1. Boolean Algebra AND gate A B | AB 0 0 | 0 0 1 | 0 1 0 | 0 1 1 | 1 OR gate A B | A + B 0 0 | 0 0 1 | 1 1 0 | 1 1 1 | 1 NOT gate _ A | A or A’ 0 | 1 1 | 0 Any digital logic circuit can be built just using these three gates as building blocks

2. Order of Operations 1.Parenthesis if any 2.NOT 3.AND 4.OR e.g.A(B + C’D) + AB’

3. Order of Operations Continued e.g.A(B + C’D) + AB’ where A = 0, B = 1, C = 1, D = 0 C’ = 0, B’ = 0NOTs first C’D = 0, AB’ = 0ANDs B + C’D = 1parenthesis A(B + C’D) = 0AND A(B + C’D) + AB’ = 0 OR

4. Truth Table Draw a truth table for Z = AB’ + AC + A’B’C ABCA’B’AB’ACA’B’CZ 000110000 001110011 010100000 011100000 100011001 101011101 110000000 111000101

5. Derive a Relationship from a Truth Table #ABCQ 00000 10011 20100 30111 41001 51011 61100 71110 SOP(sum of products or minterm) form: Q = 1 + 3 + 4 + 5 = Σm(1,3,4,5) Q = A’B’C + A’BC + AB’C’ + AB’C POS(product of sums or maxterm) form: Q = 0·2·6·7 = ΠM(0,2,6,7) Q = (A+B+C)(A+B’+C)(A’+B’+C)(A’+B’+C’)

6. NameAnd FormOr Form Identity1A = A0 + A = A Null Law0A = 01 + A = 1 Idempotent LawAA = AA + A = A Inverse or Complement LawA’A = 0A’ + A = 1 Commutative LawAB = BAA + B = B + A Associative Law(AB)C = A(BC)(A + B) + C = A + (B + C) Distributive Law(A + B)(A +C) = A + BCA(B +C) = AB + AC Absorption LawA(A + B) = A and A(A‘ + B) = AB A + AB = A and A + A’B = A + B Consensus(A+B)(A’ + C)(B + C) = (A + B)(A’ + C)AB + A’C + BC = AB + A’C DeMorgan’s Law___ _ _ AB = A + B _____ _ _ A + B = A B

7. Duality of Boolean Expressions A dual of a Boolean expression is obtained by replacing each ‘+’ with ‘·’ and each ‘·’ with ‘+’, each 1 by 0 and each 0 by 1. If a Boolean equation is true then its dual is also true. e.g. given: A + BC = (A + B)C then its dual is: A(B + C) = AB + C

8. Other Universal Gates NAND gate Any digital logic circuit can be built just using this gate A B|(AB)’ 0 0 | 1 0 1 | 1 1 0 | 1 1 1 | 0 NOR gate Any digital logic circuit can be built just using this gate A B|(A+B)’ 0 0 | 1 0 1 | 0 1 0 | 0 1 1 | 0

9. Other useful gates to build adders and subtractors XOR (exclusve OR) gate A B| A B 0 0 | 0 0 1 | 1 1 0 | 1 1 1 | 0 XNOR (exclusve NOR) gate A B| (A B)’ 0 0 | 1 0 1 | 0 1 0 | 0 1 1 | 1

10. Develop a Boolean Expression from a Boolean Circuit

11. R1 = Y’ R2 = XY’ R3 = Z’ R4 = (R2)+(R3)=XY’Z’ R5 = X(R3)=XZ’ Output = R4 + R5 = XY’Z’ + XZ’

12. Synthesis of Combinational Circuits Build a circuit to implement the following truth table: XYZP 0000 0010 0101 0111 1000 1011 1100 1111

13. Synthesis of Combinational Circuits Build a circuit to implement the following truth table: #XYZP 00000 10010 20101 30111 41000 51011 61100 71111 Do SOP form: P = Σm(2,3,5,7) = X’YZ’ + X’YZ + XY’Z + XYZ

14. Synthesis of Combinational Circuits Build a circuit to implement the following truth table: Do SOP form: P = Σm(2,3,5,7) = X’YZ’ + X’YZ + XY’Z + XYZ

15. Synthesis of Combinational Circuits Build a circuit to implement the following truth table: Do SOP form: P = Σm(2,3,5,7) = X’YZ’ + X’YZ + XY’Z + XYZ Try to simplify circuit: P = X’Y(Z’+Z) + XZ(Y’+Y) P = X’Y + XZ

16. Synthesis of Combinational Circuits Build a circuit to implement the following truth table: #XYZP 00000 10010 20101 30111 41000 51011 61100 71111 Do POS form: P = ΠM(0,1,4,6) = (X+Y+Z)(X+Y+Z’)(X’+Y+Z)(X’+Y’+Z)

17. Synthesis of Combinational Circuits Build a circuit to implement the following truth table: Do POS form: P = ΠM(0,1,4,6) = (X+Y+Z)(X+Y+Z’)(X’+Y+Z)(X’+Y’+Z)

18. Synthesis of Combinational Circuits Build a circuit to implement the following truth table: Try to simplify circuit: P=(XX+XY+XZ’+YX+YY+YZ’+ZX+ZY+ZZ’)· (X’X’+X’Y’+X’Z+YX’+YY’+YZ+ZX’+ZY’+ZZ) P=(X+XY+XZ’+Y+YZ’+XZ+YZ+0)(X’+X’Y’+X’Z+X’Y+0+YZ+X’Z+Z) P=(X(1+Y+Z’+Z)+Y(1+Z))(X’(1+Y’+Z+Y+Z)+Z(Y+1)) P=(X+Y)(X’+Z) Do POS form: P = ΠM(0,1,4,6) = (X+Y+Z)(X+Y+Z’)(X’+Y+Z)(X’+Y’+Z)