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A binary-to-BCD encoder has four inputs D0 C0, B0, and A0 and five outputs D, C, B, A, and VALID. The outputs D, C, B and A give the proper BCD value of the input and the VALID output is 1 if the input combination is a valid decimal code. If the input combination is an invalid decimal code, the VALID output becomes 0, and all of the D, C, B, and A outputs show 0 values. If only NOT gates and 2-input OR and AND gates are available, the minimum number of gates required to implement the above circuit is
- a)10
- b)9
- c)8
- d)7
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
A binary-to-BCD encoder has four inputs D0 C0, B0, and A0 and five out...
Solution:
Given that a binary-to-BCD encoder has four inputs D0 C0, B0, and A0 and five outputs D, C, B, A, and VALID.
The VALID output is 1 if the input combination is a valid decimal code. If the input combination is an invalid decimal code, the VALID output becomes 0, and all of the D, C, B, and A outputs show 0 values.
We need to design a circuit using NOT gates, 2-input OR and AND gates to implement the above functionality.
Minimum number of gates required to implement the above circuit is 8.
Explanation:
The binary-to-BCD encoder can be implemented using the following steps:
Step 1: Implement the conditions for VALID output to be 1.
VALID output is 1 if the input combination is a valid decimal code. This condition can be implemented using the following Boolean expression:
VALID = D0'C0'B0'A0 + D0'C0'B0'A0' + D0'C0'B0A0' + D0'C0B0'A0' + D0C0'B0'A0' + D0C0B0'A0' + D0C0B0A0
This can be implemented using 7 gates as shown below:
Step 2: Implement the BCD output.
The BCD output can be implemented using the following Boolean expressions:
A = D0'C0B0'A0' + D0C0B0A0' + D0C0B0'A0 + D0'C0'B0A0' + D0'C0B0A0 + D0C0'B0'A0'
B = D0'C0B0'A0' + D0C0B0A0' + D0C0B0'A0 + D0C0'B0A0' + D0'C0B0A0 + D0'C0'B0'A0'
C = D0'C0B0'A0' + D0C0B0A0' + D0C0B0'A0' + D0'C0'B0A0' + D0'C0B0A0' + D0C0'B0A0
D = D0'C0B0'A0' + D0C0B0A0' + D0C0B0'A0' + D0'C0'B0A0' + D0'C0B0A0' + D0C0'B0'A0'
This can be implemented using 8 gates as shown below:
Therefore, the minimum number of gates required to implement the above circuit is 8.
Given that a binary-to-BCD encoder has four inputs D0 C0, B0, and A0 and five outputs D, C, B, A, and VALID.
The VALID output is 1 if the input combination is a valid decimal code. If the input combination is an invalid decimal code, the VALID output becomes 0, and all of the D, C, B, and A outputs show 0 values.
We need to design a circuit using NOT gates, 2-input OR and AND gates to implement the above functionality.
Minimum number of gates required to implement the above circuit is 8.
Explanation:
The binary-to-BCD encoder can be implemented using the following steps:
Step 1: Implement the conditions for VALID output to be 1.
VALID output is 1 if the input combination is a valid decimal code. This condition can be implemented using the following Boolean expression:
VALID = D0'C0'B0'A0 + D0'C0'B0'A0' + D0'C0'B0A0' + D0'C0B0'A0' + D0C0'B0'A0' + D0C0B0'A0' + D0C0B0A0
This can be implemented using 7 gates as shown below:
Step 2: Implement the BCD output.
The BCD output can be implemented using the following Boolean expressions:
A = D0'C0B0'A0' + D0C0B0A0' + D0C0B0'A0 + D0'C0'B0A0' + D0'C0B0A0 + D0C0'B0'A0'
B = D0'C0B0'A0' + D0C0B0A0' + D0C0B0'A0 + D0C0'B0A0' + D0'C0B0A0 + D0'C0'B0'A0'
C = D0'C0B0'A0' + D0C0B0A0' + D0C0B0'A0' + D0'C0'B0A0' + D0'C0B0A0' + D0C0'B0A0
D = D0'C0B0'A0' + D0C0B0A0' + D0C0B0'A0' + D0'C0'B0A0' + D0'C0B0A0' + D0C0'B0'A0'
This can be implemented using 8 gates as shown below:
Therefore, the minimum number of gates required to implement the above circuit is 8.
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Community Answer
A binary-to-BCD encoder has four inputs D0 C0, B0, and A0 and five out...
Decimal to BCD
BCD is represented with four digits.
The valid decimal numbers for BCD conversion are from 0 to 9 only.
BCD is represented with four digits.
The valid decimal numbers for BCD conversion are from 0 to 9 only.
Calculation:
According to the given statement, 4 inputs and 5 outputs are there.
Inputs are D0, C0, B0, A0
Outputs are D, C, B, A, Valid
Logic implementation is shown in the below table
For valid output logic 1 is taken.
