NH – 67, Karur – Trichy Highways, Puliyur C.F, 639 114 Karur District

DEPARTMENT OF INFORMATION TECHNOLOGY

CS 2202 DIGITAL PRINCIPLES AND SYSTEM DESIGN

*The NPTEL material on digital circuits for

the IT branch-3rd Semester is available

here.

UNIT 1

BOOLEAN ALGEBRA AND LOGIC GATES

Review of binary number systems – Binary arithmetic – Binary codes

– Boolean algebra and theorems – Boolean functions – Simplification

of Boolean functions using Karnaugh map and tabulation methods –

Logic gates

INTRODUCTION

The world of electronics is divided into two areas: analog and

digital. Analog circuits consist mainly of amplifiers for voltage or

current variations that are smooth and continuous.

Digital circuits provide electronic switching of voltage pulses.

A pulse has abrupt changes between two extreme amplitude levels

(i.e.: 5 volt = high level and 0 volt = low level).

Since the digital signal has only two significant levels, either

high or low, it is useful to represent the pulses in a binary number

system with the digits 1 and 0.

Digital electronics are electronics systems that use digital

signals. Digital electronics are used in computers, mobile phones,

and other consumer products.

Digital electronics or any digital circuits are usually made from

large assemblies of logic gates, simple electronic representations of

Boolean logic functions.

Advantages of digital system

1. Digital systems interface well with computers and are easy to

control with software. New features can often be added to a

digital system without changing hardware. Often this can be

done outside of the factory by updating the product's software.

So, the product's design errors can be corrected after the product

is in a customer's hands.

2. Information storage can be easier in digital systems than in

analog ones. The noise-immunity of digital systems permits data

to be stored and retrieved without degradation. In an analog

system, noise from aging and wear degrade the information

stored. In a digital system, as long as the total noise is below a

certain level, the information can be recovered perfectly.

1.1 NUMBER SYSTEM

There are four important number systems which you should

become familiar with. These are decimal, binary, octal and

hexadecimal. The decimal system, which is the one you are most

familiar with, utilizes ten symbols to represent each digit. These are

0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. This system is referred to as base 10,

or radix 10 system. Similarly, the binary system is base 2, octal is

base 8, and hexadecimal is base 16 as shown in the following Table

Number System Radix SymbolsBinary 2 0 1Octal 8 0 1 2 3 4 5 6 7Decimal 10 0 1 2 3 4 5 6 7 8 9Hexadecimal 16 0 1 2 3 4 5 6 7 8 9 A B C D E F

Table below shows an example of counting from 0 to 18 in each of the four number

systems.

Decimal Binary Octal Hexadecimal 0 0000 0 0 1 0001 1 1 2 0010 2 2 3 0011 3 3 4 0100 4 4 5 0101 5 5 6 0110 6 6 7 0111 7 7 8 1000 10 8 9 1001 11 9 10 1010 12 A 11 1011 13 B 12 1100 14 C 13 1101 15 D 14 1110 16 E 15 1111 17 F 16 10000 20 10 17 10001 21 11 18 10010 22 12

Number Conversion

Working in binary can become very cumbersome and prone to

errors. It is important to learn how to convert from one system to

another. Let us take the binary pattern 1011101 as an example. If

we are working predominantly in the octal system, we would break

this into groups of three bits, starting from the right

1011101 = 1 011 101 = 1358

The subscript 8 is used to indicate that it is the octal representation.

After converting to octal, the conversion to base 10 is simplified.

Mathematically, the decimal value can be computed as (1 ×

82) + (3 × 81) + (5 × 80) = 93. However, in a computer program it

would be more efficient to use an iterative algorithm. Thus the

result is computed as:

((1 × 8) + 3) × 8 + 5 = 93.

Conversion to Binary

Converting an octal or hexadecimal number to its binary

representation is straight forward. For example:

3618 = 011 110 0012

1A416 = 0001 1010 01002

Converting a decimal number to its binary representation is

slightly more complex. It is simplified if you convert the decimal

number to its octal or hexadecimal equivalent first. For example, let

us find the binary representation of 144910. First, we shall convert it

to its octal representation.

