Understanding the 2-Bit Comparator Condition

The problem asks for a minimized Boolean function, S, that is true (S=1) if the 2-bit binary number X is strictly greater than the 2-bit binary number Y.

Let X be $X_1X_0$, where $X_1$ is the Most Significant Bit (MSB) and $X_0$ is the Least Significant Bit (LSB).

Let Y be $Y_1Y_0$, where $Y_1$ is the MSB and $Y_0$ is the LSB.

The condition $X > Y$ is satisfied in two main situations:

• Situation 1: The MSB of X is greater than the MSB of Y. This happens only when $X_1=1$ and $Y_1=0$. In this case, X is always greater than Y, regardless of the values of $X_0$ and $Y_0$.
• Situation 2: The MSBs are equal ($X_1 = Y_1$), and the LSB of X is greater than the LSB of Y ($X_0 > Y_0$). This occurs when:
  • $X_1=0$, $Y_1=0$, $X_0=1$, $Y_0=0$.
  • $X_1=1$, $Y_1=1$, $X_0=1$, $Y_0=0$.

Deriving the Unminimized Boolean Function

Translating these conditions into Boolean logic:

• Situation 1 ($X_1=1, Y_1=0$) corresponds to the Boolean expression: $X_1 \overline{Y_1}$.
• Situation 2a ($X_1=0$, $Y_1=0$, $X_0=1$, $Y_0=0$) corresponds to: $\overline{X_1} X_0 \overline{Y_1} \overline{Y_0}$.
• Situation 2b ($X_1=1$, $Y_1=1$, $X_0=1$, $Y_0=0$) corresponds to: $X_1 X_0 Y_1 \overline{Y_0}$.

The function S is the logical OR of these terms:

$S = X_1 \overline{Y_1} + \overline{X_1} X_0 \overline{Y_1} \overline{Y_0} + X_1 X_0 Y_1 \overline{Y_0}$

Minimizing the Function using Karnaugh Map (K-Map)

To simplify the expression, we identify the specific input combinations (minterms) where S=1:

• The term $X_1 \overline{Y_1}$ covers minterms where $X_1=1$ and $Y_1=0$. These are 1000, 1001, 1100, 1101.
• The term $\overline{X_1} X_0 \overline{Y_1} \overline{Y_0}$ covers minterm 0100.
• The term $X_1 X_0 Y_1 \overline{Y_0}$ covers minterm 1110.

In summary, S=1 for minterms: 0100, 1000, 1001, 1100, 1101, 1110.

We use a 4-variable Karnaugh map (K-map) with variables $X_1, X_0, Y_1, Y_0$. The cells corresponding to S=1 are filled:

Grouping the adjacent 1s in the K-map to achieve the minimum number of terms:

• A group of four 1s covers the cells corresponding to minterms 0100, 1100, 1000, 1001. This group simplifies to $X_1 \overline{Y_1}$.
• A group of two 1s covers the cells corresponding to minterms 0100 and 1100. This group simplifies to $X_0 \overline{Y_1} \overline{Y_0}$.
• A group of two 1s covers the cells corresponding to minterms 1101 and 1110. This group simplifies to $X_1 X_0 \overline{Y_0}$.

The minimized function S is the OR of these simplified terms:

$S = X_1 \overline{Y_1} + X_0 \overline{Y_1} \overline{Y_0} + X_1 X_0 \overline{Y_0}$

Comparing the Minimized Function with Option 2

Option 2 is given as $X_1\overline{Y_1} + X_0\overline{Y_0}\overline{Y_1} + X_0\overline{Y_0}X_1$.

Let's simplify Option 2. We can factor the last two terms:

$X_0\overline{Y_0}\overline{Y_1} + X_0\overline{Y_0}X_1 = X_0\overline{Y_0}(\overline{Y_1} + X_1)$

So, Option 2 can be rewritten as: $S_{opt2} = X_1\overline{Y_1} + X_0\overline{Y_0}(\overline{Y_1} + X_1)$

Expanding this expression:

$S_{opt2} = X_1\overline{Y_1} + X_0\overline{Y_0}\overline{Y_1} + X_0\overline{Y_0}X_1$

Let's further expand the term $X_0\overline{Y_0}X_1$ using the identity $A = A(B+\overline{B})$:

$X_0\overline{Y_0}X_1 = X_0\overline{Y_0}X_1(Y_1 + \overline{Y_1})$

$X_0\overline{Y_0}X_1 = X_1 X_0 Y_1 \overline{Y_0} + X_1 X_0 \overline{Y_1} \overline{Y_0}$

Substituting this back yields:

$S_{opt2} = X_1\overline{Y_1} + X_0\overline{Y_0}\overline{Y_1} + X_1 X_0 Y_1 \overline{Y_0} + X_1 X_0 \overline{Y_1} \overline{Y_0}$

Now, let's verify our K-map result $S = X_1 \overline{Y_1} + X_0 \overline{Y_1} \overline{Y_0} + X_1 X_0 \overline{Y_0}$. Expanding the last term $X_1 X_0 \overline{Y_0}$:

$X_1 X_0 \overline{Y_0} = X_1 X_0 \overline{Y_0} (Y_1 + \overline{Y_1}) = X_1 X_0 Y_1 \overline{Y_0} + X_1 X_0 \overline{Y_1} \overline{Y_0}$

So, the K-map result is equivalent to:

$S = X_1 \overline{Y_1} + X_0 \overline{Y_1} \overline{Y_0} + X_1 X_0 Y_1 \overline{Y_0} + X_1 X_0 \overline{Y_1} \overline{Y_0}$

Comparing the expanded forms, we see that $S_{opt2}$ is identical to the minimized expression derived from the K-map. Therefore, Option 2 is the correct minimized Boolean function.