📚 Channels of Communication | 通信信道：物理原理与CIE考点解析
Communication systems transfer information from a transmitter to a receiver through a physical channel. The channel may be a pair of wires, a coaxial cable, an optical fibre, or free space carrying radio waves or microwaves. In CIE A-Level Physics, "Channels of communication" focuses on how these channels behave, how signals are attenuated and distorted, and how modulation and digital techniques improve reliability. This revision guide keeps every point of the original article and expands it with the physics you must be able to quote, calculate and explain in the examination.
通信系统通过物理信道将信息从发射端传输到接收端。信道可以是一对导线、同轴电缆、光纤，也可以是传播无线电波或微波的自由空间。在 CIE A-Level 物理中，"通信信道"这一主题重点考察这些信道的特性、信号的衰减与失真，以及调制和数字技术如何提高通信可靠性。本文保留原文全部要点，并补充你在考试中必须能够引用、计算和解释的物理原理。
1. The Communication System Block Diagram | 通信系统框图
tor/decoder → destination. The channel is the medium that carries the signal; it may introduce attenuation, noise and distortion. A carrier wave is often used because the original information signal has a low frequency and cannot travel efficiently through many channels.
每个通信系统都可以简化为一条链路：信息源 → 编码/调制器 → 发射器 → 信道 → 接收器 → 解调/解码器 → 接收端。信道是承载信号的介质，它可能引入衰减、噪声和失真。通常需要使用载波，因为原始信息信号频率较低，无法高效通过许多信道传输。
The transmitter modulates the information onto a high-frequency carrier. The receiver demodulates the signal to recover the original information. In this topic, we mainly study the channel itself and the physical limitations it imposes.
发射器将信息调制到高频载波上，接收器再解调信号以恢复原始信息。在本主题中，我们主要研究信道本身及其产生的物理限制。
length of the wave. Since λ = c / f, a high carrier frequency means a short wavelength and therefore a short, practical aerial.
使用高频载波还有第二个原因：天线的长度只有与波的波长可比时才能有效辐射。由于 λ = c / f，载波频率越高，波长越短，天线也就可以做得短而实用。
λ = c / f and v = f λ
2. Types of Communication Channels | 通信信道的类型
Common channels include wire pairs, coaxial cables, optical fibres, and electromagnetic waves in free space, such as radio waves and microwaves. Each channel has different bandwidth, attenuation, noise immunity, cost and installation constraints. The choice of channel depends on distance, data rate, mobility and environment.
常见的信道包括双绞线、同轴电缆、光纤，以及自由空间中的电磁波（如无线电波和微波）。每种信道的带宽、衰减、抗噪能力、成本和安装条件各不相同。信道的选择取决于传输距离、数据速率、移动性以及使用环境。
A guiding rule is that guided media (wires, cables, fibres) are used where the route is fixed and security or reliability matters, whereas unguided media (free space) are used where mobility or long-distance broadcasting is needed.
一条基本准则是：在有固定线路、对安全性或可靠性有要求时，使用导引型介质（导线、电缆、光纤）；在需要移动性或长距离广播时，使用非导引型介质（自由空间）。
| Channel 信道 | Attenuation 衰减 | Bandwidth 带宽 | Typical use 典型用途 |
| Twisted wire pair 双绞线 | High 较高 | Low–medium 低到中 | Telephone, LAN 电话、局域网 |
| Coaxial cable 同轴电缆 | Medium 中等 | Medium–high 中到高 | TV, broadband 电视、宽带 |
| Optical fibre 光纤 | Very low 极低 | Very high 极高 | Internet backbone 互联网主干 |
| Free space 自由空间 | Depends on distance 随距离变化 | High (high f) 高（高频） | Broadcast, satellite 广播、卫星 |
3. Attenuation and the Decibel | 衰减与分贝
Attenuation is the reduction in the power (or amplitude) of a signal as it travels along a channel. It happens because energy is absorbed by the medium, radiated away, or lost as heat in the resistance of conductors. Attenuation increases with the length of the channel.
