Explainer · 10.1 + 10.2

讲解 · 10.1 + 10.2

The six logic gates

六种逻辑门

A logic gate takes one or two binary inputs (0 or 1) and produces one binary output. The 0478 syllabus uses six, each with its own standard symbol and its own rule. NOT has one input; the other five have two.

逻辑门接收一个或两个二进制输入(0 或 1),并产生一个二进制输出。0478 考纲使用六种逻辑门,每一种都有自己的标准符号和规则。NOT 有一个输入;其余五种都有两个。

!Markers report the XOR symbol drawn wrongly more than any other, and NOR / XOR drawn with straight sides so they look like a NOT gate. Each gate must keep its own distinctive shape: AND has a flat back, OR and XOR a curved back and a pointed front, XOR has the extra curved line behind it, and a small circle on the output means the result is inverted (NAND, NOR, NOT).

!考官报告称,XOR 符号被画错的次数比其他任何符号都多,而 NOR / XOR 有时被画成直边,看起来像 NOT 门。每种门都必须保持自己独特的形状:AND 后背是平的,OR 和 XOR 后背是弧形、前端是尖的,XOR 在后面多了一条弧线,输出端的小圆圈表示结果被取反(NAND、NOR、NOT)。

Explainer · 10.3(b)

讲解 · 10.3(b)

Building a truth table from a circuit or expression

从电路或表达式构建真值表

For anything bigger than one gate, work it in parts. Give each gate’s output a letter (P, Q), fill a column for each, then combine. Here is X = (A OR NOT B) AND (B NAND C) worked in full.

对于超过一个门的电路,要分部分处理。给每个门的输出取一个字母(P、Q),为每个字母填一列,然后再组合。下面是 X = (A OR NOT B) AND (B NAND C) 的完整推导。

The intermediate columns can be dropped once you are confident, leaving just A, B, C and X. Leaving them in is how you earn method marks and catch your own errors.

一旦有把握,中间列就可以省略,只留下 A、B、C 和 X。保留它们是你获得方法分、并发现自己错误的方式。

Explainer · 10.3(a) + 10.3(c)

讲解 · 10.3(a) + 10.3(c)

Moving between statement, expression and circuit

在陈述、表达式与电路之间转换

Problem statement → expression

问题陈述 → 表达式

A greenhouse fan turns on (X = 1) if the temperature is high (A) and the window is closed (B); or if the humidity is high (C) and the window is closed (B).

温室风扇会开启(X = 1),如果温度高(A)且窗户关闭(B);或者湿度高(C)且窗户关闭(B)。

Then the circuit is drawn straight from that: an AND gate for each bracket, both feeding one OR gate. Do not tidy it to B AND (A OR C) first: the syllabus and the mark scheme want the circuit drawn for the statement as given, without simplification.

然后直接依据它画出电路:每个括号对应一个 AND 门,两者都输入到同一个 OR 门。不要先把它简化为 B AND (A OR C):考纲和评分标准都要求按给定的陈述画出电路,不要先化简。

Circuit → expression

电路 → 表达式

Read the gates touching the inputs first, then work outwards. A circuit with an OR gate on A, B and a NAND gate on B, C, both feeding a final AND gate:

先读懂与输入直接相连的门,再向外推进。一个电路中,A、B 接入一个 OR 门,B、C 接入一个 NAND 门,两者都输入到最后的 AND 门:

Truth table → expression

真值表 → 表达式

Look only at the rows where the output is 1. Write each as a set of inputs joined by AND (using NOT for a 0), then join those with OR.

只看输出为 1 的那些行。把每一行写成一组由 AND 连接的输入(0 用 NOT 表示),再用 OR 把这些连接起来。

The rows where X is 0 are ignored. Every row where X is 1 adds one bracket. This one happens to be equivalent to just NOT A, but write out both output-1 rows: the syllabus wants the expression taken straight from the table, without simplification.

X 为 0 的行被忽略。每一行 X 为 1 就增加一个括号。这个例子恰好等价于 NOT A,但仍要把两个输出为 1 的行都写出来:考纲要求表达式直接从表中得出,不要先化简。