AppendixAUnit8_2Sol

AppendixAUnit8_2Sol - 8.2. Unit 2 - Boolean Algebra, Logic...

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298 8.2. Unit 2 - Boolean Algebra, Logic Functions, and Canonical Representation, 2-Level Implementations and Circuit Design w/ Karnaugh Maps 1. Probably the easiest method is perfect induction (i.e. a truth table) F = X + X’ = 1 XX ’F 0 1 1 1 0 1 2. a. F = WXYZ • (WXYZ' + WX'YZ + W'XYZ +WXY'Z) = WWXXYYZZ’ • WWXX’YYZZ + W’WXXYYZZ + W W X X Y Y Z Z T 8 = WXYZZ’ + WXX’YZ + WW’XYZ + WXYY’Z T3’ = 0 + 0 + 0 + 0 + 0 T5’ = 0 A 4 b. F = AB + ABC’D + ABDE’ + ABC’E + C’D = AB • (1 + C’D + DE’ + C’E) + C’D T8 = AB • (1 + DE’ + C’E) + C’D T6 = A B ( 1 ) + C D T 2 = A B + C D T 1 3. a. F = X'Y + X'Y'Z’ XYZ X’ X’ • Y Y’ Z’ X’• Y’ • Z’ F 000 1 0 1 1 1 1 001 1 0 1 0 0 0 010 1 1 0 1 0 1 011 1 1 0 0 0 1 100 0 0 1 1 0 0 101 0 0 1 0 0 0 110 0 0 0 1 0 0 111 0 0 0 0 0 0
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299 b. F = W’ + X' • (Y' + Z) WXYZ W’ Y’ Y’+Z X’ X’ • (Y’+Z) F 0000 1 1 1 1 1 1 0001 1 1 1 1 1 1 0010 1 0 0 1 0 1 0011 1 0 1 1 1 1 0100 1 1 1 0 0 1 0101 1 1 1 0 0 1 0110 1 0 0 0 0 1 0111 1 0 1 0 0 1 1000 0 1 1 1 1 1 1001 0 1 1 1 1 1 1010 0 0 0 1 0 0 1011 0 0 1 1 1 1 1100 0 1 1 0 0 0 1101 0 1 1 0 0 0 1110 0 0 0 0 0 0 1111 0 0 1 0 0 0 c. F = (A' + B' + CD) • (B + C' + D') ABCD A’ B’ CD A' + B' + CD C’ D’ B + C' + D’ F 0000 1 1 0 1 1 1 1 1 0001 1 1 0 1 1 0 1 1 0010 1 1 0 1 0 1 1 1 0011 1 1 1 1 0 0 0 0 0100 1 0 0 1 1 1 1 1 0101 1 0 0 1 1 0 1 1 0110 1 0 0 1 0 1 1 1 0111 1 0 1 1 0 0 1 1 1000 0 1 0 1 1 1 1 1 1001 0 1 0 1 1 0 1 1 1010 0 1 0 1 0 1 1 1 1011 0 1 1 1 0 0 0 0 1100 0 0 0 0 1 1 1 0 1101 0 0 0 0 1 0 1 0 1110 0 0 0 0 0 1 1 0 1111 0 0 1 1 0 0 1 1
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300 d. F = (((A’ + B)' + C' )' + D)' = ((A’ + B)’ + C') • D’ DEMORGANS = (A • B’ + C’) • D’ DEMORGANS = AB’D’ + C’D’ T8 ABCD B’ D’ A•B’•D’ C’ C’•D’ F 0000 1 1 0 1 1 1 0001 1 0 0 1 0 0 0010 1 1 0 0 0 0 0011 1 0 0 0 0 0 0100 0 1 0 1 1 1 0101 0 0 0 1 0 0 0110 0 1 0 0 0 0 0111 0 0 0 0 0 0 1000 1 1 1 1 1 1 1001 1 0 0 1 0 0 1010 1 1 1 0 0 1 1011 1 0 0 0 0 0 1100 0 1 0 1 1 1 1101 0 0 0 1 0 0 1110 0 1 0 0 0 0 1111 0 0 0 0 0 0 4. a. F = Σ XYZ (0,2,3) = m0 + m2 + m3 = X’Y’Z’ + X’YZ’ + X’YZ = XYZ (1,4,5,6,7) = M1 • M4 • M5 • M6 • M7 = (X+Y+Z’)•(X’+Y+Z)•(X’+Y+Z’)•(X’+Y’+Z)•(X’+Y’+Z’) b. F = ABC (1,2,4,6) = M1 • M2 • M4 • M6 = (A + B + C’) • (A + B’ + C) • (A’ + B + C) • (A’ + B’ + C) = Σ A,B,C (0,3,5,7) = m0 + m3 + m5 + m7 = A’B’C’ + A’BC + AB’C + ABC
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301 c. F = Σ ABCD (1,2,5,7) = m1 + m2 + m5 + m7 = A’B’C’D + A’B’CD’ + A’BC’D +A’BCD = ABCD (0,3,4,6,8,9,10,11,12,13,14,15) = M0 • M3 • M4 • M6 • M8 • M8• M9 • M10 • M11 • M12 • M13 • M14 • M15 = (A + B + C +D) • (A + B + C’ +D’) • (A + B’ + C +D) • (A + B’ + C’ +D) • (A’ + B + C +D) • (A’ + B + C +D’) • (A’ + B + C’ +D) • (A’ + B + C’ +D’) • (A’ + B’ + C +D) • (A’ + B’ + C +D’) • (A’ + B’ + C’ +D) • (A’ + B’ + C’ +D’) d. F = X' + YZ' +YZ' = X’ + YZ’ T3 = X’(Y + Y’)(Z + Z’) + (X+X’)(YZ’) T1’/T5 = X’Y’Z’ + X’Y’Z + X’YZ’ + X’YZ + XYZ’ + X’YZ’ T8 = m0 + m1 + m2 + m3 + m6 = Σ XYZ (0,1,2,3,6) = X,Y,Z (4,5,7) = M4 • M5 • M7 = (X’ + Y + Z) • (X’ + Y + Z’) • (X’ + Y’ + Z’) e. F = A'B +B'C +AC’ = (A’BC + A’BC’) + (AB’C + A’B’C) + (ABC’ + AB’C’) T10 = m3 + m2 + m5 + m1 + m6 + m4 = Σ ABC (1,2,3,4,5,6) = ABC (0,7) = M0 • M7 = (A + B + C) • (A’ + B’ + C’) 5.
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AppendixAUnit8_2Sol - 8.2. Unit 2 - Boolean Algebra, Logic...

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