衰减是信号沿信道传播时功率（或振幅）的减小。其原因包括介质吸收能量、能量向外辐射，以及导体电阻造成的热损耗。衰减随信道长度的增加而增大。
Because attenuation spans many orders of magnitude, it is measured on a logarithmic scale in decibels (dB). For power ratios:
由于衰减范围可跨越多个数量级，因此使用对数标度的分贝（dB）来量度。对于功率比：
attenuation (dB) = 10 log₁₀ (Pin / Pout)
For amplitude or voltage ratios the factor is 20, because power is proportional to the square of amplitude:
对于振幅或电压比，系数为 20，因为功率与振幅的平方成正比：
attenuation (dB) = 20 log₁₀ (Vin / Vout)
A negative dB value means a loss; a positive dB value means a gain, as produced by an amplifier. CIE questions often give a loss per kilometre (for example 0.2 dB km⁻¹ for a good optical fibre) and ask you to find the total loss or the maximum length before the signal becomes unusable.
dB 为负值表示损耗，正值表示增益（由放大器产生）。CIE 题目常给出每公里的损耗（例如优质光纤约为 0.2 dB km⁻¹），要求计算总损耗，或求信号在变得不可用之前的最长传输距离。
4. Noise, Distortion and Signal-to-Noise Ratio | 噪声、失真与信噪比
Noise is any unwanted random signal that is added to the wanted signal. Thermal noise arises from the random motion of electrons in conductors and increases with temperature; interference comes from external sources such as electric motors, lightning or other transmitters; crosstalk is unwanted coupling between neighbouring channels.
噪声是叠加在有用信号上的任何无用随机信号。热噪声来自导体中电子的无规则运动，并随温度升高而增大；干扰来自外部源，如电动机、闪电或其他发射机；串扰则是相邻信道之间不希望的耦合。
Distortion means that the shape of the signal is changed. A common cause is dispersion, in which different frequency components (or different modes, or different wavelengths of light) travel at different speeds and therefore arrive at different times. Distortion cannot be removed simply by amplifying the signal.
失真指信号波形发生改变。常见原因是色散：不同频率分量（或不同模式、不同波长的光）传播速度不同，因而到达时间不同。失真不能仅靠放大信号来消除。
Signal-to-noise ratio (SNR) compares the power of the signal with the power of the noise:
信噪比（SNR）把信号功率与噪声功率进行比较：
SNR (dB) = 10 log₁₀ (Psignal / Pnoise)
A high SNR means the signal stands out clearly from the background noise and the data is more likely to be received without error. In analogue systems, noise is amplified together with the signal by every amplifier, so the SNR degrades along the link; in digital systems, regeneration restores the SNR.
信噪比高意味着信号在背景噪声中清晰可辨，数据更可能被无误接收。在模拟系统中，每个放大器都会把噪声与信号一起放大，因此信噪比沿链路不断恶化；在数字系统中，再生可以恢复信噪比。
5. Wire Pairs and Coaxial Cables | 双绞线与同轴电缆
A twisted pair consists of two insulated copper wires twisted together. The twisting ensures that the two wires pick up almost equal amounts of external interference, so the induced voltages cancel at the receiver. Twisted pairs are cheap, flexible and easy to install, but they have relatively high attenuation and are susceptible to crosstalk, so their usable bandwidth and range are limited.
双绞线由两根互相缠绕的绝缘铜线组成。绞合使两根导线拾取的外界干扰几乎相等，因而感应电压在接收端相互抵消。双绞线便宜、柔韧、易于安装，但衰减相对较大且易受串扰，因此可用带宽和传输距离有限。
A coaxial cable has a central copper conductor surrounded by a plastic dielectric, a braided metal shield, and an outer plastic jacket. The signal travels in the field between the inner conductor and the shield. The shield prevents the signal from radiating outwards and blocks external interference, giving lower attenuation and a wider bandwidth than a twisted pair, up to the MHz and GHz range.
同轴电缆中心是一根铜质内导体，外包塑料介质、编织金属屏蔽层和塑料外护套。信号在内外导体之间的场中传播。屏蔽层阻止信号向外辐射并阻挡外界干扰，因而比双绞线衰减更低、带宽更宽，可达到 MHz 乃至 GHz 量级。
In both cables the attenuation per unit length rises with frequency (approximately proportional to √f for a coaxial cable), which is why high-frequency signals cannot be sent over very long cables without repeaters or amplifiers.
