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Question
Draw truth table and write Boolean function for following circuit.
Very Long Answer
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Solution
(i) Boolean expression at different stages:
X = (A′ + B′)′
After NOT gate:
X′ = A′ + B′
Lower AND gate:
AB
Therefore,
Y = (A′ + B′)AB
Since AB requires A = 1, B = 1, while A′ + B′ = 0 for A = 1, B = 1,
Y = 0
| A | B | A' | B' | A'+B' | AB | Y |
|---|---|---|---|---|---|---|
| 0 | 0 | 1 | 1 | 1 | 0 | 0 |
| 0 | 1 | 1 | 0 | 1 | 0 | 0 |
| 1 | 0 | 0 | 1 | 1 | 0 | 0 |
| 1 | 1 | 0 | 0 | 0 | 1 | 0 |
(ii) The first gate is an XOR gate.
X = A′ ⊕ B
The output is ANDed with B and then complemented:
Y = [(A′ ⊕ B)B]'
Simplified:
Y = (AB)′
| A | B | A' | A′ ⊕ B | (A′ ⊕ B)B | Y |
|---|---|---|---|---|---|
| 0 | 0 | 1 | 1 | 0 | 1 |
| 0 | 1 | 1 | 0 | 0 | 1 |
| 1 | 0 | 0 | 0 | 0 | 1 |
| 1 | 1 | 0 | 1 | 1 | 0 |
(iii) First NAND gate:
X = (A′B)′
The output of this gate and CC are given to a NOR gate:
Y = [(A′B)′ + C]′
Simplifying,
Y = A′BC′
| A | B | C | A' | A'B | (A′B)′ | Y |
|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 1 | 0 | 1 | 0 |
| 0 | 0 | 1 | 1 | 0 | 1 | 0 |
| 0 | 1 | 0 | 1 | 1 | 0 | 1 |
| 0 | 1 | 1 | 1 | 1 | 0 | 0 |
| 1 | 0 | 0 | 0 | 0 | 1 | 0 |
| 1 | 0 | 1 | 0 | 0 | 1 | 0 |
| 1 | 1 | 0 | 0 | 0 | 1 | 0 |
| 1 | 1 | 1 | 0 | 0 | 1 | 0 |
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