📚 Year 8 AQA Computer Science: Formula & Theorem Quick Reference Handbook | Year 8 AQA 计算机：公式定理速查手册
Welcome to your quick reference guide for the essential formulas and theorems covered in Year 8 AQA Computer Science. Understanding these building blocks will help you solve problems involving binary data, images, sound, and logic gates. Keep this handbook handy when revising or completing your classwork.
欢迎使用 Year 8 AQA 计算机科学必备公式定理速查手册。掌握这些基础知识将帮助你解决涉及二进制数据、图像、声音和逻辑门的问题。在复习或完成课堂作业时，请随时参考本手册。
1. Binary to Decimal Conversion | 二进制转十进制
To convert a binary number to decimal, multiply each bit by its place value (powers of 2) and sum the results. The rightmost bit has place value 2⁰, the next 2¹, then 2², and so on.
要将二进制数转换为十进制数，将每一位乘以其位权（2 的幂）并求和。最右边的位权为 2⁰，下一位为 2¹，然后是 2²，以此类推。
Decimal = (bₙ × 2ⁿ) + ... + (b₁ × 2¹) + (b₀ × 2⁰)
Example: Binary 1011₂ = (1×2³) + (0×2²) + (1×2¹) + (1×2⁰) = 8 + 0 + 2 + 1 = 11₁₀.
示例：二进制 1011₂ = (1×2³) + (0×2²) + (1×2¹) + (1×2⁰) = 8 + 0 + 2 + 1 = 11₁₀。
2. Decimal to Binary Conversion | 十进制转二进制
Divide the decimal number repeatedly by 2, recording the remainder at each step. The binary number is the sequence of remainders read from bottom to top (last remainder is the most significant bit).
将十进制数重复除以 2，记录每次的余数。二进制数是从下往上读取的余数序列（最后的余数是最高有效位）。
6 ÷ 2 = 3 remainder 0; 3 ÷ 2 = 1 remainder 1; 1 ÷ 2 = 0 remainder 1. Read upwards: 1101₂.
示例：将 13₁₀ 转换为二进制。13 ÷ 2 = 6 余 1；6 ÷ 2 = 3 余 0；3 ÷ 2 = 1 余 1；1 ÷ 2 = 0 余 1。从下往上读：1101₂。
3. Binary Addition Rules | 二进制加法规则
Binary addition follows four basic rules. When the sum of two bits is 2, write 0 and carry 1 to the next column.
二进制加法遵循四条基本规则。当两个位相加为 2 时，写 0 并向下一列进 1。
| 0 + 0 = 0 | 0 + 1 = 1 |
| 1 + 0 = 1 | 1 + 1 = 0 (carry 1) |
When adding three 1s (1+1+1), the result is 1 with a carry of 1 to the next column.
当三个 1 相加时 (1+1+1)，结果为 1，并向下一列进 1。
4. Hexadecimal to Binary and Denary | 十六进制与二进制、十进制的转换
Each hexadecimal digit represents a group of four binary bits. To convert hex to binary, replace each hex digit with its 4‑bit binary equivalent. To convert hex to denary, multiply each digit by 16ⁿ (where n is the position from the right, starting at 0).
每个十六进制数字代表一组四位二进制。将十六进制转换为二进制时，用其 4 位二进制等效值替换每个十六进制数字。将十六进制转换为十进制时，将每位数字乘以 16ⁿ（n 为从右开始的位置，从 0 开始）。
Hex digit → 4‑bit binary | 十六进制数字 → 4 位二进制
Example: 2F₁₆ = 0010 1111₂. Denary: (2×16¹) + (15×16⁰) = 32 + 15 = 47₁₀.
示例：2F₁₆ = 0010 1111₂。十进制：(2×16¹) + (15×16⁰) = 32 + 15 = 47₁₀。
5. Data Units Conversion | 数据单位换算
powers of 2. Always remember that 1 byte = 8 bits, 1 kilobyte (KB) = 1024 bytes, not 1000.
数据存储以位和字节为单位。换算基于 2 的幂。始终记住 1 字节 = 8 位，1 千字节 (KB) = 1024 字节，而不是 1000。
- 1 byte = 8 bits
- 1 KB = 1024 bytes
- 1 MB = 1024 KB = 1024 × 1024 bytes
- 1 GB = 1024 MB
- 1 TB = 1024 GB
To convert bits to bytes, divide by 8. To convert bytes to KB, divide by 1024, and so on.
将位转换为字节，除以 8。将字节转换为 KB，除以 1024，以此类推。
6. Image File Size Formula | 图像文件大小公式
The size of an uncompressed bitmap image depends on its width, height, and colour depth (bits per pixel). The formula gives the file size in bits.
未压缩位图图像的大小取决于其宽度、高度和颜色深度（每像素位数）。该公式给出以位为单位的文件大小。
File size (bits) = Width (pixels) × Height (pixels) × Colour Depth (bpp)
To convert to bytes, divide by 8; to KB, divide by 1024 again. When answering exam questions, show all steps of unit conversion.
