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MIDW07_ANS - ECSE-323 Department of Electrical and Computer...

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McGILL UNIVERSITY Department of Electrical and Computer Engineering ECSE-323 Winter 2007 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. Your Name_______________________________________________________ Question 1 :Boolean Logic Theory (10 points)
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The prime implicants of the function F(A,B,C,D) = m(0,4,6,7,10,11,15) are given below. (5 points) a) Produce the covering table of the minterms of F and give the essential prime implicants. (5 ponts) b) Apply the Petrick function to the reduced covering table and give ALL minimal two-level sum-of-products expressions of F. ___________________________________________________________________________ ANSWER: Prime implicants Covering table A B C D P1: (7,15) - 1 1 1 P2: (11,15) 1 - 1 1 P3: (10,11) 1 0 1 - P4: (0,4) 0 - 0 0 P5: (4,6) 0 1 - 0 P6: (6,7) 0 1 1 - Essential prime implicants: P3 = A f8e5 B C P4 = f8e5 A f8e5 C f8e5 D Reduced covering table Petrick function corresponding to the reduced table P R = (P5 + P6)(p1 + P6)(P1 + P2) = (P1P5 + P6) (P1 + P2P6) = P1P5 + P1P6 + P2P6 + P1P2P5P6. Minimal two-level sum-of-products expressions: F min = P3 + P4 + P1 + P5 ; F min = P3 + P4 + P1 + P6; F min = P3 + P4 + P2 + P6 F min = A f8e5 B C + f8e5 A f8e5 C f8e5 D + B C D + f8e5 A B f8e5 D F min = A f8e5 B C + f8e5 A f8e5 C f8e5 D + B C D + f8e5 A B C F min = A f8e5 B C + f8e5 A f8e5 C f8e5 D + A C D + f8e5 A B C. Your Name_______________________________________________________ Question 2 : Application of Boolean Logic Theory (10 points) A B C D 0 4 6 7 10 11 15 P1: (7,15) - 1 1 1 x x P2: (11,15) 1 - 1 1 x x P3: (10,11) 1 0 1 - x x P4: (0,4) 0 - 0 0 x x P5: (4,6) 0 1 - 0 x x P6: (6,7) 0 1 1 - x x A B C D 0 4 6 7 10 11 15 P1: (7,15) - 1 1 1 x x P2: (11,15) 1 - 1 1 x x P3: (10,11) 1 0 1 - x x P4: (0,4) 0 - 0 0 x x P5: (4,6) 0 1 - 0 x x P6: (6,7) 0 1 1 - x x A B C D 6 7 15 P1: (7,15) - 1 1 1 x x P2: (11,15) 1 - 1 1 x P5: (4,6) 0 1 - 0 x P6: (6,7) 0 1 1 - x x
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(4 points) a)
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  • Winter '07
  • redacka
  • Logic gate, Combinational circuit design, Name_______________________________________________________ Question

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