ECE314sp08_HW3 - ECE/CS 314 Spring 2008 Homework 3 Due...

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1 ECE/CS 314 Spring 2008 Homework 3 Due Tuesday, March 4, 2007 before 10:00pm EST HUNG DANG (hvd2) Problem 1. (10 points) Consider the following Boolean expression F = WX’Y’ + X’YZ’ + WXZ + WY’Z + X’Y’Z’ + W’X’Y’ (a) 3 pts. Construct the truth table for F W X Y Z WX’Y’ X’YZ’ WXZ WY’Z X’Y’Z’ W’X’Y’ F 0 0 0 0 0 0 0 0 1 1 1 0 0 0 1 0 0 0 0 0 1 1 0 0 1 0 0 1 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 1 1 0 0 1 1 0 0 1 0 0 1 1 0 1 0 0 1 0 0 0 0 1 1 0 1 1 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 1 1 0 1 0 0 1 1 0 0 1 1 1 1 0 0 0 0 0 0 0 0 1 1 1 1 0 0 1 0 0 0 1 (b) 3 pts. Draw the Karnaugh map, circle the prime implicants, and derive the minimized sum of product form for F. F = X’Z’ + X’Y’ + WXZ (c) 4 pts. Show that the first expression for F is equivalent to the minimized sum of product form for F using Boolean algebra. For each step of the derivation, mention which law of Boolean algebra you applied. Y Z \W X0 00 11 11 0 00 1 0 0 1 01 1 0 1 1 11 0 0 1 0
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2 1) F = WX’Y’ + X’YZ’ + WXZ + WY’Z + X’Y’Z’ + W’X’Y’ 2) F = (W+W’)X’Y’ + (Y+Y’)X’Z’ + WXZ + WY’Z (Distributive) 3) F = X’Y’ + X’Z’ + WXZ +WY’Z (Identity A + A’ =1 and 1*A = A) 4) F = X’Y’ + X’Z’ + WZ(X+Y’) (Distributive) 5) F = X’Y’ + X’Z’ + WZ(X(Y+Y’) + Y’(X+X’)) (Identity A+A’ = 1 and 1*A = A) 6) F = X’Y’ + X’Z’ + WZ(XY+XY’+XY’+X’Y’) (Distributive) 7) F = X’Y’ + X’Z’ + WZ(XY+XY’+X’Y’) (Identity A + A = A) 8) F = X’Y’ + X’Z’ + WZ(X(Y+Y’)+X’Y’) (Distributive) 9) F = X’Y’ + X’Z’ + WZ(X + X’Y’) (Identity A + A’ = 1) 10) F = X’Y’ + X’Z’ + WZ(X + X + Y) (DeMorgan’s Theorem) 11)
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This note was uploaded on 09/03/2008 for the course ECE 3140 taught by Professor Mckee/long during the Spring '07 term at Cornell University (Engineering School).

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ECE314sp08_HW3 - ECE/CS 314 Spring 2008 Homework 3 Due...

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