这两种电缆的单位长度衰减都随频率升高而增大（同轴电缆约正比于 √f），因此高频信号若不使用中继器或放大器，就无法在很长的电缆中传输。
6. Optical Fibres: Total Internal Reflection and Dispersion | 光纤：全反射与色散
An optical fibre is a thin thread of glass or plastic with a core of higher refractive index n₁ surrounded by a cladding of lower refractive index n₂. Light entering within the acceptance cone strikes the core–cladding boundary at an angle greater than the critical angle and is totally internally reflected, so it zig-zags along the fibre with very little loss.
光纤是由玻璃或塑料拉成的细丝，其纤芯折射率 n₁ 较高，外包折射率 n₂ 较低的包层。进入接收锥角内的光以大于临界角的角度射到纤芯–包层界面，发生全反射，于是光在光纤中曲折前进，损耗极小。
sin θ_c = n₂ / n₁ (n₁ > n₂)
Optical fibres carry infrared or visible light at frequencies of the order of 10¹⁴ Hz, giving an enormous bandwidth, and their attenuation can be as low as 0.2 dB km⁻¹ at a wavelength of 1550 nm. They are immune to electrical interference and to crosstalk, and they are very difficult to tap, so they are more secure than copper cables. Their disadvantages are that they are fragile, that they cannot be bent too sharply (bending causes light to escape), and that joints and connectors are expensive.
光纤传输频率约为 10¹⁴ Hz 的红外光或可见光，带宽极大，在波长 1550 nm 处衰减可低至 0.2 dB km⁻¹。光纤不受电磁干扰和串扰影响，且很难被窃听，因此比铜缆更安全。缺点是易碎、不能过度弯折（弯曲会使光逸出），而且接头和连接器成本较高。
Pulse broadening limits the bit rate. Modal dispersion occurs in multimode fibres because rays following different paths have different lengths; it is reduced in graded-index fibres and eliminated in single-mode fibres, whose core is so narrow that only one path is possible. Material dispersion occurs because the refractive index depends slightly on wavelength, so different spectral components of the same pulse travel at different speeds.
脉冲展宽限制了比特率。多模光纤中，沿不同路径传播的光线走过的路程不同，产生模间色散；渐变折射率光纤可减小它，而单模光纤纤芯极细、只允许一条传播路径，因而可消除模间色散。材料色散则是由于折射率随波长略有变化，同一脉冲中不同光谱成分传播速度不同而产生的。
7. Free Space Channels: Radio Waves and Microwaves | 自由空间信道：无线电波与微波
When the channel is free space, the information is carried by electromagnetic waves. Radio waves (roughly 30 kHz to 3 GHz) have long wavelengths and diffract strongly around hills and buildings, so they can be received beyond the horizon; long and medium waves also reflect from the ionosphere, which makes worldwide short-wave broadcasting possible.
当信道是自由空间时，信息由电磁波承载。无线电波（约 30 kHz 至 3 GHz）波长较长，能明显绕过山丘和建筑物发生衍射，因此可在地平线以外收到；长波和中波还会被电离层反射，使短波广播可以覆盖全球。
Microwaves (roughly 1 GHz to 300 GHz) have short wavelengths and travel in nearly straight lines, so a microwave link requires a clear line of sight between the transmitting and receiving dishes. They are absorbed by water and by water vapour, so heavy rain weakens the signal, and they cannot pass through solid obstacles. Microwave links are used for point-to-point communication, satellite links and mobile phone networks.
微波（约 1 GHz 至 300 GHz）波长较短，近似直线传播，因此微波链路要求发射与接收天线之间视线通畅。微波会被水和水蒸气吸收，所以大雨会使信号减弱，而且它无法穿透固体障碍物。微波链路用于点对点通信、卫星链路和移动电话网络。
For free-space propagation the intensity of the wave spreads over a sphere, so intensity falls as the inverse square of the distance from the transmitter, and the received power falls far below what a cable would deliver over the same distance.
