Question

If the outputs of XNOR and XOR gates with inputs a, b are connected to an AND gate, then the output is:

• A. \(a'b + ab'\)
• B. \(ab + a'b'\)
• C. \(0\)
• D. \(1\)

Hint:

Any Boolean function directly ANDed with its own inverted function always yields a static logical 0 (\(A \cdot A' = 0\)), regardless of how complex the underlying expression is.

Answer 1

The Correct Option is C

Concept: Boolean algebra provides the mathematical framework for analyzing and designing digital logic circuits. To find the final expression for this composite logic network, we must analyze each gate individually using foundational Boolean laws.
• XOR Gate (Exclusive OR): This gate yields a high output (\(1\)) if and only if its two inputs are distinct. For inputs \(a\) and \(b\), its algebraic representation is: \[ Y_{\text{XOR}} = a \oplus b = a'b + ab' \]
• XNOR Gate (Exclusive NOR): This gate is the logical complement of the XOR operation, yielding a high output only when both inputs are identical. Its algebraic representation is: \[ Y_{\text{XNOR}} = \overline{a \oplus b} = ab + a'b' \]
• AND Gate: This gate performs logical multiplication (conjunction) on its inputs, requiring all inputs to be high to output a \(1\).
• Law of Complementarity: In Boolean algebra, any variable or complex expression combined with its absolute complement via an AND operation always results in a logical zero: \[ A \cdot A' = 0 \]

Step 1: Setting up the composite output equation.
The outputs of the XOR gate and the XNOR gate are routed as the primary inputs to a standard 2-input AND gate. Let the final output of the AND gate be denoted by \(Y\). Mathematically, we can express this conjunction as: \[ Y = (Y_{\text{XOR}}) \cdot (Y_{\text{XNOR}}) \] Substituting the operational definitions for both individual gates into this expression gives: \[ Y = (a \oplus b) \cdot (\overline{a \oplus b}) \]

Step 2: Simplifying the expression using Boolean properties.
Let us substitute the complex Boolean sub-expression \((a \oplus b)\) with a single dummy variable, \(W\): \[ W = a \oplus b \] Consequently, its logical complement can be expressed as: \[ W' = \overline{a \oplus b} \] Now, substitute these terms back into our output equation for \(Y\): \[ Y = W \cdot W' \] According to the foundational Boolean Law of Complementarity, a variable ANDed with its own inverse can never be true simultaneously, as one state will always be \(0\). Therefore: \[ W \cdot W' = 0 \quad \Rightarrow \quad Y = 0 \]

Step 3: Direct algebraic expansion verification.
To ensure complete mathematical thoroughness, let us expand the expressions fully using basic Boolean multiplication rules to double-check the result: \[ Y = (a'b + ab') \cdot (ab + a'b') \] Distribute the terms across the brackets: \[ Y = (a'b \cdot ab) + (a'b \cdot a'b') + (ab' \cdot ab) + (ab' \cdot a'b') \] Rearranging the individual variables within each product term: \[ Y = (a'a \cdot bb) + (a'a' \cdot bb') + (aa \cdot b'b) + (aa' \cdot b'b') \] Using the Boolean properties \(x \cdot x = x\) and \(x \cdot x' = 0\), analyze each term:
• First term: \(a'a \cdot bb = 0 \cdot b = 0\)
• Second term: \(a' \cdot bb' = a' \cdot 0 = 0\)
• Third term: \(a \cdot b'b = a \cdot 0 = 0\)
• Fourth term: \(aa' \cdot b' = 0 \cdot b' = 0\) Summing these results gives: \[ Y = 0 + 0 + 0 + 0 = 0 \] Both analytical approaches show that the output value remains consistently at \(0\), verifying Option (C).