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For a particular design of multiplication unit with 6 bit multiplicand and 3 bit multiplier What is the worst case number of adders required for the design and what would be the size of those adders ? options are : A) 9 Adders; Size 3 bits B) 3 Adders; Size 9 bits C) 6 Adders; Size 6 bits D) 3 Adders; Size 6 bits E) 6 Adders; Size 9 bits F) 6 Adders; Size 3 bits |
4 days, 15 hours ago |
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Is that really a question? Anyhow, 2^6 times 2^3 gives 2^9, so the size must be 9. Now, since we can naively add the number up to 6 times to itself (the multiplier is a number from 0 to 7) and use a selector to get the result: if the multiplier is 1, select the number, if the multiplier is 2, select the output of the first adder, if the multiplier is 3, select the output of the second adder, etc. Inefficient, though, totally (ironic) inefficient. But the question is about the worse. More efficient would be to shift (or clear) accordingly to each bits of the multiplier and then to add the three shifted/cleared results, so two adders (if "registers" are free). I assume that the question makes more sense within its context, context that we don't have. |
1 hour, 34 minutes ago |
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