在自由空间传播中，波的强度分布在球面上，因此强度按距发射端距离的平方反比减小，接收功率远低于同样距离下电缆所能提供的功率。
I ∝ 1 / r² (free-space spreading)
8. Modulation: AM, FM and Digital Modulation | 调制：调幅、调频与数字调制
Modulation means varying a property of a high-frequency carrier wave in step with the information signal. With amplitude modulation (AM) the amplitude of the carrier varies; with frequency modulation (FM) the frequency varies while the amplitude stays constant. FM has a much larger bandwidth than AM but is far more resistant to noise, because most noise changes the amplitude of the wave, and that amplitude variation can be removed in the receiver by a limiter.
调制是指使高频载波的某一属性随信息信号变化。调幅（AM）中载波振幅随信号变化；调频（FM）中频率变化而振幅保持不变。FM 的带宽比 AM 大得多，但抗噪能力强得多，因为大多数噪声改变的是波的振幅，而接收机中的限幅器可以消除这种振幅变化。
In digital modulation the carrier is switched between discrete states: amplitude shift keying (ASK), frequency shift keying (FSK) and phase shift keying (PSK). Changing phase is the most robust method because noise affects amplitude and frequency much more than it affects the timing of the zero crossings, so PSK (and its variants such as QPSK) allows more bits per symbol and higher data rates.
数字调制中，载波在离散状态之间切换：幅移键控（ASK）、频移键控（FSK）和相移键控（PSK）。改变相位是最稳健的方法，因为噪声对振幅和频率的影响远大于对波形过零点位置的影响，因此 PSK（及其变体如 QPSK）可实现每符号更多比特和更高的数据速率。
Multiplexing and bandwidth are closely tied to modulation: the bandwidth required by an AM signal is twice the highest audio frequency, whereas an FM signal needs roughly 2(Δf + fm), which is why commercial FM stations are spaced much further apart in the frequency spectrum than AM stations.
多路复用和带宽与调制密切相关：AM 信号所需带宽是最高音频频率的两倍，而 FM 信号大约需要 2(Δf + fm)，这就是商业调频电台在频谱中占用的间隔远大于调幅电台的原因。
9. Bandwidth, Data Rate and Channel Capacity | 带宽、数据速率与信道容量
Bandwidth is the range of frequencies a channel can transmit without unacceptable attenuation or distortion. A larger bandwidth allows more information per second: data rate is limited by bandwidth, not by signal power alone. Digital signals are sent as pulses, and if pulses are sent too fast they overlap, an effect called inter-symbol interference, which causes errors at the receiver.
带宽是信道能够在不产生不可接受衰减或失真的情况下传输的频率范围。带宽越大，每秒可传输的信息越多：数据速率受带宽限制，而不只取决于信号功率。数字信号以脉冲形式发送，若发送过快，脉冲会互相重叠，即码间干扰，从而在接收端引起误码。
An analogue signal must be sampled and quantised before it can be transmitted digitally. The sampling rate must satisfy the Nyquist criterion, and quantisation introduces a small quantisation error:
模拟信号必须先采样和量化，然后才能数字传输。采样频率必须满足奈奎斯特准则，而量化会引入一个小的量化误差：
fs ≥ 2 fmax (Nyquist sampling theorem)
ls, the maximum data rate is C = 2B log₂M bits per second. For a noisy channel the Shannon–Hartley theorem gives:
关于信道容量，有两个有用的结论。对于具有 M 个信号电平的无噪声信道，最大数据速率为 C = 2B log₂M 比特每秒。对于有噪声信道，香农–哈特定理给出：
C = B log₂ (1 + S / N) bits per second
These equations show the physical trade-off at the heart of this topic: you can obtain a higher data rate by using a wider bandwidth, by increasing the signal-to-noise ratio (more power, lower noise), or by using more signal levels, but more levels reduce the margin between them and make errors more likely.
