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EXAM 2-solution

# EXAM 2-solution - EE348EXAM2NAME:SOLUTION 19.07.2011SURNAME...

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EE348 EXAM 2 NAME: SOLUTION 19.07.2011 SURNAME: Q1) (15 pts) Given F(a,b,c,d)= (1,3,7,8,13), d=(0,2,4,5,10,14,15). Assuming complement of the input variables are available a) Draw the circuit for minimal two level AND‐OR implementation F= bd+b’d’ +a’b’ or F= bd+b’d’ +a’d b) Draw the circuit for minimal two level OR‐AND implementation F’=bd’+ab’d F=(b’+d)(a’+b+d’) c) Draw the circuit for minimal implementation using one EX‐OR gate (no restriction on the number of inputs) and other gates necessary F 1 =bd+b’d’=b d’=b’ d F 2 =a’b’ F 1 +F 2 =(b’ d)+a’b’ F=F 1 +F 2 =(b d’)+a’b’ or F=((b d’)’(a’b’)’)’ =( (b d)(a+b))’ Instead F 2 here F 3 =a’d can be used (other equivalent or simpler solutions are accepted) F: d c x 1 1 x x x 1 1 x x 1 x b a ab cd 01 00 11 10 00 01 11 10 F F or b d b’ d’ a’ b’ b d b’ d’ a’ d F’: d c x x x x 1 1 x x 1 1 x b a ab cd 01 00 11 10 00 01 11 10 F b' d a’ b d’ F 1 : F 2 : F 1 +F 2 : d c 1 1 1 1 b a ab cd 01 00 11 10 00 01 11 10 d c 1 1 1 1 1 1 1 1 b a ab cd 01 00 11 10 00 01 11 10 d c 1 1 1 1 1 1 1 1 1 1 b a ab cd 01 00 11 10 00 01 11 10 F 1 F 2 F F 1 F 2 F F 1 F 3 F F 1 F 3 F b d’ a’ b’ b d a b b d’ a’ d b d a d’

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EE348 EXAM 2 NAME: SOLUTION 19.07.2011
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EXAM 2-solution - EE348EXAM2NAME:SOLUTION 19.07.2011SURNAME...

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