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 |...
TRANSCRIPT
Boolean Algebra
AND gate
A B | AB0 0 | 00 1 | 01 0 | 01 1 | 1
OR gate
A B | A + B0 0 | 00 1 | 11 0 | 11 1 | 1
NOT gate
_A | A or A’0 | 11 | 0
Any digital logic circuit can be built just using these three gates as building blocks
Order of Operations
1. Parenthesis if any
2. NOT
3. AND
4. OR
e.g. A(B + C’D) + AB’
Order of Operations Continuede.g. A(B + C’D) + AB’ where A = 0,
B = 1, C = 1, D = 0• C’ = 0, B’ = 0 NOTs first• C’D = 0, AB’ = 0 ANDs • B + C’D = 1 parenthesis• A(B + C’D) = 0 AND• A(B + C’D) + AB’ = 0 OR
Truth Table• Draw a truth table for
Z = AB’ + AC + A’B’C
A B C A’ B’ AB’ AC A’B’C Z
0 0 0 1 1 0 0 0 0
0 0 1 1 1 0 0 1 1
0 1 0 1 0 0 0 0 0
0 1 1 1 0 0 0 0 0
1 0 0 0 1 1 0 0 1
1 0 1 0 1 1 1 0 1
1 1 0 0 0 0 0 0 0
1 1 1 0 0 0 1 0 1
Derive a Relationship from a Truth Table
# A B C Q
0 0 0 0 0
1 0 0 1 1
2 0 1 0 0
3 0 1 1 1
4 1 0 0 1
5 1 0 1 1
6 1 1 0 0
7 1 1 1 0
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’)
Name And Form Or Form
Identity 1A = A 0 + A = A
Null Law 0A = 0 1 + A = 1
Idempotent Law AA = A A + A = A
Inverse or Complement Law A’A = 0 A’ + A = 1
Commutative Law AB = BA A + B = B + A
Associative Law (AB)C = A(BC) (A + B) + C = A + (B + C)
Distributive Law (A + B)(A +C) = A + BC A(B +C) = AB + AC
Absorption Law A(A + B) = AandA(A‘ + B) = AB
A + AB = AandA + 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
Duality of Boolean ExpressionsA 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 itsdual is: A(B + C) = AB + C
Other Universal Gates
NAND gateAny digital logic circuit can be built just using this gate
A B|(AB)’0 0 | 10 1 | 11 0 | 11 1 | 0
NOR gateAny digital logic circuit can be built just using this gate
A B|(A+B)’0 0 | 10 1 | 01 0 | 01 1 | 0
Other useful gates to build adders and subtractors
XOR (exclusve OR) gate
A B| A B0 0 | 00 1 | 11 0 | 11 1 | 0
XNOR (exclusve NOR) gate
A B| (A B)’0 0 | 10 1 | 01 0 | 01 1 | 1
Develop a Boolean Expression from a Boolean Circuit
X
Y
Z
Develop a Boolean Expression from a Boolean Circuit
R1 R4
R5
X
Y
Z
R2
R3
R1 = Y’R2 = XY’R3 = Z’R4 = (R2)+(R3)=XY’Z’R5 = X(R3)=XZ’Output = R4 + R5 = XY’Z’ + XZ’
Synthesis of Combinational CircuitsBuild a circuit to implement the following truth table:
X Y Z P
0 0 0 0
0 0 1 0
0 1 0 1
0 1 1 1
1 0 0 0
1 0 1 1
1 1 0 0
1 1 1 1
Synthesis of Combinational CircuitsBuild a circuit to implement the following truth table:
# X Y Z P
0 0 0 0 0
1 0 0 1 0
2 0 1 0 1
3 0 1 1 1
4 1 0 0 0
5 1 0 1 1
6 1 1 0 0
7 1 1 1 1
Do SOP form:P = Σm(2,3,5,7) = X’YZ’ + X’YZ + XY’Z + XYZ
Synthesis of Combinational CircuitsBuild a circuit to implement the following truth table:
Do SOP form:P = Σm(2,3,5,7) = X’YZ’ + X’YZ + XY’Z + XYZ
X Y Z
P
Synthesis of Combinational CircuitsBuild 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
X Y Z
P
Synthesis of Combinational CircuitsBuild a circuit to implement the following truth table:
# X Y Z P
0 0 0 0 0
1 0 0 1 0
2 0 1 0 1
3 0 1 1 1
4 1 0 0 0
5 1 0 1 1
6 1 1 0 0
7 1 1 1 1
Do POS form:P = ΠM(0,1,4,6) = (X+Y+Z)(X+Y+Z’)(X’+Y+Z)(X’+Y’+Z)
Synthesis of Combinational CircuitsBuild 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)
X ZY
Synthesis of Combinational CircuitsBuild 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)
X ZY