void setup() {

}

void loop() {
  digitalWrite(13, HIGH);
  delay(1);
  digitalWrite(13, LOW);
  delay(1000);
}

Fault 1: setup() is empty — it needs pinMode(13, OUTPUT);. Fault 2: delay(1) is 1 millisecond, not 1 second — it should be delay(1000).

void setup() {
  pinMode(2, INPUT);
  pinMode(8, OUTPUT);
}

void loop() {
  if (digitalRead(2) = HIGH) {
    digitalWrite(8, HIGH);
    delay(2000);
    digitalWrite(8, LOW);
  }

Fault 1: the test uses = (assignment) instead of == (comparison) — it should be if (digitalRead(2) == HIGH). Fault 2: the closing brace } for loop() is missing.

1. An analogue signal can take any value in a range and changes smoothly; a digital signal has only two values (1 or 0).
2. Any two: push button, limit switch, guard switch, infrared sensor.
3. Any two: lamp/LED, buzzer, motor, relay.
4. Near the supply voltage (about 5 V).
5. Near 0 V.
6. Logic 1.
7. The guard is not in place (0 = not detected).
8. e.g. a buzzer sounding an alarm when a fault is detected.
9. A digital signal has only two widely-separated levels, so small amounts of noise cannot change a 1 into a 0 — the signal is still read correctly.
10. Analogue.

1. AND, OR and NOT.
2. The output is 1 only when all inputs are 1.
3. The output is 1 when any input is 1.
4. The output is the opposite of the input.
5. AND: Q column reads 0, 0, 0, 1.
6. OR: Q column reads 0, 1, 1, 1.
7. AND.
8. OR.
9. 0.
10. Because its output is always the inverse (opposite) of its input.

• (4a) Q = A OR (NOT B) — X = NOT B: rows (A,B → X,Q): 0,0 → 1,1 · 0,1 → 0,0 · 1,0 → 1,1 · 1,1 → 0,1
• (4b) Q = (NOT A) AND B — X = NOT A: 0,0 → 1,0 · 0,1 → 1,1 · 1,0 → 0,0 · 1,1 → 0,0
• (Task 5) Q = (A AND B) OR C — X = A AND B: Q column in binary order (000→111): 0, 1, 0, 1, 0, 1, 1, 1 (X is 1 only for 110 and 111).

1. rows = 2ⁿ for n inputs.
2. 4 rows.
3. 8 rows.
4. Logic where the output depends only on the current combination of inputs (no memory).
5. The intermediate column works out the inner gate first, so each row becomes two simple steps instead of one complicated one.
6. Boolean: AND: Q = A · B · OR: Q = A + B · NOT: Q = NOT A.
7. Fan: Q = A · B · (NOT C) — temperature on AND master on AND safety NOT triggered.
8. Q = (A · B) + C for the diagram in question 7.
9. "On when A, OR when both B and C": Q = A + (B · C).
10. Draw the inner gate(s) first, then the final gate — check with a truth table.

1. AND — Q = A · B (both limit switches pressed).
2. OR — Q = A + B + C (any window).
3. Q = A · B · (NOT C) — start AND guard closed AND e-stop NOT pressed; diagram: NOT on C, three-way AND.
4. OR — Q = A + B (too hot OR too damp).
5. AND.
6. AND.
7. Q = A · B (hot AND lid open).
8. Inputs: tank-full sensor (A), master switch (B); output: pump. Q = (NOT A) · B.
9. Work inside-out: NOT first, then the inner gate, then the output gate.
10. Evaluate the expression for every input combination, row by row.

1. A program (or section) that repeats forever until the power is removed.
2. A section that repeats a set number of times, then the program continues.
3. A pause for a set time — the outputs stay in their current state while the program waits.
4. A yes/no decision — the program follows a different path depending on the answer.
5. Fixed loop (6 flashes, then stop).
6. Continuous loop (watches the sensor forever).

1. START → Lamp (output, pin 13) ON → wait 4 s → Lamp OFF → STOP.
2. START → Motor (output) ON → wait 5 s → Buzzer (output) ON → wait 2 s → Buzzer OFF → Motor OFF → STOP.
3. START → LED ON → wait 1 s → LED OFF → wait 1 s → arrow back to LED ON (continuous loop — no STOP).
4. START → read button (input) → decision "button pressed?" — No: loop back to read button; Yes: Buzzer ON → wait 2 s → Buzzer OFF → loop back to read button.
5. START → set counter = 0 → LED ON → wait 0.5 s → LED OFF → wait 0.5 s → add 1 to counter → decision "counter = 5?" — No: loop back to LED ON; Yes: STOP.

1. setup() runs once when the board switches on — used to set which pins are inputs or outputs.
2. loop() runs forever after setup() finishes — the main control program goes here.
3. digitalWrite(13, HIGH);
4. pinMode(4, INPUT);
5. 2500 ms = 2.5 seconds.
6. Pseudocode: for count = 1 to 4 / high LED / pause 0.5 seconds / low LED / pause 0.5 seconds / next count / end.
7. setup(): pinMode(2, INPUT); pinMode(8, OUTPUT); — loop(): if (digitalRead(2) == HIGH) { digitalWrite(8, HIGH); delay(3000); digitalWrite(8, LOW); }
8. Put inside void loop(): digitalWrite(pin, HIGH); delay(1000); digitalWrite(pin, LOW); delay(1000); — with pinMode in setup().

• Debug 1: setup() is missing pinMode(13, OUTPUT); and delay(1) should be delay(1000) (1 ms vs 1 s).
• Debug 2: the if test needs == not =, and the closing brace } of loop() is missing.