 cos(2t) V，因此角频率 ω = 2 rad/s。各元件为：R₁ = 4 Ω，R₂ = 2 Ω，R₃ = 6 Ω，L = 1 H，C = 0.25 F。网孔电流 i₁(t) 和 i₂(t) 按顺时针方向定义。公共电阻 R₃ 将两个回路耦合在一起。我们的目标是先用相量法求出正弦稳态电流和电容电压，然后分析在 t = 0 时刻突然移除电压源后的自然响应。

2. Complex Impedance and Phasor Representation | 复数阻抗与相量表示

In the frequency domain, sinusoidal voltages and currents are represented by phasors. The source becomes V_s = 10∠0°. The impedance of the inductor is Z_L = iωL = i·2·1 = 2i Ω. The impedance of the capacitor is Z_C d. We denote the phasor mesh currents as I₁ and I₂. All analysis now takes place in terms of these complex quantities.

在频域中，正弦电压和电流用相量表示。电源变为 V_s = 10∠0°。电感的阻抗为 Z_L = iωL = i·2·1 = 2i Ω。电容的阻抗为 Z_C = 1/(iωC) = 1/(i·2·0.25) = 1/(i·0.5) = -2i Ω。电阻保持不变。我们将相量网孔电流记为 I₁ 和 I₂。此刻起，所有分析都基于这些复数量。

3. Mesh Current Equations in the Frequency Domain | 频域网孔电流方程

Applying Kirchhoff's voltage law to mesh 1 gives: (R₁ + Z_L + R₃)I₁ - R₃ I₂ = V_s )I₁ - 6 I₂ = 10, which simplifies to (10 + 2i)I₁ - 6 I₂ = 10. For mesh 2: -R₃ I₁ + (R₂ + Z_C + R₃)I₂ = 0, i.e., -6 I₁ + (2 - 2i + 6)I₂ = 0, giving -6 I₁ + (8 - 2i)I₂ = 0.

对网孔 1 应用基尔霍夫电压定律得：(R₁ + Z_L + R₃)I₁ - R₃ I₂ = V_s。代入数值：(4 + 2i + 6)I₁ - 6 I₂ = 10，简化为 (10 + 2i)I₁ - 6 I₂ = 10。网孔 2：-R₃ I₁ + (R₂ + Z_C + R₃)I₂ = 0，即 -6 I₁ + (2 - 2i + 6)I₂ = 0，得 -6 I₁ + (8 - 2i)I₂ = 0。

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4. Matrix Formulation of Linear System | 线性方程组的矩阵表示

The two equations can be written in matrix form Z I = V:

[ 10 + 2i -6 ] [ I₁ ] [ 10 ]
[ -6 8 - 2i ] [ I₂ ] = [ 0 ]

两个方程可写成矩阵形式 Z I = V。

This is a 2 x 2 complex matrix equation. Because the determinant is non‑zero, a unique solution for I₁ and I₂ exists. The next step is to solve for the phasor currents using Gaussian elimination or by computing the inverse matrix.

这是一个 2×2 的复数矩阵方程。由于行列式非零，I₁ 和 I₂ 存在唯一解。下一步我们将用高斯消元法或通过求逆矩阵来解出相量电流。

5. Solving the Complex Matrix Equation | 求解复数矩阵方程

icients, but it is simpler to compute the determinant Δ = (10 + 2i)(8 - 2i) - (-6)(-6). Expanding: (10)(8) + (10)(-2i) + (2i)(8) + (2i)(-2i) - 36 = 80 - 20i + 16i - 4i² - 36. Since i² = -1, -4i² = 4. Thus Δ = 80 - 4i + 4 - 36 = (48 - 4i).

Determinant is 48 - 4i. Then I₁ = (1/Δ) ∣ V_s ; 0 8-2i ∣? Instead, use Cramer' / (48 - 4i). Multiply numerator and denominator by the conjugate of denominator (48 + 4i):

2304 + 16) = (3840 - 640i + 80) / 2320 = (3920 - 640i) / 2320 = 1.6897 - 0.2759i (approx).

用高斯消元法：先算行列式 Δ = (10+2i)(8-2i) - 36 = 80 -20i +16i -4i² -36。因 i²=-1，得 Δ = 80 -4i +4 -36 = 48 -4i。然后用 Cramer 法则：I₁ = det([10, -6; 0, 8-2i]) / Δ = 10(8-2i)/(48-4i) = (80-20i)/(48-4i)。分子分母同乘分母的共轭 (48+4i)： ) = (3840+320i-960i-80i²)/2320 = (3840-640i+80)/2320 = (3920-640i)/2320 ≈ 1.6897 - 0.2759i。

/ (48-4i). Multiply by conjugate: 60(48+4i)/2320 = (2880 + 240i)/2320 ≈ 1.2414 + 0.1034i.

对于 I₂：det([10+2i,10; -6,0]) / Δ = (0 - (-60)) / (48-4i) = 60/(48-4i)。乘共轭：60(48+4i)/2320 = (2880+240i)/2320 ≈ 1.2414 + 0.1034i。

6. Interpretation: Magnitudes and Phase Angles | 解释：幅值与相角

The phasor I₁ = 1.6897 - 0.2759i has magnitude |I₁| = √1.6897² + 0.2759² 27°. Hence i₁(t) = 1.712 cos(2t - 9.27°) A in the steady state. For I₂ = 1.2414 + 0.1034i, |I₂| ≈ 1.246 A, φ₂ = arctan(0.1034/1.2414) ≈ 4.76°, so i₂(t) = 1.246 cos(2t + 4.76°) A.

相量 I₁ = 1.6897 - 0.2759i 的模 |I₁| ≈ √1.6897²+0.2759² = 1.712 A，相角 φ₁ = arctan(-0.2759/1.6897) ≈ -9.27°。因此稳态下 i₁(t) = 1.712 cos(2t - 9.27°) A。I₂ 的模 ≈1.246 A，相角 φ₂ ≈ 4.76°，故 i₂(t) = 1.246 cos(2t + 4.76°) A。

The capacitor voltage phasor is V_C = Z_C I₂ = (-2i)(1.2414 + 0.1034i) = -2.4828i - 0.2068i² = 0.2068 - 2.4828i. Its magnitude is √0.2068² + 2.4828² ≈ 2.491 V, and the phase is arctan(-2.4828/0.2068) ≈ -85.2°. So v_C(t) = 2.491 cos(2t - 85.2°) V.

电容电压相量 V_C = Z_C I₂ = (-2i)(1.2414+0.1034i) ≈ 0.2068 - 2.4828i，幅值约 2.491 V，相角约 -85.2°，因此 v_C(t) = 2.491 cos(2t - 85.2°) V。

7. Time-Domain Analysis: Transient Behaviour after Source Removal | 时域

Published by TutorHao | Year 12 进阶数学 Revision Series | aleveler.com

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