1. 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)*)*

2. 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

3. 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

4. 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

5. Minterms and Maxterms & Binary representationsC
Min-terms
Max-terms 1

6. 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)

7. 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

8. 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’)

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

15. 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

16. 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

17. 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