📚 KS3 Edexcel Computer Science: Formula and Theorem Quick Reference | KS3 Edexcel 计算机：公式定理速查手册
This quick reference handbook summarises the essential formulas and theorems you need for KS3 Edexcel Computer Science, covering binary conversions, storage calculations, sound and image file sizes, Boolean logic, and data transmission. Keep this guide handy to reinforce your understanding of how digital data is represented, processed, and transferred.
本速查手册总结了 KS3 Edexcel 计算机课程所需的基本公式和定理，涵盖二进制转换、存储计算、声音和图像文件大小、布尔逻辑以及数据传输。将本指南放在手边，有助于巩固你对数字数据表示、处理和传输方式的理解。
1. Binary to Decimal Conversion | 二进制转十进制
To convert a binary number to decimal, multiply each bit (0 or 1) by 2 raised to the power of its position, starting from 0 at the rightmost bit. Then sum all the products. The place values from right to left are 2⁰, 2¹, 2², 2³, and so on.
要将二进制数转换为十进制，将每个位（0 或 1）乘以 2 的该位次方，从最右边的位以 0 次方开始。然后将所有乘积相加。从右到左的位值依次为 2⁰，2¹，2²，2³，以此类推。
For example, 1011₂ = 1×2³ + 0×2² + 1×2¹ + 1×2⁰ = 8 + 0 + 2 + 1 = 11₁₀.
例如，1011₂ = 1×2³ + 0×2² + 1×2¹ + 1×2⁰ = 8 + 0 + 2 + 1 = 11₁₀。
Decimal = dₙ₋₁×2ⁿ⁻¹ + ... + d₁×2¹ + d₀×2⁰
2. Decimal to Binary Conversion | 十进制转二进制
To convert a positive decimal integer to binary, repeatedly divide the number by 2 and record the remainders. The binary result is the sequence of remainders read from last to first (bottom to top).
要将一个十进制正整数转换为二进制，请反复将该数除以 2 并记下余数。二进制结果就是将余数从后向前（由下至上）读出的序列。
ainder 0; 3 ÷ 2 = 1 remainder 1; 1 ÷ 2 = 0 remainder 1. Read remainders backwards: 11001₂.
示例：将 25₁₀ 转换为二进制。25 ÷ 2 = 12 余 1；12 ÷ 2 = 6 余 0；6 ÷ 2 = 3 余 0；3 ÷ 2 = 1 余 1；1 ÷ 2 = 0 余 1。反向读取余数得到 11001₂。
3. Hexadecimal Conversions | 十六进制转换
Hexadecimal (base‑16) uses digits 0–9 and letters A–F to represent values 10–15. Each hex digit corresponds to exactly 4 bits (a nibble), making conversion between hex and binary straightforward.
十六进制（基数为 16）使用数字 0–9 和字母 A–F 表示 10–15 的值。每个十六进制位恰好对应 4 个比特（半个字节），使得十六进制与二进制之间的转换非常直接。
6 raised to its position (rightmost is 16⁰). E.g., 2A₁₆ = 2×16¹ + 10×16⁰ = 32 + 10 = 42₁₀.
十六进制转十进制：将每个十六进制位乘以 16 的该位次方（最右边为 16⁰）。例如，2A₁₆ = 2×16¹ + 10×16⁰ = 32 + 10 = 42₁₀。
Hex to binary: Replace each hex digit with its 4‑bit binary equivalent. For instance, 3F₁₆ = 0011 1111₂.
十六进制转二进制：将每个十六进制位替换为对应的 4 位二进制数。例如，3F₁₆ = 0011 1111₂。
4. Data Storage Units & Conversion | 数据存储单位与换算
Digital data is measured in bits and bytes. Common storage units and their relationships:
数字数据以比特和字节为单位。常见的存储单位及其换算关系：
= 1024 B; 1 megabyte (MB) = 1024 KB; 1 gigabyte (GB) = 1024 MB; 1 terabyte (TB) = 1024 GB.
1 字节（B） = 8 比特（b）；1 千字节（KB） = 1024 B；1 兆字节（MB） = 1024 KB；1 吉字节（GB） = 1024 MB；1 太字节（TB） = 1024 GB。
When converting between units, multiply or divide by 1024 (2¹⁰) depending on direction. File sizes are often expressed in bytes, while network speeds use bits per second.
在单位之间转换时，根据方向乘以或除以 1024（2¹⁰）。文件大小通常以字节表示，而网络速度则使用比特每秒。
5. Image File Size Calculation | 图像文件大小计算
The file size of a bitmap image depends on its pixel dimensions and colour depth. The basic formula is:
位图图像的文件大小取决于其像素尺寸和颜色深度。基本公式为：
File Size (bits) = Width (pixels) × Height (pixels) × Colour Depth (bits per pixel)
esult by 8. Colour depth determines the number of available colours: Number of colours = 2colour depth.
要获得以字节为单位的大小，请将结果除以 8。颜色深度决定了可用颜色的数量：颜色数 = 2颜色深度。
s per pixel has a file size of 200 × 300 × 24 = 1,440,000 bits = 180,000 bytes ≈ 175.8 KB.
