Presentation is loading. Please wait.

Presentation is loading. Please wait.

COMBINATIONAL LOGIC CIRCUITS C.L. x1 x2 xn Z Z = F (x1, x2, ……., Xn) F is a Binary Logic (BOOLEAN ) Function Knowing F Allows Straight Forward Direct Implementation.

Similar presentations


Presentation on theme: "COMBINATIONAL LOGIC CIRCUITS C.L. x1 x2 xn Z Z = F (x1, x2, ……., Xn) F is a Binary Logic (BOOLEAN ) Function Knowing F Allows Straight Forward Direct Implementation."— Presentation transcript:

1 COMBINATIONAL LOGIC CIRCUITS C.L. x1 x2 xn Z Z = F (x1, x2, ……., Xn) F is a Binary Logic (BOOLEAN ) Function Knowing F Allows Straight Forward Direct Implementation of the C.L. Circuit. OBJECTIVES Learn Binary Logic and BOOLEAN Algebra Learn How To Manipulate Boolean Expressions and Simplify Them Learn How to Map a Boolean Expression into Logic Circuit Implementation

2 Binary Logic & Logic Gates Elements of Binary Logic / Boolean Algebra 1- Constants, 2- Variables, and 3- Operators. Constant Values are either 0 or 1 Binary Variables  { 0, 1} 3 Possible Operators AND, OR, NOT Physically 1- Constants  Power Supply Voltage ( Logic 1)  Ground Voltage ( Logic 0) 2- Variables  Signals (High = 1, Low = 0) 2- Operators  Electronic Devices (Logic Gates) AND - Gate OR - Gate NOT - Gate (Inverter)

3 Logic Gates & Logic Operations

4 Multi-Input Gates Boolean Expression Combination of Boolean Variables, AND- operators, OR-operators, and NOT operators. Boolean Expressions (Functions) Can be Expressed as a Truth Table Boolean Algebra

5 ExampleF = X + Y. Z Boolean Algebra

6 Properties of Boolean Algebra 1- Parentheses 2- Not operator (Complement) 3- AND operator, 4- OR operator Operator Precedence

7 Example X + XY = X Proof: X + XY = X. (1 + Y) = X.1 = X Example X + X`Y = X + Y Proof: (1) X + X`Y = (X+ X`) (X + Y) = 1.(X + Y) = X + Y (2) X + X`Y = X.1 + X`Y = X.(1+Y) + X`Y = X +XY + X`Y (XY +X`Y) + X = X + Y Example ``Consensus Theory`` XY + X`Z + YZ = XY + X`Z Proof: XY + X`Z + YZ = XY + X`Z + YZ(X +X`)= XY(1 + Z) +X`Z(1 + Y) = XY + X`Z Algebraic Manipulation

8 Canonical & Standard Forms For 3- Variables X, Y, and Z Define the Following: (A) MinTerms m0 =X`Y`Z` = 1 IFF XYZ = 000 m1 =X`Y`Z = 1 IFF XYZ = 001 m2 =X`Y Z` = 1 IFF XYZ = 010 ………………………………………. m7 =X Y Z = 1 IFF XYZ = 111 (A) MaxTerms M0 = X + Y + Z = 0 IFF XYZ = 000 M1 = X + Y + Z` = 0 IFF XYZ = 001 M2 = X + Y` + Z = 0 IFF XYZ = 010 ………………………………………. M7 =X`+ Y`+ Z` = 0 IFF XYZ = 111 M i = m i (DeMorgan / Truth Table) 2 n minterms 2 n MaxTerms

9 Expressing Functions as a Sum of Minterms or a Product of MaxTerms Example: Consider the Function F Given By its Truth Table F = m 2 + m 4 + m 5 +m 7 =  (2, 4, 5, 7) = Sum of minterms F = m 0 + m 1 + m 3 +m 6 =  (0, 1, 3, 6) ORing Complementary MinTerms

10 Expressing Functions as a Sum of Minterms or a Product of MaxTerms F = m 2 + m 4 + m 5 +m 7 =  (2, 4, 5, 7) = Sum of minterms F = m 0 + m 1 + m 3 +m 6 =  (0, 1, 3, 6) F = (F) = m 0.m 1.m 3.m 6 = M 0.M 1.M 3.M 6 =  (0, 1, 3, 6) = Product of Maxterm F = M 2.M 4.M 5.M 7 =  (2, 4, 5, 7) ANDing

11 Expressing Functions as a Sum of Minterms or a Product of MaxTerms F =  (2, 4, 5, 7) =  (0, 1, 3, 6) F =  (0, 1, 3, 6) =  (2, 4, 5, 7) Example Implement SOP F = XZ + Y`Z + X`YZ` Two-Level Implementation Level-1: AND-Gates ; Level-2: One OR-Gate Standard Forms: Sum of Products (SOP) and Product of Sums (POS)

12 Expressing Functions as a Sum of Minterms or a Product of MaxTerms Example Implement POS F = (X+Z )(Y`+Z)(X`+Y+Z`) Two-Level Implementation Level-1: OR-Gates ; Level-2: One AND-Gate

13 Example Implement POS F = (X+Z). (Y`+Z). (X`+Y+Z`) Two-Level Implementation Level-1: OR-Gates ; Level-2: One AND-Gate


Download ppt "COMBINATIONAL LOGIC CIRCUITS C.L. x1 x2 xn Z Z = F (x1, x2, ……., Xn) F is a Binary Logic (BOOLEAN ) Function Knowing F Allows Straight Forward Direct Implementation."

Similar presentations


Ads by Google