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A logic circuit takes 4-BCD inputs {A, B, C and D) to give an output F. Output F is '1' if the input is an invalid BCD-code. The number of two input NAND gates required to implement the output Y is
- a)3
- b)4
- c)5
- d)6
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A logic circuit takes 4-BCD inputs {A, B, C and D) to give an output F...
Invalid BCD inputs are from decimal 10 to 11. The K-map is shown below:
Thus, output F = AB + AC, which can be implemented using three 2-input NAND gates as shown below,.
Thus, output F = AB + AC, which can be implemented using three 2-input NAND gates as shown below,.
Most Upvoted Answer
A logic circuit takes 4-BCD inputs {A, B, C and D) to give an output F...
Explanation:
To implement the output Y, which is true if the input is an invalid BCD code, we need to check if any of the BCD inputs (A, B, C, or D) is greater than 9.
BCD Code:
BCD code represents decimal digits using a 4-bit binary code. In BCD, each decimal digit is represented by a 4-bit binary number. The valid BCD code ranges from 0000 to 1001, which represents decimal digits 0 to 9. Any other combination is considered invalid.
Implementation:
To implement this logic, we can use a series of NAND gates.
Steps:
1. Convert each BCD input to its binary representation using 4 NAND gates each.
2. Compare each binary representation with the binary representation of 9 using 4 NAND gates each.
3. If any of the comparisons result in a logic 1 output, it means the BCD input is greater than 9 and is invalid.
4. Combine all the comparison outputs using a NAND gate to get the final output Y.
Calculating the Number of NAND Gates:
- Converting each BCD input to binary representation: 4 BCD inputs * 4 NAND gates each = 16 NAND gates.
- Comparing each binary representation with the binary representation of 9: 4 BCD inputs * 4 NAND gates each = 16 NAND gates.
- Combining all the comparison outputs using a NAND gate: 4 NAND gates.
Total Number of NAND Gates = 16 + 16 + 4 = 36
Therefore, the correct answer is option 'A' as it requires 36 NAND gates to implement the output Y.
To implement the output Y, which is true if the input is an invalid BCD code, we need to check if any of the BCD inputs (A, B, C, or D) is greater than 9.
BCD Code:
BCD code represents decimal digits using a 4-bit binary code. In BCD, each decimal digit is represented by a 4-bit binary number. The valid BCD code ranges from 0000 to 1001, which represents decimal digits 0 to 9. Any other combination is considered invalid.
Implementation:
To implement this logic, we can use a series of NAND gates.
Steps:
1. Convert each BCD input to its binary representation using 4 NAND gates each.
2. Compare each binary representation with the binary representation of 9 using 4 NAND gates each.
3. If any of the comparisons result in a logic 1 output, it means the BCD input is greater than 9 and is invalid.
4. Combine all the comparison outputs using a NAND gate to get the final output Y.
Calculating the Number of NAND Gates:
- Converting each BCD input to binary representation: 4 BCD inputs * 4 NAND gates each = 16 NAND gates.
- Comparing each binary representation with the binary representation of 9: 4 BCD inputs * 4 NAND gates each = 16 NAND gates.
- Combining all the comparison outputs using a NAND gate: 4 NAND gates.
Total Number of NAND Gates = 16 + 16 + 4 = 36
Therefore, the correct answer is option 'A' as it requires 36 NAND gates to implement the output Y.
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