Consider 4-bit gray code representation of numbers. Let h3 h2 h1 h0 be the gray code representation of a number n and g1g2g3g0 be the gray code representation of the number (n + 1) modulo 16. Which one of the following functions is correct?
g2 (h3 h2 h1 h0) = Σ(2,4,5,6,7,12,13,15)
The question asks us to identify the correct Boolean function that maps the 4-bit Gray code representation of a number n, denoted as $h_3 h_2 h_1 h_0$, to a specific bit ($g_2$) of the Gray code representation of the next number $(n + 1) \pmod{16}$. We need to find which of the given functions accurately represents $g_2$. The notation $\Sigma$ represents the sum of minterms for which the function evaluates to 1, based on the decimal value of the input Gray code $h_3 h_2 h_1 h_0$.
To solve this, we first list the standard 4-bit Binary-Reflected Gray Code (BRGC) sequence. We construct a truth table that shows the Gray code for n ($H = h_3h_2h_1h_0$), its corresponding decimal value (which serves as the minterm index), the next number $(n+1) \pmod{16}$, and the Gray code for that next number ($G = g_3g_2g_1g_0$). We specifically track the value of the $g_2$ bit.
| n | Binary Representation of n | Gray Code of n ($H = h_3h_2h_1h_0$) | Decimal Value of $H$ (Minterm) | n + 1 (mod 16) | Binary Representation of n + 1 | Gray Code of n + 1 ($G = g_3g_2g_1g_0$) | $g_2$ |
|---|---|---|---|---|---|---|---|
| 0 | 0000 | 0000 | 0 | 1 | 0001 | 0001 | 0 |
| 1 | 0001 | 0001 | 1 | 2 | 0010 | 0011 | 0 |
| 2 | 0010 | 0011 | 3 | 3 | 0011 | 0010 | 0 |
| 3 | 0011 | 0010 | 2 | 4 | 0100 | 0110 | 1 |
| 4 | 0100 | 0110 | 6 | 5 | 0101 | 0111 | 1 |
| 5 | 0101 | 0111 | 7 | 6 | 0110 | 0101 | 1 |
| 6 | 0110 | 0101 | 5 | 7 | 0111 | 0100 | 1 |
| 7 | 0111 | 0100 | 4 | 8 | 1000 | 1100 | 1 |
| 8 | 1000 | 1100 | 12 | 9 | 1001 | 1101 | 1 |
| 9 | 1001 | 1101 | 13 | 10 | 1010 | 1111 | 1 |
| 10 | 1010 | 1111 | 15 | 11 | 1011 | 1110 | 1 |
| 11 | 1011 | 1110 | 14 | 12 | 1100 | 1010 | 0 |
| 12 | 1100 | 1010 | 10 | 13 | 1101 | 1011 | 0 |
| 13 | 1101 | 1011 | 11 | 14 | 1110 | 1001 | 0 |
| 14 | 1110 | 1001 | 9 | 15 | 1111 | 1000 | 0 |
| 15 | 1111 | 1000 | 8 | 0 | 0000 | 0000 | 0 |
By examining the truth table, we identify the rows where the output bit $g_2$ is equal to 1. The function $g_2$ should evaluate to 1 for the input Gray codes $H$ corresponding to the minterms (decimal values of $H$) where $g_2$ is 1.
From the table, $g_2 = 1$ when the input Gray code's decimal value is 2, 4, 5, 6, 7, 12, 13, or 15.
Therefore, the function $g_2$ expressed as a sum of minterms is:
$$ g_2(h_3 h_2 h_1 h_0) = \Sigma(2, 4, 5, 6, 7, 12, 13, 15) $$Comparing this derived function $g_2 = \Sigma(2, 4, 5, 6, 7, 12, 13, 15)$ with the provided options, we find that it exactly matches the function stated in Option 3.
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