Year 7 Cambridge Computing: Formula & Theorem Quick Reference Guide | Year 7 剑桥计算机：公式定理速查手册

📚 Year 7 Cambridge Computing: Formula & Theorem Quick Reference Guide | Year 7 剑桥计算机：公式定理速查手册

Welcome to your quick reference handbook for the essential formulas and theorems you will meet in Year 7 Cambridge Computing. This guide collects the key conversions, rules, and relationships you need to work fluently with binary, data units, logic gates, and digital representation. Keep it handy when you practise calculations or revise for assessments – each entry includes a worked example and a brief explanation.

欢迎阅读这份 Year 7 剑桥计算机必备公式与定理速查手册。这里整理了你将在课程中遇到的关键换算公式、运算规则和数学关系，帮助你熟练掌握二进制、数据单位、逻辑门和数字表示等内容。请在练习计算或考前复习时随手翻阅 – 每一条都配有实例和简要说明。

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1. Binary to Decimal Conversion | 二进制转十进制

To convert a binary number to decimal, write the place values as powers of 2 from right to left (2⁰, 2¹, 2², 2³ …). Multiply each binary digit by its place value and add the results. The formula can be summarised as:
Decimal = dₙ×2ⁿ + ... + d₁×2¹ + d₀×2⁰

将二进制数转换为十进制时，从右向左依次写出 2 的幂次作为位值（2⁰, 2¹, 2², 2³ …）。将每一位二进制数字乘以对应的位值，然后全部相加。公式概括为：十进制 = dₙ×2ⁿ + ... + d₁×2¹ + d₀×2⁰

Example: Convert 1101₂ to decimal.
1×2³ + 1×2² + 0×2¹ + 1×2⁰ = 8 + 4 + 0 + 1 = 13₁₀

示例：将 1101₂ 转换为十进制。
1×2³ + 1×2² + 0×2¹ + 1×2⁰ = 8 + 4 + 0 + 1 = 13₁₀

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2. Decimal to Binary Conversion | 十进制转二进制

Divide the decimal number repeatedly by 2, keeping track of the remainders. Read the remainders from bottom to top to obtain the binary equivalent. This is often called the division-by-2 algorithm.
Step: number ÷ 2 → quotient & remainder; repeat until quotient becomes 0.

将十进制数不断除以 2，记录每一步的余数。从下往上读出余数，即得到对应的二进制数。这种方法通常称为“除 2 取余法”。
步骤：数 ÷ 2 → 商与余数；重复直至商为 0。

Example: Convert 25 to binary.
25 ÷ 2 = 12 remainder 1
12 ÷ 2 = 6 remainder 0
6 ÷ 2 = 3 remainder 0
3 ÷ 2 = 1 remainder 1
1 ÷ 2 = 0 remainder 1
Reading remainders upwards: 11001₂

示例：将 25 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
从下往上读出余数：11001₂

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3. Binary Addition Rules | 二进制加法规则

Binary addition follows four simple rules, similar to decimal addition but with a carry to the next column when the sum reaches 2.

二进制加法遵循四条简单规则，与十进制加法类似，但当和达到 2 时需向高位进位。

Rule	Operation	Carry
0 + 0	0	0
0 + 1	1	0
1 + 0	1	0
1 + 1	0	1 (carry to next column)
1 + 1 + 1	1	1

Example: Add 1011₂ and 1101₂.
1011
+ 1101
————
11000₂ (carry occurs at the 2³ and 2⁴ positions)

示例：计算 1011₂ + 1101₂。
1011
+ 1101
————
11000₂ （在 2³ 和 2⁴ 位产生进位）

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4. Data Storage Units Conversion | 数据存储单位换算

Digital information is measured in bits and bytes. Larger units follow powers of 2 (binary) or powers of 10 (decimal, often used by manufacturers). In Cambridge Computing, you mainly work with binary multiples.

数字信息以比特和字节为基本单位。更大的单位遵循 2 的幂次（二进制）或 10 的幂次（十进制，常用于制造商标注）。在剑桥计算机课程中，主要采用二进制倍数。

1 byte = 8 bits
1 KiB = 2¹⁰ bytes = 1024 bytes
1 MiB = 2²⁰ bytes = 1,048,576 bytes
1 GiB = 2³⁰ bytes

To convert between units, multiply or divide by 1024. Always check whether the question expects binary or decimal interpretation.

单位换算时乘以或除以 1024。务必留意题目要求的是二进制还是十进制解释。

Example: Convert 3 MiB to bytes.
3 × 1024 × 1024 = 3,145,728 bytes.