这些公式揭示了本主题核心的物理权衡：要获得更高的数据速率，可以增大带宽、提高信噪比（更大功率、更低噪声），或使用更多信号电平；但电平越多，各电平之间的余量越小，越容易出错。
10. Multiplexing and Digital Techniques | 多路复用与数字技术
Multiplexing allows several signals to share one channel. Frequency-division multiplexing (FDM) gives each signal its own band of frequencies; time-division multiplexing (TDM) gives each signal a short repeating time slot; wavelength-division multiplexing (WDM) does the same job with different light wavelengths in a single optical fibre, which is why one fibre can carry tens of terabits per second.
多路复用使多个信号共用一条信道。频分复用（FDM）给每个信号分配自己的频带；时分复用（TDM）给每个信号分配一个短促而重复的时隙；波分复用（WDM）则在单根光纤中用不同波长的光实现同样目的，这就是一根光纤可承载每秒数十太比特的原因。
Digital transmission has important advantages over analogue transmission: the signal can be regenerated rather than merely amplified, so noise does not accumulate along the link; errors can be detected and corrected using check bits or parity; data can be compressed, encrypted and stored; and digital signals are easily multiplexed with computer data.
与模拟传输相比，数字传输有重要优势：信号可以再生而不只是放大，因此噪声不会沿链路累积；可用校验位或奇偶校验检测并纠正错误；数据可以压缩、加密和存储；数字信号也易于与计算机数据复用。
A digital signal can be regenerated because only two states must be distinguished, so as long as the receiver can still identify a 1 or a 0, the original clean pulse train is re-created. An analogue signal, by contrast, contains an infinite number of possible values and can never be restored to exactly its original shape.
数字信号之所以能够再生，是因为只需区分两种状态：只要接收端仍能辨认出 1 或 0，就能重建原来干净的脉冲序列。相反，模拟信号包含无限多个可能取值，永远无法完全恢复其原始波形。
11. Repeaters, Loss Budgets and Practical Links | 中继器、损耗预算与实际链路
Over a long link the signal must be boosted. Analogue links use amplifiers, which increase the signal amplitude but also amplify the accumulated noise. Digital links use repeaters, which detect the incoming pulses, decide whether each is a 1 or a 0, and transmit a fresh, clean pulse; the noise is therefore not passed on.
在长距离链路中，必须对信号进行增强。模拟链路使用放大器，它提高信号振幅，但同时放大了累积的噪声。数字链路使用中继器，它检测到达的脉冲，判断每个脉冲是 1 还是 0，然后发送全新的干净脉冲，因此噪声不会继续传递。
m and the receiver needs at least −28 dBm, then the maximum fibre length is 28 / 0.2 = 140 km before a repeater is needed (ignoring joint losses).
损耗预算是计算信号能传输多远的常用方法。例如，若光纤衰减为 0.2 dB km⁻¹，发射功率为 0 dBm，而接收端至少需要 −28 dBm，则在中继器之前光纤的最大长度为 28 / 0.2 = 140 km（忽略接头损耗）。
In optical links the wavelength chosen matters: attenuation has a minimum near 1550 nm, but material dispersion is smallest near 1310 nm, so engineers compromise or use dispersion-shifted fibre. In copper links the limiting factors are attenuation and crosstalk, while in radio links they are distance, obstacles and rain.
在光链路中，波长的选择很重要：衰减在 1550 nm 附近最小，而材料色散在 1310 nm 附近最小，因此工程师需要折中，或使用色散位移光纤。在铜缆链路中，限制因素是衰减和串扰；在无线电链路中，限制因素则是距离、障碍物和降雨。
12. CIE Exam Technique and Common Pitfalls | CIE 考试技巧与常见失分点
Typical CIE questions ask you to define attenuation, calculate a loss in dB, explain why a repeater is better than an amplifier, describe total internal reflection in an optical fibre, draw or interpret a block diagram, compare AM with FM, and list the advantages of digital transmission. Command words matter: "state" needs a short answer, "describe" needs a sequence of points, and "explain" requires a physical reason linked by because or so that.
CIE 常见题型包括：定义衰减、计算以 dB 表示的损耗、解释中继器为何优于放大器、描述光纤中的全反射、画出或解释框图、比较 AM 与 FM，以及列举数字传输的优点。指令词很重要："state"（写出）要求
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