Web_Answers_2 - Unit 2 Problem Solutions 2.1 2.2 (a) See...

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(A + B + C + D) (A + B + C + E) (A + B + C + F) = A + B + C + DEF Apply second distributive law (Th. 8D) twice See FLD p. 626 for solution. 2.1 Unit 2 Problem Solutions 2.2 (a) In both cases, if X = 0, the transmission is 0, and if X = 1, the transmission is 1. 2.2 (b) In both cases, if X = 0, the transmission is YZ, and if X = 1, the transmission is 1. For the answer to 2.3, refer to FLD p. 626 2.3 F = [(A· 1 ) + (A· 1 )] + E + BCD = A + E + BCD 2.4 (a) Y = (AB' + (AB + B)) B + A = (AB' + B) B + A = (A + B) B + A = AB + B + A = A + B 2.4 (b) (A + B) (C + B) (D' + B) (ACD' + E) = (AC + B) (D' + B) (ACD' + E) By Th. 8D = (ACD' + B) (ACD' + E) By Th. 8D = ACD' + BE By Th. 8D 2.5 (a) (A' + B + C') (A' + C' + D) (B' + D') = (A' + C' + BD) (B' + D') {By Th. 8D with X = A' + C' } = A'B' + B'C' + B'BD + A'D' + C'D' + BDD' = A'B' + A'D' + C'B' + C'D' 2.5 (b) AB + C'D' = (AB + C') (AB + D') = (A + C') (B + C') (A + D') (B + D') 2.6 (a) WX + WY'X + ZYX = X(W + WY' + ZY) = X(W + ZY) {By Th. 10}
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This note was uploaded on 09/03/2008 for the course EE 316 taught by Professor Brown during the Fall '08 term at University of Texas at Austin.

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Web_Answers_2 - Unit 2 Problem Solutions 2.1 2.2 (a) See...

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