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

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McGILL UNIVERSITY Department of Electrical and Computer Engineering ECSE-323 Winter 2008 MIDTERM EXAM Question Maximum Points Points Attained 1 10 2 10 3 15 4 15 5 15 6 10 Total 75 points Please write down your name: ANSWER KEY Please write your student ID: ______________________________________ Instructions/Please read carefully! This is a close book exam. No books or notes are allowed. You may use a standard calculator. All work is to be done on the attached sheets and under no circumstance are booklets or loose sheets to be used. Write your name at the top of every sheet. Read the question carefully. If something appears ambiguous, write down your assumption. The points have been assigned according to the formula that 1 point = 1 exam minute, so please pace yourself accordingly.
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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 ______________________________________________________________ ____________________________________________________________________________
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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.
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  • Winter '07
  • redacka
  • Boolean Algebra, Boolean function, Digital Systems Design, Combinational circuit design, ECSE-323 Digital Systems

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