ITPEC FE Morning April 2018 Question 3
ITPEC FE Morning April 2018 — Question 3 of 80
NAND Gate — Building OR from NAND — expressing X OR Y using only NAND operations.
NAND is a universal gate: X NAND Y = NOT(X AND Y).
To build OR from NAND, apply De Morgan's law: X OR Y = NOT(NOT X AND NOT Y).
Step-by-step construction:
- NOT X = X NAND X (since NOT(X AND X) = NOT X)
- NOT Y = Y NAND Y
- X OR Y = (X NAND X) NAND (Y NAND Y), which computes NOT(NOT X AND NOT Y) = X OR Y
Verification with truth table:
| X | Y | X NAND X | Y NAND Y | Result NAND | X OR Y |
|---|---|---|---|---|---|
| 0 | 0 | 1 | 1 | 0 | 0 |
| 0 | 1 | 1 | 0 | 1 | 1 |
| 1 | 0 | 0 | 1 | 1 | 1 |
| 1 | 1 | 0 | 0 | 1 | 1 |
All rows match — confirmed.
Why not others:
- (a) ((X NAND Y) NAND X) NAND Y — fails for X=0, Y=0 (gives 1 instead of 0)
- (c) (X NAND Y) NAND (X NAND Y) — this is NOT(NAND(X,Y)) = X AND Y, not OR
- (d) X NAND (Y NAND (X NAND Y)) — fails for X=1, Y=1 (gives 0 instead of 1)
Key rule: OR from NAND = negate each input with itself, then NAND the results.
AI-generated — may contain errors
The original exam layout is preserved in the image so diagrams, formulas, tables, and code remain accurate.
This question comes from an official ITPEC past paper. ITPEC Practice is an independent study tool and is not affiliated with ITPEC. See the official FE past-paper collection or Report an issue.