Remember These:
- Logic gates most commonly used are AND, OR. NOT. NAND, NOR, XOR. XNOR.
- NAND and NOR are universal gates.
- Output of AND gate is low even if one input is low (Y = A.B) where A and B are inputs and Y is the output.
- Output of OR gate is high if any one input is high (Y = A + B)
- In NOT gate, when a high is applied as input, a low appears at output and vice versa.
- NAND gate has output high when any one of its input is low.
- The output of NOR gate is high when any input is low.
- Output of XOR gate is high if one an only one input is high.
- The output of XNOR gate is high when all inputs are high.
- NAND and NOR can be used to realize any gate.
- SOP involves sum of given product terms and these terms are known as minterms (m).
- POS involves product of given sum terms of these are known as maxterms (M).
- A karnaugh map is simply a graphical method for representing a boolean function. It is used to simplify a logic equation or to convert a truth table to its logic circuit.
- Types at K-map are 2 variable, 3 variable, 4 variable, 5 variable and 6 variable.
- = M formula is used for the calculation of total number of squares in a K-map. Here, n=number of variables and M = number of squares.
- For representing SOP form for K map: enter 1 for each minterm and 0 otherwise.
- To minimize the boolean expression using K-map, pair, Quad and octet are formed in increasing priority.
- Q-M method or Quine Mc-Clusky method or Tabular minimization method is used for large number of variables if increases from 6 variable i.e. for 7, 8, 9 or oven 10 variables. K-map method fails for large number of variables.
- Incompletely specified functions are don’t care conditions. In these cases output level in rot defined, it can be ‘high’ or low’ i.e. ‘1’ or ‘0’.
- Don’t care conditions are marked by ‘d’ or x’ or
Q.1 What is K-map? Why we need K-maps? Give the various types of K-map.
Ans. K-map i.e. Karnaugh map is simply a graphical method for representing a Boolean
fraction. The Karnaugh
map is a systematic method for simplify and manipulating Boolean
expression. It is used to simplify a logic expression or to convert a
truth table to its corresponding logic circuit. It is used for the minimization of switching functions but upto ‘six’ variables. For more than ‘six’ variable it becomes complex or cubersome.
The K-map for n-Boolean variable switching function consists of
squares. Here square represents the normal or standard term i.e. one minterm or maxterm.
Need of K-maps: We need K-map for representing
Boolean function through graphical method. Because K-map simplify and
manipulates a Boolean expression. So to solve or simplify a Boolean
expression, we use K-map. K-map can be used for problems involving any number of input variables (upto six variables) which is not easily solve by Boolean Algebra. Types of K-map :
Types of K-maps commonly used are
1. Two variable K-map
2. Three variable K-map
3. Four variable K-map
4. Five variable K-map
5. Six variable K-map
1 Two variable K-map
= M formula is used where, n = Number of variables and M = Number of squares.
2. Three Variable K-map:
3. Four Variable K-map:
Formula used is = M
Where, n = number of variables, M = Number Of squares
n = 4
So, ‘4’ variable K-map consists of ‘16’ squares.
4. Five and Six Variable K-maps: Five variable K-map has 
squares and five
variables are A, B, C, D and E. It is shown in figure.
Grouping of 2 variable K-map.
Q 2. Solve following using K-map and Boolean algebra:
Solution.
Using boolean,
So Y = B, is the simplified expression.
Using boolean,
It is the simplified expression.
Grouping of 3 variable K-map
Q 3 Solve following using K-map and boolean algebra:
Solution.
Using boolean,
Using boolean,
Minimized output of K-map is V = C
Using boolean,
Grouping of 4 variable K-map.
Q 4 Solve the following using K-map and verify by using boolean algebra:
(i) F (A, B, C, D) = 
(3, 4, 5,7, 9, 13, 14, 15)
Solution
(ii) F (A, B, C, D) = (0, 1, 2,3, 6, 8, 9, 10, 11, 12, 13)
Solution.
