MIDTERM-W09-ANS

MIDTERM-W09-ANS - McGILL UNIVERSITY Department of...

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McGILL UNIVERSITY Department of Electrical and Computer Engineering ECSE-323 Winter 2009 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 closed 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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COURSE: ECSE – 323 MIDTERM EXAM WINTER 2009 2 YOUR NAME: ANSWER KEY ________________________________________________________________________ Question 1: Booelan Logic Theory (10 points) The prime implicants of the function F(A,B,C,D) = m(0,1,5,7,8,9,12,14,15) with “Don’t cares” D = (3,11,13) are given in the following table Group of Minterms Prime implicant A B C D P1:(1,3,9,11,7,15,5,13) - - - 1 P2:(0,1,8,9) - 0 0 - P3:(1,3,5,7) 0 - - 1 P4:(8,12,9,13) 1 - 0 - P5:(12,13,14,15) 1 1 - - (5 points) a) Using the covering table produce ALL minimal sum-of-products forms of F. (5 points) b) Repeat part a) but using this time the Petrick function. ________________________________________________________________________ ANSWER a) Group of Minterms Prime implicant A B C D 0 1 5 7 8 9 12 14 15 essential essential P1:(1,3,9,11,7,15,5,13) - - - 1 XXX X X P2:(0,1,8,9) - 0 0 - XX P3:(1,3,5,7) 0 - - 1 XXX P4:(8,12,9,13) 1 - 0 - X X X P5:(12,13,14,15) 1 1 - - XXX P2, P5 are essential prime-implicant and , therefore, they have to appear in every minimal form These prime implicants cover all minterms expect 5 and 7 which can be covered with either P1 or P3. P1 should be prefered because it contains only one variable, while P3 requires is product term with two variables.Therefore, there is only one minimal sum-of- products form, namely , F = P2 + P5 + P1 = B C + A B + D b) The Petrick function is P = (P2)(P1+P2+P3)(P1+ P3)(P1+P3)(P2 + P4) )(P1+P2+P4) (P4 + P5)(P5)(P1 + P5) = (P2)(P5)(P1 + P3) = P2 P5 P1 + P2 P5 P3 Apparently, the Petrick function gives us two minimal forms, each one containing three prime-implicants. Closer analysis of each one of these solutions shows that the solution P2P5P1 contains fewer number of variables than P2P5P3, therefore, P2P5P1 forms the minimal sum-of-products of F,namely, F = P2 + P5 + P1 = B C + A B + D, as we found in part a)
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COURSE: ECSE – 323 MIDTERM EXAM WINTER 2009 3 YOUR NAME: ANSWER KEY ________________________________________________________________________ Question 2: CMOS Circuit Technology (10 points) Given the function F (A,B,C,D) = A B D + A C D + A C D + A B D + A B D (5 points) a) Produce a CMOS fully complementary circuit producing F. Show clearly the Pull-Up-Network and the Pull-Down-Network.
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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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MIDTERM-W09-ANS - McGILL UNIVERSITY Department of...

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