Question
Identify the logic gate represented by the following circuit by writing its truth table:
✓
Answer
The output of OR gate, C = A + B
The output of NAND gate, $\text{D}=\overline{\text{AB}}$
The input of AND gate are C and D, so it's output is $\text{Y}=\text{C}.\text{D}=(\text{A}+\text{B})\overline{\text{AB}}$
When $\text{A}=0,\text{ B}=0,$
$\text{ Y}=(0+0)(0\bar{\cdot}0)=(\text{}0\cdot\bar0)=0\cdot1=0$
When $\text{A}=1,\text{ B}=0,$
$\text{ Y}=(1+0)(1\bar\cdot0)=1\cdot\bar0=1\cdot1=1$
When $\text{A}=0,\text{ B}=1,$
$\text{ Y}=(0+1)(0\bar\cdot1)=1\cdot\bar0=1\cdot1=1$
When $\text{A}=1,\text{ B}=1,$
$\text{ Y}=(1+1)(1\bar\cdot1)=(1\cdot\bar1)=1\cdot0=0$
Thus, truth table of given circuit is:
Thia ia the truth table of XOR gate, hence the given circuit represents XOR gate.
The output of NAND gate, $\text{D}=\overline{\text{AB}}$
The input of AND gate are C and D, so it's output is $\text{Y}=\text{C}.\text{D}=(\text{A}+\text{B})\overline{\text{AB}}$
When $\text{A}=0,\text{ B}=0,$
$\text{ Y}=(0+0)(0\bar{\cdot}0)=(\text{}0\cdot\bar0)=0\cdot1=0$
When $\text{A}=1,\text{ B}=0,$
$\text{ Y}=(1+0)(1\bar\cdot0)=1\cdot\bar0=1\cdot1=1$
When $\text{A}=0,\text{ B}=1,$
$\text{ Y}=(0+1)(0\bar\cdot1)=1\cdot\bar0=1\cdot1=1$
When $\text{A}=1,\text{ B}=1,$
$\text{ Y}=(1+1)(1\bar\cdot1)=(1\cdot\bar1)=1\cdot0=0$
Thus, truth table of given circuit is:
| A | B | Y |
| 0 | 0 | 0 |
| 1 | 0 | 1 |
| 0 | 1 | 1 |
| 1 | 1 | 0 |
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