MIDW08_ANS

# MIDW08_ANS - ECSE-323 Department of Electrical and Computer...

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Question 1 : Boolean Logic Theory (10 points) The Majority function M(A,B,C) is equal to 1 when two or three of its variables assume the value 1. That is M(A,B,C) = A B + A C + C B (5 points) a) Find the algebraic expression of F(A,B,C,D) produced by the following network A 1 B !A !D D C !C M M M M F(A,B,C,D) (5 points) b) Draw a network producing the function F(A,B,C,D) = ABC + ABD +ACD + BCD and using only Majority functions M(X,Y,Z) = XY + XZ + YZ as building blocks. Note: A solution with 5 or less M blocks gets full marks and a correct solution gets partial marks. ___________________________________________________________________________ ANSWER a) F (A,B,C,D) = f8e5 A B + f8e5 C A + f8e5 D A. b) 1 D 0 B C D A 0 F(A,B,C,D) M M M M B C 0 M 1 ECSE-323 Digital Systems Design - Midterm Exam - Winter 2008 Your Name ______________________________________________________________ ____________________________________________________________________________
Question 2 : Application of Boolean Theory (10 points) (5 points) a) Give the prime implicants of the function F(A,B,C,D) = m(4,10,11,13) + Dm (0,2,5,15), where Dm denotes the don’t care minterms (5 points) b) Give ALL two-level AND-OR minimal expressions of F. _________________________________________________________________________ ANSWER First and only reduction a) ID A B C D 0 0 0 0 0 4 0 1 0 0 2 0 0 1 0 5 0 1 0 1 10 1 0 1 0 11 1 0 1 1 13 1 1 0 1 15 1 1 1 1 The prime implicants are shown in the first column of the covering table below Covering Table Primeimplicant s A B C D 4 10 11 13 P1: 0 – 0 0 x P2: 0 1 0 - x P3: - 1 0 1 x P4: 1 1 - 1 x P5: 1 – 1 1 x P6: 1 0 1 - x x P7: - 0 1 0 x P8: 0 0 - 0 There are no essential prime implicants. Groups A B C D (0,4) 0 – 0 0 (0,2) 0 0 - 0 (4,5) 0 1 0 - (5,13) - 1 0 1 (13,15) 1 1 - 1 (15,11) 1 – 1 1 (10,11) 1 0 1 - (2,10) - 0 1 0 2 ECSE-323 Digital Systems Design - Midterm Exam - Winter 2008 Your Name ______________________________________________________________ ____________________________________________________________________________

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Using the Petrick function P = (P1 + P2) (P6 + P7) (P5 + P6) (P3 + P4)
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## This note was uploaded on 11/22/2010 for the course ECSE ecse 323 taught by Professor Redacka during the Winter '07 term at McGill.

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MIDW08_ANS - ECSE-323 Department of Electrical and Computer...

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