Each decimal digit is coded separately by its 4-bit binary value

The whole decimal number is converted to binary

Each digit is coded as digit + 3

A code where successive values differ by one bit

Which equations give the 3-bit binary to Gray converter?

G2 = B2, G1 = B2 XOR B1, G0 = B1 XOR B0

G2 = B2 XOR B1, G1 = B1 XOR B0, G0 = B0

G2 = B2 AND B1, G1 = B1 AND B0, G0 = B0

2081 Bhadra · 2+4 marks

BCD code and 3-bit binary-to-Gray converter

Short answer

BCD encodes each decimal digit in 4-bit binary (8421), valid 0000–1001. A 3-bit binary to Gray converter: G2 = B2, G1 = B2⊕B1, G0 = B1⊕B0, implemented with two XOR gates. Example: binary 101 yields Gray 111.

What is a BCD code? Design a 3-bit binary to Gray code converter.

BCD code

BCD (Binary Coded Decimal) represents each decimal digit separately by its 4-bit binary value (8421 weights). Only 0000–1001 are valid; 1010–1111 are invalid. Example: $(93)_{10}$ = 1001 0011 (BCD), while in binary $93 = 1011101_2$ . BCD is used in calculators, clocks and 7-segment displays.

3-bit binary to Gray code converter

Inputs: binary $B_2B_1B_0$ ( $B_2$ = MSB). Outputs: Gray $G_2G_1G_0$ . In Gray code, successive values differ in only one bit.

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 $B_1B_0$ ):

        B1B0
 B2    00  01  11  10
  0  |  0 | 0 | 1 | 1 |
  1  |  1 | 1 | 0 | 0 |

Logic circuit

 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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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