HW6_sol

HW6_sol - __|O|1|0|0| A |O|0|0|1| | O | You will find that...

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HW 6 Due Friday, May 22nd, 4:45pm in the homework box K-map simplification For each problem: a) Write out the truth table b) Use a K-map to simplify the above equation c) Write out the final equation 1. Determine the logic for the overflow bit on an adder There are two ways to do this. I will present one way - the overflow bit is determined by looking at the top bits of the inputs as well as the top bit of the output. If both inputs are negative and the output positive, this is an overflow. That is, if the top bit of the two inputs are both 1 and the output is 0, there is an overflow. Likewise, if both inputs are positive and the output negative, it is an overflow. Truth table: A B O | Overflow _______________ 0 0 0 | 0 0 0 1 | 1 0 1 0 | 0 0 1 1 | 0 1 0 0 | 0 1 0 1 | 0 1 1 0 | 1 1 1 1 | 0 K-map: B ____|____

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Unformatted text preview: __|O|1|0|0| A |O|0|0|1| | O | You will find that with this K-map, there is no simplification to be done. Therefore, the final equation is: /A/BO + AB/O 2. f(A,B,C) = /A/B/C + A/B/C + /A/BC A B C | O 0 0 0 | 1 0 0 1 | 1 0 1 0 | 0 0 1 1 | 0 1 0 0 | 1 1 0 1 | 0 1 1 0 | 0 1 1 1 | 0 K-map: B ____|____ __|1|1|0|0| A |1|0|0|0| | C | We can circle two pairs, giving us: /A/B + /B/C 3. f(A,B,C,D) = (AD + /AC)(/B(C + B/D)) A B C D | O 0 0 0 0 | 0 0 0 0 1 | 0 0 0 1 0 | 1 0 0 1 1 | 1 0 1 0 0 | 0 0 1 0 1 | 0 0 1 1 0 | 0 0 1 1 1 | 0 1 0 0 0 | 0 1 0 0 1 | 0 1 0 1 0 | 0 1 0 1 1 | 1 1 1 0 0 | 0 1 1 0 1 | 0 1 1 1 0 | 0 1 1 1 1 | 0 K-map: C ____|____ |0|0|1|1|__ __|0|0|0|0| B A |0|0|0|0|__ |0|0|1|0| | D | We can draw two circles, each with two 1's in them. /A/BC + /BCD...
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This note was uploaded on 12/27/2011 for the course CMPSC 30 taught by Professor Franklin during the Fall '09 term at UCSB.

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HW6_sol - __|O|1|0|0| A |O|0|0|1| | O | You will find that...

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