| English | 의미 |
|---|---|
| logic circuit/ˈlɒdʒɪk ˈsɜːkɪt/ | 논리 회로 |
| Boolean expression/ˈbuːlɪən ekˈspreʃn/ | 부울 식 |
| problem statement/ˈprɒbləm ˈsteɪtmənt/ | 문제 서술 |
| truth table/truːθ ˈteɪbl/ | 진표 |
| sum of products/sʌm ɒv ˈprɒdʌkts/ | 적의 합 |
| half adder/hɑːf ˈædə/ | 반 더러 |
Three sensors, two gates, one decision
- A microwave oven heats only when the door is closed and the start button has been pressed and the timer has not reached zero.
- Three sensors give three 0/1 signals. Two AND gates combine them into one signal that switches the magnetron. Open the door and the output drops to 0 at once.
- Every safety interlock, every alarm and every processor instruction is a decision like this, built from the six gates of the last lesson.
- This lesson is about building and reading those circuits, and moving between the four ways of writing the same decision.
Four views of one function
- A logic circuit 逻辑电路 is a network of gates that carries out a Boolean expression 布尔表达式.
- The same function can be written as a problem statement 问题陈述 in English, as an expression, as a circuit diagram, or as a truth table 真值表. The exam asks you to move in every direction between them.
- The paper writes expressions in words,
X = (A AND NOT B) OR (B AND C), and accepts the algebraic form $X = A\overline{B} + BC$, where a dot or nothing is AND, a plus is OR and a bar is NOT. Use whichever the question uses.
Gates wired together to carry out one Boolean expression
Expression to circuit
- Draw one gate per operator, starting from the innermost brackets.
- For
X = (A AND B) OR (NOT C): an AND gate on A and B, a NOT gate on the C wire, and an OR gate combining the two results. - Inputs on the left, the single output on the right, every line ending at a gate input or the output, and the output labelled X.
One gate for each operator in the expression
논리 회로
게이트가 결합하여 회로를 만듭니다
각 게이트에는 고정 규칙이 있으며,它們을 체인하면 모든 회로를 구축할 수 있습니다. 하나의 게이트부터 시작하십시오.
$X = (A \cdot B) + \overline{C}$가 $A=1, B=1, C=1$일 때, $X$은 무엇입니까?
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
$A \cdot B = 1$, 그리고 $\overline{C} = 0$. $X = 1 + 0 = 1$.
Worked example: circuit to expression
- Work forwards from the inputs and label every intermediate output.
- In the circuit below, B passes through a NOT gate. A and NOT B feed an AND gate: call its output P, so
P = A AND NOT B. B and C feed a second AND gate:Q = B AND C. - P and Q feed the OR gate, so
X = P OR Q = (A AND NOT B) OR (B AND C). - Labelling P and Q is not decoration. It is what lets you fill the truth table one gate at a time.
Label every intermediate output before you write the expression
작업된 회로에서 두 번째 AND 게이트의 출력은 Q = B AND ____입니다.
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
B와 C가 하단 AND 게이트에 공급됩니다. 그 출력을 Q로 표시하면 진리표를 게이트마다 채울 수 있습니다.
Circuit to truth table
- For $n$ inputs there are $2^{n}$ rows: two inputs give 4, three give 8, four give 16. List them in binary counting order.
- Give the table a column for each intermediate output as well as the final one, so every row is checked one gate at a time.
- Fill the columns left to right: first the NOT, then each AND, then the OR.
입력이 3개인 회로의 진리표에는 몇 개의 행이 있습니까?
난이도 2/5 · 140 코인 + 오늘 첫 완료 시 소액 보너스
$2^n$행; 입력이 3개인 경우, $2^3 = 8$.
입력이 4개인 경우 몇 행입니까?
난이도 2/5 · 140 코인 + 오늘 첫 완료 시 소액 보너스
$2^4 = 16$행.
Worked example: the eight rows
- The circuit from the worked example above, with columns for NOT B, P, Q and X:
| A | B | C | NOT B | P | Q | X |
|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 1 | 0 | 0 | 0 |
| 0 | 0 | 1 | 1 | 0 | 0 | 0 |
| 0 | 1 | 0 | 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 0 | 0 | 1 | 1 |
| 1 | 0 | 0 | 1 | 1 | 0 | 1 |
| 1 | 0 | 1 | 1 | 1 | 0 | 1 |
| 1 | 1 | 0 | 0 | 0 | 0 | 0 |
| 1 | 1 | 1 | 0 | 0 | 1 | 1 |
- X is 1 in four of the eight rows. Check any one against the expression: row 1 0 0 has A = 1 and NOT B = 1, so P = 1, so X = 1.
