dsdMIDTERM-W09

# dsdMIDTERM-W09 - McGILL UNIVERSITY ECSE-323 Department of...

This preview shows pages 1–5. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 1 2 1 4 1 5 essential essential P1:(1,3,9,11,7,15,5,13) - - - 1 X X X X X P2:(0,1,8,9) - 0 0 - X X X X P3:(1,3,5,7) 0 - - 1 X X X P4:(8,12,9,13) 1 - 0 - X X X P5:(12,13,14,15) 1 1 - - X X X 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 = f8e5 B f8e5 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
COURSE: ECSE – 323 MIDTERM EXAM WINTER 2009 3 YOUR NAME: ANSWER KEY _______________________________________________________________________ _ minimal sum-of-products of F,namely, F = P2 + P5 + P1 = f8e5 B f8e5 C + A B + D, as we found in part a)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
COURSE: ECSE – 323 MIDTERM EXAM WINTER 2009 4 YOUR NAME: ANSWER KEY _______________________________________________________________________ _ Question 2:
This is the end of the preview. Sign up to access the rest of the document.

## 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.

### Page1 / 16

dsdMIDTERM-W09 - McGILL UNIVERSITY ECSE-323 Department of...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online