示例：将 3 MiB 转换为字节。
3 × 1024 × 1024 = 3,145,728 字节。

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5. File Size Calculation | 文件大小计算

File size can be estimated by multiplying the data capacity of each element by the number of elements. The general formula is:
File size (bits) = number of elements × bits per element

文件大小可通过元素数量乘以每个元素所占位数来估算。通用公式为：
文件大小（比特）= 元素数量 × 每元素比特数

el has a file size of 10,000 bits. For colour images, multiply by colour depth (e.g. 24 bits per pixel).

例如，一张 100×100 像素、每像素 1 比特的黑白图像，文件大小为 10,000 比特。彩色图像则需乘以颜色深度（如每像素 24 比特）。

If the size is to be expressed in bytes, divide by 8. To express in KiB, divide again by 1024.

若需以字节为单位，则除以 8；若以 KiB 为单位，再除以 1024。

Example: 300×200 pixel image at 16-bit colour depth.
Bits = 300 × 200 × 16 = 960,000 bits
Bytes = 960,000 ÷ 8 = 120,000 bytes ≈ 117.2 KiB.

示例：一张 300×200 像素、16 位颜色深度的图像。
比特数 = 300 × 200 × 16 = 960,000 比特
字节数 = 960,000 ÷ 8 = 120,000 字节 ≈ 117.2 KiB。

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6. Network Speed Conversion | 网络速度换算

Network speeds are often measured in bits per second (bps) and its multiples. Download times can be calculated using:
Time (seconds) = File size (bits) ÷ Transfer speed (bps)

网络速度通常以比特每秒（bps）及其倍数表示。下载时间可以通过以下公式计算：
时间（秒）= 文件大小（比特） ÷ 传输速率（bps）

Make sure to express file size in bits, not bytes. If speed is given in Mbps, convert to bps by multiplying by 1,000,000 (for decimal interpretation) or 1,048,576 for binary, but Cambridge at this stage usually uses decimal.

务必确保文件大小以比特为单位，而非字节。如果速率以 Mbps 给出，需要转换为 bps——乘以 1,000,000（十进制记法）；Cambridge 当前阶段通常使用十进制。

Example: A 5 MB file is downloaded at 2 Mbps. How long does it take?
5 MB = 5 × 8 = 40 Mb = 40,000,000 bits
2 Mbps = 2,000,000 bps
Time = 40,000,000 ÷ 2,000,000 = 20 seconds.

示例：一个 5 MB 的文件以 2 Mbps 的速度下载，需要多长时间？
5 MB = 5 × 8 = 40 Mb = 40,000,000 比特
2 Mbps = 2,000,000 bps
时间 = 40,000,000 ÷ 2,000,000 = 20 秒。

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7. Logic Gate Truth Tables | 逻辑门真值表

Logic gates are the building blocks of digital circuits. Their behaviour is defined by truth tables. The three basic gates are AND, OR, and NOT. Here the inputs are labelled A and B, and the output is Q.

逻辑门是数字电路的基本构件，其行为由真值表定义。三种基本逻辑门为与门（AND）、或门（OR）和非门（NOT）。输入用 A 和 B 表示，输出用 Q 表示。

AND gate: Output is 1 only when both inputs are 1.

与门：仅当两个输入都为 1 时，输出才为 1。

A	B	Q = A AND B
0	0	0
0	1	0
1	0	0
1	1	1

OR gate: Output is 1 if at least one input is 1.

或门：若至少有一个输入为 1，则输出为 1。

A	B	Q = A OR B
0	0	0
0	1	1
1	0	1
1	1	1

NOT gate: Only one input; output is the inverse.

非门：只有一个输入；输出为输入的反相。

A	Q = NOT A
0	1
1	0

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8. Image Representation – Colour Depth Formula | 图像表示 – 颜色深度公式

Colour depth is the number of bits used to represent the colour of one pixel. The maximum number of colours that can be displayed is given by:
Number of colours = 2ⁿ, where n = colour depth in bits

颜色深度是表示一个像素颜色所用的比特数。可显示的最大颜色数由以下公式给出：
颜色数 = 2ⁿ，其中 n 表示以比特为单位的颜色深度

For a monochrome image, n = 1 → 2 colours. For 8‑bit colour, the range is 2⁸ = 256 colours. Modern ‘true colour’ uses 24 bits, giving 2²⁴ ≈ 16.7 million colours.

对于单色图像，n = 1 → 2 种颜色。8 位颜色可显示 2⁸ = 256 种颜色。现代“真彩色”采用 24 位，可呈现 2²⁴ ≈ 1670 万种颜色。

Example: How many colours can be shown with a 6‑bit colour depth?
2⁶ = 64 colours.

