FINAL_F08SOL - 1 McGill University Faculty of Engineering...

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1 McGill University DIGITAL SYSTEM DESIGN Faculty of Engineering ECSE-323 FINAL EXAMINATION FALL 2008 (December2008) STUDENT NAME McGILL I.D. NUMBER Examiner: Prof. J. Clark Associate Examiner: Prof. Miguel Marin Signature: Signature: Date: December 4, 2008 Time: 2:00 pm INSTRUCTIONS: SEE NEXT PAGE.
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2 McGILL UNIVERSITY Electrical and Computer Engineering Department ECSE-323 Fall 2008 FINAL EXAM Question Maximum Points Points Attained 1 15 2 10 3 20 4 15 5 15 6 15 7 15 8 15 9 15 10 20 11 15 12 10 Total 180 Please write down your name: Please write your student ID:ANSWERKEY Instructions/Please read carefully! This is a closed book exam. No books or notes are allowed. You may use a faculty standard calculator. All work is to be done on the attached sheets and under no circumstances 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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ECSE-323-Digital Systems Design Final Exam Fall 2008 Your name: ANSWER KEY Question 1 :Boolean Logic Theory (15 points) The following function, F(A,B,C,D), is given by its k-map representation. K-MAP OF FUNCTION F(A,B,C,D) AB CD 00 01 11 10 00 1 1 1 0 01 1 0 0 0 11 1 1 1 0 10 0 0 1 0 There are two minimal sum-of-products forms of F. (5 points) a) Give the two minimal sum-of-products forms of F, showing the answer on the given k-maps. (5 points) b) Using the Quine-McCluskey method, determine the set of prime implicants of F. (5 points) c) Using the covering table method and the corresponding Petrick function, find the two minimal sum-of-products forms of F. ____________________________________________________________________________ ANSWER FIRST MINIMAL FORM OF F SECOND MINIMAL FORM OF F AB CD 00 01 11 10 00 1 1 1 0 01 1 0 0 0 11 1 1 1 0 10 0 0 1 0 F min = f8e5 A f8e5 B f8e5 C + f8e5 A C D + B f8e5 C f8e5 D + A B C F min = f8e5 A f8e5 C f8e5 D + f8e5 A f8e5 B D + A B f8e5 D + B C D b) First and only reduction AB CD 00 01 11 10 00 1 1 1 0 01 1 0 0 0 11 1 1 1 0 10 0 0 1 0 ID A B C D 0 0 0 0 0 1 0 0 0 1 4 0 1 0 0 3 0 0 1 1 12 1 1 0 0 7 0 1 1 1 14 1 1 1 0 15 1 1 1 1 Prime implicant GROUP A B C D P1 (0,1) 0 0 0 - P2 (0,4) 0 - 0 0 P3 (1,3) 0 0 - 1 P4 (4,12) - 1 0 0 P5 (3,7) 0 - 1 1 P6 (12,14) 1 1 - 0 P7 (7,15) - 1 1 1 P8 (14,15) 1 1 1 - 3
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ECSE-323-Digital Systems Design Final Exam Fall 2008 Your name: ANSWER KEY The prime implicants of F are Prime implicants A B C D P1 0 0 0 - f8e5 A f8e5 B f8e5 C P2 0 - 0 0 f8e5 A f8e5 C f8e5 D P3 0 0 - 1 f8e5 A f8e5 B D P4 - 1 0 0 B f8e5 C f8e5 D P5 0 - 1 1 f8e5 A C D P6 1 1 - 0 A B f8e5 D P7 - 1 1 1 B C D P8 1 1 1 - A B C c) COVERING TABLE Minterms Prime implicant GROUP A B C D 0 1 3 4 7 12 14 15 P1 (0,1) 0 0 0 - P2 (0,4) 0 - 0 0 P3 (1,3) 0 0 - 1 P4 (4,12) - 1 0 0 P5 (3,7) 0 - 1 1 P6 (12,14) 1 1 - 0 P7 (7,15) - 1 1 1 P8 (14,15) 1 1 1 - 4
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ECSE-323-Digital Systems Design Final Exam Fall 2008 Your name: ANSWER KEY Petrick function : P = (P1 + P2) (P1 + P3) (P3 + P5) (P2 + P4)(P5 + P7) (P4 + P6) (P6 + P8)(P7 + P8) P = (P1 + P2P3)(P5 + P3P7)(P4 + P2P6)(P8 +P6P7) P = P1P5P4P8 + P2P3P6P7 The two minimal sum-of-products forms are F min = f8e5 A f8e5 B f8e5 C + f8e5 A C D + B f8e5 C f8e5 D + A B C F min = f8e5 A f8e5 C f8e5 D + f8e5 A f8e5 B D + A B f8e5 D + B C D 5
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