Analysis of Logic Circuits Example 1 Consider the logic circuit comprising of NOT gates, AND gates, NAND gate and NOR gate. The circuit can be represented by a Boolean Expression Let us define the Boolean expression that represents the Logic circuit. Starting from the left hand side of the circuit Output of NOT gate 1 is B* Output of NOT gate 2 is A* Output of two input AND gate 3 is AB* Output of two input AND gate 4 is CD Output of three input NAND gate 5 is (A*B*CD)* Output of two input NOR gate 6 is (AB*+(A*B*CD)*)*

Evaluating Boolean Expression The expression Assume and Expression Conditions for output = 1 X=0 & Y=0 Since X=0 when A=0 or B=1 Since Y=0 when A=0, B=0, C=1 and D=1 Now consider the expression which has two terms The term AB* and the term (A*B*CD)* Assume that the term AB* is represented by literal X Also assume that the term (A*B*CD)*is represented by the literal Y Thus the expression is now represented as (X+Y)* For which condition does the expression output a 1? The expression shows that a NOR operation is being performed between terms X and Y. Therefore when terms X and Y are both 0s only then the output is a 1. The term X=AB* is 0 when literal A is 0 or literal B* is zero or B is 1. The term Y=(A*B*CD)* is 0 when the term A*B*CD =1 the bar has been removed. The term A*B*CD is 1 when literals A*=1, B*=1, C=1 and D=1 Or when A=0, B=0, C=1 and D=1

Evaluating Boolean Expression & Truth Table Conditions for o/p =1 A=0, B=0, C=1 & D=1 Input Output A B C D F 1 Conditions for output =1 are For the term AB* A=0 OR B=1 For the term (A*B*CD)* A=0 AND B=0 AND C=1 AND D=1 For both terms A has to be 0 otherwise the output is not equal to 1 B has to be 0 otherwise the second term is equal to 1 and the output 0 Similarly C and D variables have to be equal to 1s otherwise the second term is equal to 1 and the output 0 The truth table describes the function of the logic circuit and the expression where all outputs are equal to 0 except for the input combination of A=0, B=0, C=1 & D=1

Minterms and Maxterms Minterms: Product terms in Standard Sum Of Product (SOP) form Maxterms: Sum terms in Standard Product of Sum (POS) form The following truth table represent SOP and POS terms

Minterms and Maxterms & Binary representations C Min-terms Max-terms 1

Sum of products (SoP): OR of ANDs e.g. F = Y + XYZ + XY Product of sums (PoS): AND of ORs 6 e.g. G = X(Y + Z)(X + Y + Z)

Converting from PoS (or any form) to SoP Converting from PoS (or any form) to SoP Just multiply through and simplify, e.g., G = X(Y + Z)(X + Y + Z) = XYX + XYY + XYZ + XZX + XZY + XZZ = XY + XY + XYZ + XZ + XZY + XZ = XY + XZ

Converting from SoP to PoS Complement, multiply through, complement via DeMorgan, e.g., F = Y’Z’ + XY’Z + XYZ’ F' = (Y+Z)(X’+Y+Z’)(X’+Y’+Z) = YZ + X’Y + X’Z (after lots of simplification) F = (Y’+Z’)(X+Y’)(X+Z’)

e.g., Minterms for 3 variables A,B,C

Karnaugh Map Simplification of Boolean Expressions Doesn’t guarantee simplest form of expression Terms are not obvious Skills of applying rules and laws K-map provides a systematic method An array of cells Used for simplifying 2, 3, 4 and 5 variable expressions

3-Variable K-map Used for simplifying 3-variable expressions K-map has 8 cells representing the 8 minterms and 8 maxterms K-map can be represented in row format or column format

4-Variable K-map Used for simplifying 4-variable expressions K-map has 16 cells representing the 16 minterms and 8 maxterms A 4-variable K-map has a square format