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Question

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?

The correct answer is

g2 (h3 h2 h1 h0) = Σ(2,4,5,6,7,12,13,15)

Gray Code Problem: Analysis

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 $h_3 h_2 h_1 h_0$.

Gray Code Generation: n and n+1

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

Function Derivation: g2

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:

$$ g_2(h_3 h_2 h_1 h_0) = \Sigma(2, 4, 5, 6, 7, 12, 13, 15) $$

Result Verification: Function Comparison

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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Important Questions from Combinational Circuits

  1. Which of the following represents the borrow in a half subtraction?

  2. Carry input is present in which of the following?

  3. The number of control lines for 8-to-1 multiplexer is:

  4. X/Y (where X represents the input code and Y represents the output code) as a qualifying symbol does NOT relate to:

  5. The time delay in a look-ahead carry adder is independent of

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