(X = (A AND NOT B) OR (B AND C))에서 X = 1가 되는 입력 행은 무엇입니까? 적용되는 모든 것을 선택하십시오.
난이도 2/5 · 140 코인 + 오늘 첫 완료 시 소액 보너스
행 011은 Q = 1을给出; 행 100은 P = 1을给出합니다. 행 110에서는 B = 1이 P를, C = 0이 Q를 소멸시킵니다; 행 001에서는 두 항이 모두 0입니다.
Truth table to expression: sum of products
- Sum of products 积之和: for each row whose output is 1, write an AND of the inputs, putting NOT on any input that is 0 in that row. Then OR those terms together.
- A table that is 1 only on (A = 0, B = 1) and (A = 1, B = 0) gives
(NOT A AND B) OR (A AND NOT B), which is exactly A XOR B. - Simplify only if the question asks; the unsimplified sum of products earns the marks.
항의 곱합법에서 출력이 1인 각 행에 대해 다음과 같이 씁니다:
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
각 1행은 AND 항(0인 입력 제외)이 되며, 이들을 모두 OR합니다.
항의 곱합법에서 출력이 1인 각 행에 대한 AND 항을 모두 OR합니다.
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
각 출력-1 행은 AND 항(0인 입력에 NOT)이 되며, 이들을 OR하면 정확히 해당 행에서만 1이 되는 식이 됩니다.
The half adder
- Adding two bits gives a sum bit and a carry bit: 1 + 1 = 10 in binary.
- The sum column is 0, 1, 1, 0, which is XOR. The carry column is 0, 0, 0, 1, which is AND. Two gates make a half adder 半加器.
- It is the standard example of reading gates straight off a truth table, and the first piece of every processor's arithmetic unit.
Each input pair flows through the gates: XOR gives the sum, AND gives the carry, so 1 + 1 = 10
하위 가산기
XOR 및 AND를 동일한 두 입력에 연결하십시오: XOR은 합 비트를给出, AND는 카리를给出합니다. A와 B를 클릭하십시오.
하위 가산기에서 카리 비트를 생성하는 게이트는 무엇입니까?
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
카리가 1이 되려면 두 비트가 모두 1이어야 하므로 이는 AND입니다. 합이 1이 되려면 비트가 다르므로 이는 XOR입니다.
From a problem statement
- Turn the English into Boolean one clause at a time before drawing anything.
- "A and B" is
A AND B. "A or B, or both" isA OR B. "exactly one of A and B" isA XOR B. - "neither A nor B" is
A NOR B. "not both" isA NAND B. "unless" usually means AND NOT.
각 영어 구절을 구현하는 단일 게이트와 매칭하십시오.
난이도 2/5 · 140 코인 + 오늘 첫 완료 시 소액 보너스
both = AND, differ = XOR, neither = NOR, not-both = NAND입니다.
Worked example: a machine alarm
- An alarm X sounds when the guard is open (A = 1) and either the motor is running (B = 1) or the temperature is high (C = 1).
- "Either B or C" is
B OR C. "A and that" isX = A AND (B OR C). The brackets around the OR are essential. - X = 1 needs A = 1 and at least one of B, C equal to 1: the rows (1, 0, 1), (1, 1, 0) and (1, 1, 1). Three rows out of eight, and A = 0 can never sound the alarm.
- The circuit is one OR gate on B and C feeding one AND gate with A.
경보 X = A AND (B OR C)의 경우, 경보는 A = 0일 때 B와 C가 모두 1이라면 울릴 수 있습니다.
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
A와의 AND는 X = 1인 모든 행에서 A = 1이어야 함을 의미합니다. A = 0일 때 B와 C가 어떻게 되더라도 출력은 0입니다.
Marks that slip away
A AND B OR Cwithout brackets is ambiguous, and the examiner reads it as you did not intend. Bracket the OR before ANDing it.- A wire that goes nowhere, or an output with no label, loses the circuit mark even when the gates are right.
- A NOT on an input is a gate on the diagram, not a bar written over the letter.
- Three inputs mean eight rows. A table with fewer rows cannot be marked as complete.
You've got it
- expression → circuit: one gate per operator, innermost brackets first, inputs left, labelled output right
- circuit → expression: label each intermediate output (P, Q) and combine; circuit → truth table: $2^{n}$ rows with a column per gate
- truth table → expression: sum of products, one AND-term per output-1 row, ORed together; XOR + AND make the half adder
- problem → Boolean: exactly one = XOR, neither = NOR, not both = NAND; bracket the OR inside an AND