转换为字节时除以 8；转换为 KB 时再除以 1024。回答考试题目时，请展示所有单位转换步骤。
Example: A 200×100 pixel image with 24‑bit colour depth = 200 × 100 × 24 = 480,000 bits = 60,000 bytes ≈ 58.6 KB.
示例：一幅 200×100 像素、24 位颜色深度的图像 = 200 × 100 × 24 = 480,000 位 = 60,000 字节 ≈ 58.6 KB。
7. Sound File Size Formula | 声音文件大小公式
Uncompressed sound file size depends on sample rate, bit depth, number of channels, and duration. The formula below returns the value in bits.
未压缩声音文件大小取决于采样率、位深度、声道数和持续时间。以下公式返回以位为单位的值。
Sound file size (bits) = Sample rate (Hz) × Bit depth × Channels × Time (seconds)
channels = 1. Remember to convert bit depth and sample rate correctly before multiplying.
对于立体声，声道数 = 2；对于单声道，声道数 = 1。在相乘之前，请记住正确转换位深度和采样率。
Example: CD quality audio: sample rate 44,100 Hz, 16‑bit, stereo, 60 seconds → 44,100 × 16 × 2 × 60 = 84,672,000 bits ≈ 10.1 MB.
示例：CD 质量音频：采样率 44,100 Hz，16 位，立体声，60 秒 → 44,100 × 16 × 2 × 60 = 84,672,000 位 ≈ 10.1 MB。
8. File Transfer Time Calculation | 文件传输时间计算
The time needed to transfer a file over a network or from a storage device can be found using the data transfer time formula. Make sure units on both sides match (bits or bytes).
通过网络或从存储设备传输文件所需的时间可以使用数据传输时间公式计算。确保两侧单位一致（位或字节）。
Time (seconds) = File size / Transfer rate
If file size is given in bytes and transfer rate in bits per second (bps), convert file size to bits first (multiply by 8).
如果文件大小以字节为单位，传输速率以每秒位数 (bps) 为单位，则先将文件大小转换为位（乘以 8）。
onnection: (10 × 1024 × 1024 × 8 bits) / (2 × 1,000,000) ≈ 41.9 seconds (using 2×10⁶ bps).
示例：10 MB 文件通过 2 Mbps 连接传输：(10 × 1024 × 1024 × 8 位) / (2 × 1,000,000) ≈ 41.9 秒（使用 2×10⁶ bps）。
9. Boolean Logic Laws | 布尔逻辑定律
Boolean algebra helps simplify logic circuits. Key laws include the commutative, associative, distributive, and De Morgan's theorems. These apply to AND (∧), OR (∨), and NOT (¬).
布尔代数有助于简化逻辑电路。关键定律包括交换律、结合律、分配律和德摩根定理。这些定律适用于与 (∧)、或 (∨) 和非 (¬)。
De Morgan's Laws | 德摩根定律：
¬(A ∧ B) = ¬A ∨ ¬B
¬(A ∨ B) = ¬A ∧ ¬B
These mean: the negation of (A AND B) is (NOT A) OR (NOT B); the negation of (A OR B) is (NOT A) AND (NOT B).
这表示：(A 与 B) 的非等于 (非 A) 或 (非 B)；(A 或 B) 的非等于 (非 A) 与 (非 B)。
Double negation: ¬(¬A) = A. A double NOT cancels out.
双重否定：¬(¬A) = A。双重 NOT 相互抵消。
10. Logic Gate Truth Tables | 逻辑门真值表
Logic gates are the building blocks of digital circuits. The most common gates are NOT, AND, OR, and XOR. Their truth tables show the output for every possible input combination.
逻辑门是数字电路的基本构件。最常见的门是非门、与门、或门和异或门。它们的真值表显示了每种可能输入组合下的输出。
| Gate | Symbol | Input A, B | Output |
|---|---|---|---|
| NOT | ¬A | 0 → 1, 1 → 0 | Inverts input |
| AND | A ∧ B | (0,0)→0, (0,1)→0, (1,0)→0, (1,1)→1 | 1 only when A=1 AND B=1 |
| OR | A ∨ B | (0,0)→0, (0,1)→1, (1,0)→1, (1,1)→1 | 1 when at least one is 1 |
| XOR | A ⊕ B | (0,0)→0, (0,1)→1, (1,0)→1, (1,1)→0 | 1 when A and B are different |
NAND and NOR gates are combinations: NAND is AND followed by NOT; NOR is OR followed by NOT. They are widely used because any logic circuit can be built entirely from NAND (or NOR) gates alone.
与非门和或非门是组合：与非门是与门后接非门；或非门是或门后接非门。它们被广泛使用，因为任何逻辑电路都可以完全仅用与非门（或仅用或非门）构建。
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