Ptuece.loremate.com Digital Electronics 2

8/12/2019 Ptuece.loremate.com Digital Electronics 2

1/11

http://ptuece.loremate.com/die/node/2 March 7, 2012

Digital Electronics

Remember These:

1. Logic gates most commonly used are AND, OR. NOT.

NAND, NOR, XOR. XNOR.

2. NAND and NOR are universal gates.

3. Output of AND gate is low even if one input is low (Y = A.B) where A and B are inputs and Y isthe output.

4. Output of OR gate is high if any one input is high (Y = A + B)

5. In NOT gate, when a high is applied as input, a low appears at output and vice versa.

6. NAND gate has output high when any one of its input is low.

7. The output of NOR gate is high when any input is low.

8. Output of XOR gate is high if one an only one input is high.

9. The output of XNOR gate is high when all inputs are high.

10. NAND and NOR can be used to realize any gate.

11. SOP involves sum of given product terms and these terms are known as minterms (m).

12. POS involves product of given sum terms of these are known as maxterms (M).

13. A karnaugh map is simply a graphical method for representing a boolean function. It is used tosimplify a logic equation or to convert a truth table to its logic circuit.

14. Types at K-map are 2 variable, 3 variable, 4 variable, 5 variable and 6 variable.

15. = 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.

16. For representing SOP form for K map: enter 1 for each minterm and 0 otherwise.

17. To minimize the boolean expression using K-map, pair, Quad and octet are formed inincreasing priority.

18. Q-M method or Quine Mc-Clusky method or Tabular minimization method is used for largenumber of variables if increases from 6 variable i.e. for 7, 8, 9 or oven 10 variables. K-mapmethod fails for large number of variables.

19. Incompletely specified functions are dont care conditions. In these cases output level in rotdefined, it can be high or low i.e. 1 or 0.

20. Dont 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 Booleanexpression. It is used to simplify a logic expression or to convert a truth table to its correspondinglogic circuit. It is used for the minimization of switching functions but upto six variables. For more
http://www.printfriendly.com/print?url=http%3A%2F%2Fptuece.loremate.com%2Fdie%2Fnode%2F2&partner=a2ahttp://www.printfriendly.com/print?url=http%3A%2F%2Fptuece.loremate.com%2Fdie%2Fnode%2F2&partner=a2a


2/11

than six variable it becomes complex or cubersome.

The K-map for n-Boolean variable switching function consists of squares. Here squarerepresents 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 Booleanexpression, 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 ofvariables, M = Number Ofsquares

n = 4

So, 4 variableK-map consists

of 16 squares.

4. Five and SixVariable K-maps: Fivevariable K-maphas squaresand five

variables areA, B, C, D

and E. It isshown infigure.

Grouping of2 variable K-


3/11

map.

Q 2. Solvefollowingusing K-mapand Booleanalgebra:

Solution.

Usingboolean,

So Y =B, is thesimplified expression.

Using boolean,

It is the simplifiedexpression.

Grouping of 3 variable K-

map

Q 3 Solve following usingK-map and booleanalgebra:

Solution.

Using boolean,

Using boolean,

Minimized output of K-mapis V = C

Using boolean,

Grouping of 4 variable K-

map.

Q 4 Solve the followingusing K-map and verify byusing boolean algebra:


4/11

(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


5/11

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 variablesthe total maximum number ofterms which the

expression can have willbe

Ans. A function having 6variables, then the totalmaximum number of terms willbe 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 NORgates.

Ans. (i) AND gate using NOR gates:

(ii) OR gate using NOR gates:

(iii ) NOR gate using NOR gateonly:


6/11

Q 13. Realize X-ORfunction using NORgates only.

Ans. Realization of

X-OR using NORgate 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.


7/11

(iii ) Minimize it and realize the minimized expression using NANDgates 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 withfour entities. Write the corresponding Booleanexpression.

Ans. The corresponding boolean expression is

abcd + a b c d + abc d + abc d

Also, the minimised output is cd.

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


8/11


9/11


10/11


11/11

Ptuece.loremate.com Digital Electronics 2

Documents

Transcript of Ptuece.loremate.com Digital Electronics 2