(iii) F (A, B, C, D) = (0, 1, 4, 5, 3, 2, 11, 10)
Solution
Q 5. Solve the following using k-map:
f (A, B, C, D) = (0,2, 3, 8, 11, 12) + d (1,9, 14)
Ans. K-map is as shown:
Q 6. Find the maxterms for the expression 
Ans.
The max terms are given by:
Q 7 Construct the truth table for F = 
Ans Truth table for F = 
This is the output of an XOR gate.
Truth Table :
Q 8 Convert the given expression in canonical SOP form Y = AC + AB + BC
Ans.Y=AC +AB+ BC
The canonical SOP form is given by:
Q 9. Two minterms can be adjacent only if they differ by ……
Ans. Single bit.
(i) Evaluate the expressions
(i)A+1;
(ii) A + A;
Ans. (i) A + 1 = 1
It comes under OR laws.
Oring an input with 1, always results in a high output.
A + 1 = 1
Case I, Let A = 0 0+ 1 = 1
Case II, Let A = 1 1 + 1 = 2
Hence, A + 1 = 1
(ii) A + A = A
Case I, Let A = 0 0 + 0 = 0
Case II, Let A = 1 1 + 1 = 1
Hence, A + A = A
Q 10. In a function of six variables the total maximum number of terms which the
expression can have will be ……
Ans. A function having 6 variables, then the total maximum number of terms will be given by:
Q 11. Explain the difference between Boolean operations OR and XOR. Use truth
tables to described how these operations differ.
Ans.
Q 12. Realize and OR and NOT using NOR gates.
Ans. (i) AND gate using NOR gates:
(ii) OR gate using NOR gates:
(iii) NOR gate using NOR gate only:
Q 13. Realize X-OR function using NOR gates only.
Ans. Realization of X-OR using NOR gate only
Q14. A four-variable function is given as f (A, B, C, D) = 
(0, 2, 3, 4, 5, 7, 8, 13, 15). Use a K-map to minimize the function.
Ans. f (A B, C D) = (0, 2,3, 4 5, 7, 8, 13, 15)
Q. 15. Simplify the expression 2. =AB + AC + ABC (AB + C). Implement using
minimum number of NAND gates.
Ans Z = AB + AC + ABC (AB + C)
= AB + AC + ABC. AB + ABCC
= AB + AC + ABC + ABC
= AB + AC + ABC
=AB(1 + C) + AC
= AB + AC
Implementation using minimum number & NAND gates:
Q 16. A four-variable function is given as f (A, B, C, 0) = HM (0, 3, 4, 5, 6, 7, 11, 13,
15). Use a K-map to minimize the function.
Ans.
Q 17. (i) Make a K-map for the function:
(ii) Express f in standard SOP form.
(iii) Minimize it and realize the minimized expression using NAND gates only.
Ans. (i)
(ii) Express f in standard SOP form
Circuit diagram:
Q 18. Realize OR, AND, NOT, NOR gates using NAND gates only.
Ans. (i) OR gate using NAND only:
(ii) AND gate using NAND only:
(iii) NOT gate using NAND only.:
(iv) NOR gate using NAND only:
Q 19. Minimize the function using K-map.
f (A, B, C, D) = 
(0, 1, 2, 3, 5, 7, 8, 9, 11,14)
Ans.
Q 20. Given below a four variable Karnaugh map with four entities. Write the corresponding Boolean expression.
Ans. The corresponding boolean expression is
a’b’c’d + a’ b c’ d + abc’ d + ab’c’ d
Also, the minimised output is c’d.
Q 21.Obtain the minimal SOP expression for
(0, 1, 2, 4, 6, 9. 11, 12, 13) and implement it in NAND logic.
Ans. f (A, B, C, D) = (0, 1, 2, 4, 6, 9. 11, 12, 13)
Firstly K-map
Circuit diagram using NAND only
Q 22. Obtain the minimal POS expression for 
(0, 1, 2, 4, 5, 6, 9, 11, 12, 13, 14, 15) and implement it in NOR logic.
Ans. Firstly, fill given o’s in K-map, to get the expression in POS form.
Implementation using NOR gates
