Digital logic | Code Converters – Binary to/from Gray Code

Prerequisite – Number System and base conversions

Gray Code system is a binary number system in which every successful pair of numbers differs in only one bit. It is used in applications in which the normal sequence of binary numbers generated by the hardware may produce an error or ambiguity during the transition from one number to the next.
For example, the states of a system may change from 3(011) to 4(100) as- 011 — 001 — 101 — 100. Therefore there is a high chance of a wrong state being read while the system changes from the initial state to the final state.
This could have serious consequences for the machine using the information. The Gray code eliminates this problem since only one bit changes its value during any transition between two numbers.

Converting Binary to Gray Code –

Let b_0,\:b_1,\:b_2\:,\:and\:b_3 be the bits representing the binary numbers, where b_0 is the LSB and b_3 is the MSB, and
Let g_0,\:g_1,\:g_2\:,\:and\:g_3 be the bits representing the gray code of the binary numbers, where g_0 is the LSB and g_3 is the MSB.
The truth table for the conversion is-
 \begin{tabular}{||c|c|c|c||c|c|c|c||} \hline  \multicolumn{4}{||c||}{Binary} & \multicolumn{4}{|c||}{Gray Code}\\ \hline  b_3 & b_2 & b_1 & b_0 & g_3 & g_2 & g_1 & g_0 \\ \hline \hline  0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\  \hline  0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\  \hline  0 & 0 & 1 & 0 & 0 & 0 & 1 & 1 \\  \hline  0 & 0 & 1 & 1 & 0 & 0 & 1 & 0 \\  \hline \hline  0 & 1 & 0 & 0 & 0 & 1 & 1 & 0 \\  \hline  0 & 1 & 0 & 1 & 0 & 1 & 1 & 1 \\  \hline  0 & 1 & 1 & 0 & 0 & 1 & 0 & 1 \\  \hline  0 & 1 & 1 & 1 & 0 & 1 & 0 & 0 \\  \hline \hline  1 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\  \hline  1 & 0 & 0 & 1 & 1 & 1 & 0 & 1 \\  \hline  1 & 0 & 1 & 0 & 1 & 1 & 1 & 1 \\  \hline  1 & 0 & 1 & 1 & 1 & 1 & 1 & 0 \\  \hline \hline  1 & 1 & 0 & 0 & 1 & 0 & 1 & 0 \\  \hline  1 & 1 & 0 & 1 & 1 & 0 & 1 & 1 \\  \hline  1 & 1 & 1 & 0 & 1 & 0 & 0 & 1 \\  \hline  1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 \\  \hline \hline \end{tabular}
To find the corresponding digital circuit, we will use the K-Map technique for each of the gray code bits as output with all of the binary bits as input.
K-map for b_0

K-map for b_1

K-map for b_2

K-map for b_3

Corresponding minimized boolean expressions for gray code bits –
 g_0 = b_0b_1^\prime + b_1b_0^\prime = b_0 \oplus b_1\\ g_1 = b_2b_1^\prime + b_1b_2^\prime = b_1 \oplus b_2\\ g_2 = b_2b_3^\prime + b_3b_2^\prime = b_2 \oplus b_3\\ g_3 = b_3
The corresponding digital circuit –

Converting Gray Code to Binary –

Converting gray code back to binary can be done in a similar manner.
Let b_0,\:b_1,\:b_2\:,\:and\:b_3 be the bits representing the binary numbers, where b_0 is the LSB and b_3 is the MSB, and
Let g_0,\:g_1,\:g_2\:,\:and\:g_3 be the bits representing the gray code of the binary numbers, where g_0 is the LSB and g_3 is the MSB.
Truth table-
 \begin{tabular}{||c|c|c|c||c|c|c|c||} \hline  \multicolumn{4}{||c||}{Gray Code} & \multicolumn{4}{|c||}{Binary}\\ \hline  g_3 & g_2 & g_1 & g_0 & b_3 & b_2 & b_1 & b_0\\ \hline \hline  0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\  \hline  0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\  \hline  0 & 0 & 1 & 0 & 0 & 0 & 1 & 1 \\  \hline  0 & 0 & 1 & 1 & 0 & 0 & 1 & 0 \\  \hline \hline  0 & 1 & 0 & 0 & 0 & 1 & 1 & 1 \\  \hline  0 & 1 & 0 & 1 & 0 & 1 & 1 & 0 \\  \hline  0 & 1 & 1 & 0 & 0 & 1 & 0 & 0 \\  \hline  0 & 1 & 1 & 1 & 0 & 1 & 0 & 1 \\  \hline \hline  1 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\  \hline  1 & 0 & 0 & 1 & 1 & 1 & 1 & 0 \\  \hline  1 & 0 & 1 & 0 & 1 & 1 & 0 & 0 \\  \hline  1 & 0 & 1 & 1 & 1 & 1 & 0 & 1 \\  \hline \hline  1 & 1 & 0 & 0 & 1 & 0 & 0 & 0 \\  \hline  1 & 1 & 0 & 1 & 1 & 0 & 0 & 1 \\  \hline  1 & 1 & 1 & 0 & 1 & 0 & 1 & 1 \\  \hline  1 & 1 & 1 & 1 & 1 & 0 & 1 & 0 \\  \hline \hline \end{tabular}

Using K-map to get back the binary bits from the gray code –
K-map for b_0

K-map for b_1

K-map for b_2

K-map for b_3

Corresponding Boolean expressions –

     \begin{align*} b_0 &=g_3^\prime g_2^\prime g_1^\prime g_0 + g_3^\prime g_2^\prime g_1g_0^\prime + g_3^\prime g_2g_1^\prime g_0^\prime  + g_3^\prime g_2g_1g_0 +g_3g_2^\prime g_1^\prime g_0^\prime  + g_3g_2^\prime g_1g_0 \\ &\:\:\:+g_3g_2g_1^\prime g_0 + g_3g_2g_1g_0^\prime \\ &= g_3^\prime g_2^\prime( g_1^\prime g_0 +  g_1g_0^\prime) + g_3^\prime g_2(g_1^\prime g_0^\prime  + g_1g_0) +g_3g_2^\prime(g_1^\prime g_0^\prime  +  g_1g_0 )\\ &\:\:\:+g_3g_2 (g_1^\prime g_0 + g_1g_0^\prime) \\ &= g_3^\prime g_2^\prime(g_0\oplus g_1) + g_3^\prime g_2(g_0\odot g_1)+g_3g_2^\prime(g_0\odot g_1) + g_3g_2 (g_0\oplus g_1) \\ &= (g_0\oplus g_1)(g_2\odot g_3) + (g_0\odot g_1)(g_2\oplus g_3)\\ &= g_3\oplus g_2\oplus g_1\oplus g_0\\ b_1 &= g_3^\prime g_2^\prime g_1 + g_3^\prime g_2g_1^\prime  + g_3g_2g_1 + g_3g_2^\prime g_1^\prime \\ &= g_3^\prime(g_2^\prime g_1 + g_2g_1^\prime)  + g_3(g_2g_1 + g_2^\prime g_1^\prime) \\ &= g_3^\prime(g_2\oplus g_1)  + g_3(g_2\odot g_1) \\ &= g_3\oplus g_2\oplus g_1\\ b_2 &= g_3^\prime g_2 + g_3g_2^\prime\\ &= g_3\oplus g_2\\ b_3 &= g_3 \end{align*}

Corresponding digital circuit –

References –

Digital Design, 5th edition by Morris Mano and Michael Ciletti

This article is contributed by Chirag Manwani. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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