Answers
PART (A)
As per the circuit diagram we analysis the circuit and get the output as per the table
The output C is the circuit of the NAND gate and the output D is the circuit of AND gate: So the truth table for the given circuit is
| A(V) | B(V) | C(V) | D(V) |
| 0 | 0 | 10 | 0 |
| 0 | 10 | 10 | 0 |
| 10 | 0 | 10 | 0 |
| 10 | 10 | 0 | 10 |
2. So we can find the equation of Output C and D using truth table:
The output C is the equation of NAND gate.
D=AB
The output D is the equation of AND gate.
3. A and B are the inputs, these inputs are given to CMOS circuit of AND gate.
From the boolean equation of D, we conclude that for implement of D, the AND gate are required.
PART B.
Given function :
Apply the De-morgen theorem:
So simplified boolean equation is:
............(1)
1. From equation 1, the gate-level diagram using inputs A,B and C using basic logic gate
2.From the boolean equation (1) , we can find the value of F for the inputs A, B and C
| A | B | C | F |
| 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 0 |
| 1 | 1 | 0 | 0 |
| 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 1 |
C = AB + ĀB + AB
C = Ā+B
C = AB
F = (A + B).C
F = A + B + C
F = ĀB+C
o ĀB ĀBtc te
.
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