--- title: "Gray code and 3-bit binary-to-Gray converter" subject: "Digital Logic" chapter: 1 key: "semester_3/digital_logic:gray-code-binary-gray-converter" source: https://www.bctengineer.com/q/semester_3/digital_logic/gray-code-binary-gray-converter site: BCT Engineering --- # Gray code and 3-bit binary-to-Gray converter Digital Logic · Chapter 1: Introduction - Asked in: 2080 Bhadra, 2+3+1 marks - Video: [Binary Code to Gray Code Converter: Circuit, Truth Table, and Designing](https://www.youtube.com/watch?v=Pk9YouWUwHI) — Engineering Funda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/digital_logic/gray-code-binary-gray-converter - Whole chapter: https://www.bctengineer.com/md/semester_3/digital_logic/chapter-1 ## Question Define Gray Code. Design 3-bit binary to gray code converter circuit with necessary truth table and circuit diagram. ## Short answer Gray code: successive numbers differ by one bit. A 3-bit binary to Gray converter: truth table gives G2=B2, G1=B2⊕B1, G0=B1⊕B0. Circuit uses two XOR gates. Example: binary 010 yields Gray 011. ## Answer #### Gray code **Gray code** is an unweighted, reflected binary code in which **successive numbers differ in only one bit position**. Examples: 0 = 000, 1 = 001, 2 = 011, 3 = 010. It is used in shaft encoders and K-maps because only one bit changes at each step, avoiding false intermediate values. #### 3-bit binary to Gray code converter Inputs: $B_2B_1B_0$ (binary, $B_2$ = MSB). Outputs: $G_2G_1G_0$. #### Truth table (3-bit binary to Gray) | $B_2$ | $B_1$ | $B_0$ | $G_2$ | $G_1$ | $G_0$ | |---|---|---|---|---|---| | 0 | 0 | 0 | 0 | 0 | 0 | | 0 | 0 | 1 | 0 | 0 | 1 | | 0 | 1 | 0 | 0 | 1 | 1 | | 0 | 1 | 1 | 0 | 1 | 0 | | 1 | 0 | 0 | 1 | 1 | 0 | | 1 | 0 | 1 | 1 | 1 | 1 | | 1 | 1 | 0 | 1 | 0 | 1 | | 1 | 1 | 1 | 1 | 0 | 0 | #### Simplification (K-maps) - $G_2 = \sum m(4,5,6,7) = B_2$ - $G_1 = \sum m(2,3,4,5) = \bar B_2 B_1 + B_2\bar B_1 = B_2 \oplus B_1$ - $G_0 = \sum m(1,2,5,6) = \bar B_1 B_0 + B_1\bar B_0 = B_1 \oplus B_0$ K-map for $G_1$ (rows $B_2$, columns $B_1B_0$): ```text B1B0 B2 00 01 11 10 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | ``` #### Logic circuit ```text B2 ---+------------------------ G2 | +-------+ +--->| | | XOR |----------- G1 B1 ---+--->| | | +-------+ | +-------+ +--->| | | XOR |----------- G0 B0 ------->| | +-------+ ``` The converter needs only **two XOR gates**: the MSB passes straight through and each other Gray bit is the XOR of two neighbouring binary bits. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the defining property of Gray code? A. Each bit has weights 8, 4, 2, 1 B. Each decimal digit is coded as digit + 3 C. Successive numbers differ in exactly one bit D. Successive numbers differ in exactly two bits Answer: C. Gray code is a unit-distance code: consecutive values differ in only one bit. (Key point: Gray code definition) 2. Which equations give the 3-bit binary to Gray converter? A. G2 = B2 XOR B1, G1 = B1 XOR B0, G0 = B0 B. G2 = B2, G1 = B1, G0 = B0 C. G2 = B2 AND B1, G1 = B1 AND B0, G0 = B0 D. G2 = B2, G1 = B2 XOR B1, G0 = B1 XOR B0 Answer: D. The MSB passes straight through and each other Gray bit is the XOR of neighbouring binary bits. (Key point: 3-bit converter equations) 3. Which expression does the K-map of G1 give? A. B2 XOR B1 B. B2 AND B1 C. B1 XOR B0 D. B2 OR B1 Answer: A. G1 = m(2,3,4,5) = B2'B1 + B2B1' = B2 XOR B1. (Key point: K-map of G1) 4. How many gates does the 3-bit binary to Gray converter need? A. Three XOR gates B. Two XOR gates C. Two AND gates and one OR gate D. Three NOT gates Answer: B. G2 is a direct wire; G1 and G0 each need one XOR gate. 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