示例：6 位颜色深度可以显示多少种颜色？
2⁶ = 64 种颜色。

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9. Sound Sampling Formula | 声音采样公式

When sound is digitised, the sample rate and bit depth determine the quality and file size. The bit rate (bits per second) is calculated as:
Bit rate = sample rate (Hz) × bit depth × number of channels

声音数字化时，采样率和位深度决定了音质和文件大小。比特率（每秒比特数）的计算公式为：
比特率 = 采样率（Hz）× 位深度 × 声道数

From the bit rate, the file size for a given duration can be found:
File size (bits) = bit rate × duration (seconds)

由比特率可计算特定时长的文件大小：
文件大小（比特）= 比特率 × 时长（秒）

Example: A 1‑minute stereo recording uses a 44.1 kHz sample rate and 16‑bit depth.
Bit rate = 44,100 × 16 × 2 = 1,411,200 bps
Duration = 60 s → File size = 1,411,200 × 60 = 84,672,000 bits ≈ 10.58 MB (uncompressed).

示例：一段 1 分钟的立体声录音采用 44.1 kHz 采样率和 16 位深度。
比特率 = 44,100 × 16 × 2 = 1,411,200 bps
时长 60 秒 → 文件大小 = 1,411,200 × 60 = 84,672,000 比特 ≈ 10.58 MB（未压缩）。

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10. Compression Ratio | 压缩比率

Compression reduces file size. The compression ratio compares the original size to the compressed size:
Compression ratio = Original size ÷ Compressed size

压缩可以减小文件体积。压缩比率定义为原始大小与压缩后大小之比：
压缩比 = 原始大小 ÷ 压缩后大小

A higher ratio means more compression. For instance, a ratio of 5:1 means the original file is five times larger than the compressed file. Sometimes you also calculate the percentage reduction:
Reduction (%) = ((Original – Compressed) ÷ Original) × 100

比值越高，压缩程度越大。例如，压缩比 5:1 表示原始文件大小是压缩后文件的五倍。有时还需计算缩减百分比：
缩减百分比 = ((原始 – 压缩后) ÷ 原始) × 100

Example: A 20 MB image is compressed to 4 MB.
Compression ratio = 20 ÷ 4 = 5:1
Reduction = ((20 − 4) ÷ 20) × 100 = 80%.

示例：一张 20 MB 的图像被压缩至 4 MB。
压缩比 = 20 ÷ 4 = 5:1
缩减百分比 = ((20 − 4) ÷ 20) × 100 = 80%。

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11. Memory Address Calculation | 内存地址计算

A computer’s memory is organised as a series of addressable locations. The number of addresses that can be accessed depends on the width of the address bus. The formula is:
Number of addressable locations = 2ⁿ, where n = number of address lines

计算机内存由一系列可寻址的存储单元组成。可访问的地址数量取决于地址总线的宽度。公式为：
可寻址单元数 = 2ⁿ，其中 n 为地址线数量

e total memory capacity equals the number of locations in bytes. For a processor with a 16‑bit address bus, the maximum directly addressable memory is 2¹⁶ = 65,536 bytes (64 KiB).

若每个单元存储 1 字节，则总内存容量（字节）等于地址单元数。对于具有 16 位地址总线的处理器，最大可直接寻址的内存为 2¹⁶ = 65,536 字节（64 KiB）。

Example: How many locations can a 10‑bit address bus access?
2¹⁰ = 1024 locations.

示例：10 位地址总线可访问多少个存储单元？
2¹⁰ = 1024 个单元。

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12. ASCII Character Code Range | ASCII 字符码范围

ASCII (American Standard Code for Information Interchange) originally used 7 bits to represent 128 characters. The decimal range is 0–127. Extended ASCII uses 8 bits, giving 256 characters (0–255). Common ranges are worth memorising:

ASCII（美国信息交换标准代码）最初采用 7 位来表示 128 个字符，十进制范围为 0–127。扩展 ASCII 使用 8 位，可表示 256 个字符（0–255）。以下常见区间值得记忆：

Character type	Decimal range
Digits 0–9	48–57
Uppercase A–Z	65–90
Lowercase a–z	97–122

To find the code of another letter, note that the difference between uppercase and lowercase is exactly 32 (e.g., 'A' is 65, 'a' is 97).

若要推算其他字母的编码，请注意大写与小写之间恰好相差 32（如 'A' 为 65，'a' 为 97）。

Example: What is the ASCII code of 'F'?
'A' is 65, so 'F' is 65 + (position of F – A) = 65 + 5 = 70.

示例：'F' 的 ASCII 码是多少？
'A' 为 65，所以 'F' 是 65 + (6 − 1) = 65 + 5 = 70。

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