示例：一张 200×300 像素、颜色深度为每像素 24 比特的图像，其文件大小为 200 × 300 × 24 = 1,440,000 比特 = 180,000 字节 ≈ 175.8 KB。
6. Sound File Size Calculation | 声音文件大小计算
Digital audio file size is determined by sample rate, bit depth, number of channels, and duration. The formula is:
数字音频文件的大小由采样率、位深度、声道数和持续时间决定。公式如下：
File Size (bits) = Sample Rate (Hz) × Bit Depth × Channels × Duration (s)
Sample rate is the number of samples taken per second; bit depth is the number of bits per sample. Stereo audio uses 2 channels, mono uses 1.
采样率是每秒采集的样本数；位深度是每个样本的比特数。立体声使用 2 个声道，单声道使用 1 个声道。
Example: 30 seconds of stereo audio at 44.1 kHz sample rate and 16‑bit depth gives: 44100 × 16 × 2 × 30 = 42,336,000 bits = 5,292,000 bytes ≈ 5.05 MB.
示例：30 秒的立体声音频，采样率 44.1 kHz，位深度 16 比特，则文件大小为：44100 × 16 × 2 × 30 = 42,336,000 比特 = 5,292,000 字节 ≈ 5.05 MB。
7. Sampling Theorem (Nyquist Theorem) | 采样定理（奈奎斯特定理）
The Nyquist theorem states that to accurately capture a signal, the sample rate must be at least twice the highest frequency present in the signal. This is the minimum sample rate to avoid aliasing.
奈奎斯特定理指出，为了准确地捕获信号，采样率必须至少是信号中最高频率的两倍。这是避免混叠所需的最低采样率。
Sample Rate ≥ 2 × Highest Frequency
For example, if the highest audio frequency is 20 kHz, the sample rate must be at least 40 kHz. CD‑quality audio uses 44.1 kHz to satisfy the theorem and include a small guard band.
例如，如果最高音频频率为 20 kHz，则采样率必须至少为 40 kHz。CD 音质的音频采用 44.1 kHz 的采样率，既满足了定理要求，也包含了一个小的保护频带。
8. Boolean Logic & Truth Tables | 布尔逻辑与真值表
Boolean logic deals with values that are either TRUE (1) or FALSE (0). The three fundamental gates are AND, OR, and NOT. Their truth tables summarise all possible input‑output combinations.
布尔逻辑处理的值要么为 TRUE（1），要么为 FALSE（0）。三个基本逻辑门为 AND、OR 和 NOT。它们的真值表总结了所有可能的输入输出组合。
put is 1 only if all inputs are 1. So, 0 AND 0 = 0; 0 AND 1 = 0; 1 AND 0 = 0; 1 AND 1 = 1.
与门：仅当所有输入都为 1 时，输出才为 1。因此，0 AND 0 = 0；0 AND 1 = 0；1 AND 0 = 0；1 AND 1 = 1。
utput is 1 if at least one input is 1. So, 0 OR 0 = 0; 0 OR 1 = 1; 1 OR 0 = 1; 1 OR 1 = 1.
或门：如果至少有一个输入为 1，则输出为 1。因此，0 OR 0 = 0；0 OR 1 = 1；1 OR 0 = 1；1 OR 1 = 1。
NOT gate: it has a single input and inverts it. NOT 0 = 1; NOT 1 = 0.
非门：它只有一个输入，并将其反转。NOT 0 = 1；NOT 1 = 0。
9. Boolean Algebra Laws & De Morgan's Theorems | 布尔代数定律与德摩根定理
Boolean algebra includes several laws that simplify logical expressions. De Morgan's two theorems are particularly useful for redesigning circuits and simplifying code conditions.
布尔代数包含若干用于简化逻辑表达式的定律。德摩根的两个定理在重新设计电路和简化代码条件时特别有用。
De Morgan's first theorem: NOT (A AND B) = (NOT A) OR (NOT B). In symbols: ¬(A ∧ B) = ¬A ∨ ¬B.
德摩根第一定理：并非（A 与 B）等于（非 A）或（非 B）。用符号表示为：¬(A ∧ B) = ¬A ∨ ¬B。
De Morgan's second theorem: NOT (A OR B) = (NOT A) AND (NOT B). In symbols: ¬(A ∨ B) = ¬A ∧ ¬B.
德摩根第二定理：并非（A 或 B）等于（非 A）与（非 B）。用符号表示为：¬(A ∨ B) = ¬A ∧ ¬B。
ciativity, and distributivity. These help reduce the number of gates in a digital circuit.
其他有用的定律包括交换律（A AND B = B AND A）、结合律和分配律。这些定律有助于减少数字电路中的逻辑门数量。
10. Data Transmission Time | 数据传输时间
The time required to transfer a file over a network is calculated by dividing the file size by the transmission rate. It is essential to use consistent units (bits with bits, bytes with bytes).
通过网络传输文件所需的时间是通过将文件大小除以传输速率来计算的。关键是使用一致的单位（比特与比特对应，字节与字节对应）。
Transmission Time (s) = File Size ÷ Transfer Rate
econds. For example, a 100 megabit file transferred at 50 Mbps takes 100 ÷ 50 = 2 seconds.
如果文件大小以比特为单位，传输速率以比特每秒（bps）为单位，则时间的单位为秒。例如，一个 100 兆比特的文件以 50 Mbps 的速率传输需要 100 ÷ 50 = 2 秒。
When file size is in bytes and rate in bytes/s, the same formula applies. Always check whether your values are expressed in decimal (1000‑based) or binary (1024‑based) prefixes.
当文件大小以字节为单位、速率以字节/秒为单位时，同样的公式仍然适用。务必检查你的数值采用的是十进制前缀（基于 1000）还是二进制前缀（基于 1024）。
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