csc116
Digital Logic
Hard Exam Preparation: Unknown
Question Papers (6)
Selected Question Paper: 2081 Boards
Community
FM: 60 PM: 24
Digital Logic
2081 Boards
FM: 60 PM: 24
Section A
Answer any two questions.
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Section B
Answer any eight questions.
#Answer
-
Represent A and B in Binary:
- A = 46 (decimal) = 101110 (binary)
- B = 35 (decimal) = 100011 (binary)
-
Determine 1's Complement of B:
- B = 100011
- 1's Complement of B = 011100 (invert all bits)
-
Perform Addition of A and 1's Complement of B:
A = 101110 1's Comp B = +011100 -------------------- Sum = 1001010 -
Handle End-Around Carry:
- The sum results in a 7-bit number, indicating a carry-out from the most significant bit.
- The carry-out is '1'.
- Add this carry-out back to the least significant bit of the 6-bit result:
001010 (6-bit sum, ignoring carry) + 1 (End-around carry) -------------------- 001011
-
Final Result:
- The result in binary is 001011.
- Converting back to decimal: 0*32 + 0*16 + 1*8 + 0*4 + 1*2 + 1*1 = 8 + 2 + 1 = 11.
- Therefore, A - B = 46 - 35 = 11.
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community grow.
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Answer is not available or verified. Become a contributor and help this
community grow.
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Connect with us on Discord to become a contributor.
Answer is not available or verified. Become a contributor and help this
community grow.
Connect with us on Discord to become a contributor.
Connect with us on Discord to become a contributor.
Answer is not available or verified. Become a contributor and help this
community grow.
Connect with us on Discord to become a contributor.
Connect with us on Discord to become a contributor.
Answer is not available or verified. Become a contributor and help this
community grow.
Connect with us on Discord to become a contributor.
Connect with us on Discord to become a contributor.
Answer is not available or verified. Become a contributor and help this
community grow.
Connect with us on Discord to become a contributor.
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Answer is not available or verified. Become a contributor and help this
community grow.
Connect with us on Discord to become a contributor.
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