Question

For the logic gates shown below, the correct output is

• A. $A+B+C$
• B. $\overline{A}.\overline{B}.\overline{C}$
• C. $\overline{A}+\overline{B}+\overline{C}$
• D. $\overline{A.B+B.C}$

Hint:

Identify each gate and write its output in terms of its inputs.
A NAND gate with inputs tied together acts as a NOT gate. ($ \overline{A \cdot A} = \overline{A} $).
A NOR gate with inputs tied together acts as a NOT gate. ($ \overline{A+A} = \overline{A} $).
Use De Morgan's theorems for simplification: $\overline{X+Y} = \overline{X} \cdot \overline{Y}$ $\overline{X \cdot Y} = \overline{X} + \overline{Y}$
Other Boolean algebra rules: $X \cdot \overline{X}=0$, $X+\overline{X}=1$, $X \cdot X=X$, $X+X=X$, $X+XY=X$, $X(X+Y)=X$.
If derivation does not match options, double-check algebra or consider if the question/options might be flawed. Testing with specific input combinations can help verify equivalence or find errors.

Answer 1

The Correct Option is C

Step-by-step Logic Gate Evaluation:

1. Gate 1: AND gate → \( G_1 = A \cdot B \)
2. Gate 2: AND gate → \( G_2 = B \cdot C \)
3. Gate 3: NAND gate with both inputs as \( C \) → functions as NOT gate → \( G_3 = \overline{C} \)
4. Gate 4: OR gate with inputs \( G_1, G_2 \) → \( G_4 = G_1 + G_2 = A \cdot B + B \cdot C \)
5. Gate 5: NOR gate with inputs \( G_4 \) and \( G_3 \): \[ Y = \overline{G_4 + G_3} = \overline{(A \cdot B + B \cdot C) + \overline{C}} \]

Simplifying:

\[ Y = \overline{(A \cdot B + B \cdot C) + \overline{C}} \] Using De Morgan's Theorem: \[ Y = \overline{A \cdot B + B \cdot C} \cdot \overline{\overline{C}} = \overline{A \cdot B + B \cdot C} \cdot C \] Now simplify \( \overline{A \cdot B + B \cdot C} \): \[ \overline{A \cdot B + B \cdot C} = \overline{A \cdot B} \cdot \overline{B \cdot C} = (\overline{A} + \overline{B}) \cdot (\overline{B} + \overline{C}) \] Expand: \[ (\overline{A} + \overline{B}) \cdot (\overline{B} + \overline{C}) = \overline{B} + \overline{A} \cdot \overline{C} \] Multiply by \( C \): \[ Y = (\overline{B} + \overline{A} \cdot \overline{C}) \cdot C = \overline{B}C + \overline{A} \cdot \overline{C} \cdot C \] Since \( \overline{C} \cdot C = 0 \): \[ Y = \overline{B}C \]

✅ Final Simplified Output:

\[ \boxed{Y = \overline{B} \cdot C} \]

Comparison with Options:

• (a) \( A + B + C \) — clearly not correct
• (b) \( \overline{A} \cdot \overline{B} \cdot \overline{C} \) — output only 1 when all inputs are 0 — does not match
• (c) \( \overline{A} + \overline{B} + \overline{C} \) — equivalent to \( \overline{A \cdot B \cdot C} \), not same as \( \overline{B}C \)
• (d) \( \overline{A \cdot B + B \cdot C} \) — this is \( \overline{G_4} \), and not the final output (since \( G_3 \) is also involved)

Conclusion: The correct simplified expression is:

\[ \boxed{\overline{B} \cdot C} \]

However, none of the given options match this output. Therefore, the problem appears to have a flaw or mismatch between the logic diagram and the answer options.