EXAM 1-solution - EE348EXAMINAME 12.07.2011SURNAME...

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EE348 EXAM I NAME: 12.07.2011 SURNAME: 40 minutes Q1) a) Do (18+14) 10 in signed 2’s complement system (use 8 bits) carry 1 1 1 1 +18 0 0 0 1 0 0 1 0 +14 + 0 0 0 0 1 1 1 0 0 0 1 0 0 0 0 0 b) Do (18‐14) 10 in signed 2’s complement system (use 8 bits ) carry 1 1 1 1 +18 0 0 0 1 0 0 1 0 ‐14 + 1 1 1 1 0 0 1 0 1 0 0 0 0 0 1 0 0 ignore c)Do (18+14) 10 in BCD (use 4 BCD digit) 1 1 +18 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 +14 + 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 + 0 1 1 0 1 0 0 1 0 Result 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 d) Do (18‐14) 10 in BCD (use 4 BCD digit) 1 1 1 1 1 +18 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 ‐14 + 1 0 0 1 1 0 0 1 1 0 0 0 0 1 1 0 0 0 0 0 1 0 1 0 1 0 1 0 1 1 1 0 + 0 1 1 0 + 0 1 1 0 + 0 1 1 0 + 0 1 1 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 1 0 0 ignore Result 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0
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EE348 EXAM I NAME: 12.07.2011 SURNAME: 40 minutes Q2) Given the function f(w,x,y,z)=x(xy+yz)+wxz+wxz’ a) Convert to sum of products standart form (simplified as possible) f(w,x,y,z)=x(xy+yz)+wxz+wxz’=xy+xyz+wxz+wxz’=xy(1+z)+wxz+wxz’ =xy+wxz+wxz’=xy+wx(z+z’)=xy+wx1=xy+wx b) convert it to product of sums standart form (simplified as possible) f(w,x,y,z)= x(xy+yz)+wxz+wxz’= x(xy+yz)+wx(z+z’)= x(xy+yz)+wx=xy(1+z)+wx=xy+wx=x(y+w) or f(w,x,y,z)= xy+wx=(xy+w)(xy +x)=(xy+w)x=(x+w) (y+w) x c) find minterms (in anyway you wish) f(w,x,y,z)=xy+wx=(w+w’)xy(z+z’)+wx(y+y’)(z+z’) =wxyz+w’xyz+ wxyz’+w’xyz’+wxyz+wxy’z+ wxyz’+wxy’z’ =m 15 +m 7 +m 14 +m 6 +m 15 +m 13 +m 1 4 +m 12 f(w,x,y,z)= (6,7,12,13,14,15) d) find maxterms
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EXAM 1-solution - EE348EXAMINAME 12.07.2011SURNAME...

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