The K-map for Valid is
Valid = Σ (0 to 9)
Output equation is
The K-map for D is
Valid = Σ (8, 9)
Output equation is
The K-map for C is
Valid = Σ (4, 5, 6, 7)
Output equation is
The K-map for B is
Valid = Σ (2, 3, 6, 7)
Output equation is
The K-map for A is
Valid = Σ (1, 3, 5, 7, 9)
Output equation is
Implementing the above logic circuit we require 8 minimum gates as shown:
According to the given statement, 4 inputs and 5 outputs are there.
Inputs are D0, C0, B0, A0
Outputs are D, C, B, A, Valid
Logic implementation is shown in the below table
For valid output logic 1 is taken.
The K-map for Valid is
Valid = Σ (0 to 9)
Output equation is
The K-map for D is
Valid = Σ (8, 9)
Output equation is
The K-map for C is
Valid = Σ (4, 5, 6, 7)
Output equation is
The K-map for B is
Valid = Σ (2, 3, 6, 7)
Output equation is
The K-map for A is
Valid = Σ (1, 3, 5, 7, 9)
Output equation is
Implementing the above logic circuit we require 8 minimum gates as shown:
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A binary-to-BCD encoder has four inputs D0 C0, B0, and A0 and five outputs D, C, B, A, and VALID. The outputs D, C, B and A give the proper BCD value of the input and the VALID output is 1 if the input combination is a valid decimal code. If the input combination is an invalid decimal code, the VALID output becomes 0, and all of the D, C, B, and A outputs show 0 values. If only NOT gates and 2-input OR and AND gates are available, the minimum number of gates required to implement the above circuit isa)10b)9c)8d)7Correct answer is option 'C'. Can you explain this answer?
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A binary-to-BCD encoder has four inputs D0 C0, B0, and A0 and five outputs D, C, B, A, and VALID. The outputs D, C, B and A give the proper BCD value of the input and the VALID output is 1 if the input combination is a valid decimal code. If the input combination is an invalid decimal code, the VALID output becomes 0, and all of the D, C, B, and A outputs show 0 values. If only NOT gates and 2-input OR and AND gates are available, the minimum number of gates required to implement the above circuit isa)10b)9c)8d)7Correct answer is option 'C'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A binary-to-BCD encoder has four inputs D0 C0, B0, and A0 and five outputs D, C, B, A, and VALID. The outputs D, C, B and A give the proper BCD value of the input and the VALID output is 1 if the input combination is a valid decimal code. If the input combination is an invalid decimal code, the VALID output becomes 0, and all of the D, C, B, and A outputs show 0 values. If only NOT gates and 2-input OR and AND gates are available, the minimum number of gates required to implement the above circuit isa)10b)9c)8d)7Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A binary-to-BCD encoder has four inputs D0 C0, B0, and A0 and five outputs D, C, B, A, and VALID. The outputs D, C, B and A give the proper BCD value of the input and the VALID output is 1 if the input combination is a valid decimal code. If the input combination is an invalid decimal code, the VALID output becomes 0, and all of the D, C, B, and A outputs show 0 values. If only NOT gates and 2-input OR and AND gates are available, the minimum number of gates required to implement the above circuit isa)10b)9c)8d)7Correct answer is option 'C'. Can you explain this answer?.
A binary-to-BCD encoder has four inputs D0 C0, B0, and A0 and five outputs D, C, B, A, and VALID. The outputs D, C, B and A give the proper BCD value of the input and the VALID output is 1 if the input combination is a valid decimal code. If the input combination is an invalid decimal code, the VALID output becomes 0, and all of the D, C, B, and A outputs show 0 values. If only NOT gates and 2-input OR and AND gates are available, the minimum number of gates required to implement the above circuit isa)10b)9c)8d)7Correct answer is option 'C'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A binary-to-BCD encoder has four inputs D0 C0, B0, and A0 and five outputs D, C, B, A, and VALID. The outputs D, C, B and A give the proper BCD value of the input and the VALID output is 1 if the input combination is a valid decimal code. If the input combination is an invalid decimal code, the VALID output becomes 0, and all of the D, C, B, and A outputs show 0 values. If only NOT gates and 2-input OR and AND gates are available, the minimum number of gates required to implement the above circuit isa)10b)9c)8d)7Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A binary-to-BCD encoder has four inputs D0 C0, B0, and A0 and five outputs D, C, B, A, and VALID. The outputs D, C, B and A give the proper BCD value of the input and the VALID output is 1 if the input combination is a valid decimal code. If the input combination is an invalid decimal code, the VALID output becomes 0, and all of the D, C, B, and A outputs show 0 values. If only NOT gates and 2-input OR and AND gates are available, the minimum number of gates required to implement the above circuit isa)10b)9c)8d)7Correct answer is option 'C'. Can you explain this answer?.
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