1.2 Binary number system

The native language of digital computers is inherently binary.

Thus, numbers are naturally represented as binary integers by the

computer. The range of numbers which can be represented is

dependent on the number of bits used. A common unit of storage is

a byte which is a group of eight bits. Eight bits can represent 28

unique states, i.e. 256 possible combinations. This fact can be used

to our advantage to represent many different things on the

computer. For example, a byte can be used to represent 256

colours, 256 shades of grey, 256 shapes, 256 symbols, 256 names,

or even 256 different numbers. More typically, we can use a byte to

represent 256 sequential numbers such as the numbers from 1 to

256, or the numbers from 0 to 255. Remember that 0 is not

"nothing" and has equal significance as any other number.

Signed Integers

One of the eight bit is reserved to indicate the sign of the

number. Following usual convention, the left-most bit (called the

most significant bit or MSB) is dedicated as the sign bit. For the

numbers from -128 to 127, there are 256 unique numbers in this

range. With this scheme, we can now have 128 positive and 128

negative numbers, typically the numbers from 0 to 127. This

scheme has the anomaly that there are two unique representations

for positive and negative zero. A more serious problem is that the

rules of binary arithmetic breakdown when we decrement by 1 from

positive to negative numbers.

To overcome the discontinuity at zero, a scheme of

complementary binary is used whereby negative numbers are

represented by the binary complement of the positive

representation. This system is called one's complement binary.

This system also has two unique representations for positive and

negative zero. The two's complement binary system overcomes

both problems mentioned above. In this notation, the negative

number is represented by forming the binary complement of the

positive number and adding one.

To summarize, integers are commonly represented by the

unsigned binary and the 2's complement binary

representations. The range of integer numbers can be extended by

utilizing more bits. For example, 16 bits will allow us to represent 216

unique items, i.e. 65536 possible states. These can be the numbers

from 0 to 65535 for unsigned integers or -32768 to 32767 for 2's

complement signed integers

00011001 ------------ +25

10011001 ------------ -25 SIGN- MAGNITUDE form

11100110 ----------- -25 1s compliment form

11100111 ----------- -25 2s compliment form

1.3 Binary Arithmetic

It shows the way in which the manipulation of bits inside the

logic units or devices being carried out.

Binary Addition

The rules for binary integer arithmetic follow the same rules

as for decimal numbers.

1+0 = 1

1+1 = 10 Sum = 0 ; Carry = 1

1+1+1 = 11 Sum = 1 ; Carry = 1

Binary Subtraction

0-0 = 0

1-1 = 0

1-0 = 1

0-1 = 1 Borrowing 1 makes 0 to 10.

(Ex)

Binary Multiplication

Usually, two other logic functions, left shift and right shift, are

provided since these are also easy to implement in hardware. When

a number is shifted by one digit to the left, for example, 123 left

shifted becomes 1230, this is equivalent to multiplying the number

by the radix, whatever it may be. Similarly, 123 shifted to the right

is 12.3 and this is the same as dividing by the radix. In binary,

shifting left by one bit is equivalent to multiplying by 2. Shifting

right by one bit is the same as dividing by 2.

To find 2941 x 318, the first number is called the multiplicand and

the second is the multiplier, there are two ways depending on

which end of the multiplier we begin with. The following should look

familiar:

  2941   × 318

2941 × 8 = 23528 29410 × 1 = 29410

294100 × 3 = 882300  935238

A less familiar method is the following where we start with the left digit of the multiplier:

2941 × 3 = 8823 × 100 = 882300 2941 × 1 = 2491 ×10 = 24910 2941 × 8 = 23528 × 1 = 23528

      935238

In both cases, the product can be formally stated as the sum

of partial products as follows:

2941 × 318 = (2941 × 3 × 102 ) + (2941 × 1 × 101 ) + (2941 × 8 ×

100 )

Multiplication in binary (or any other radix) can be performed

using the same technique as for decimal. Both methods shown are

easily implemented in binary on a digital computer. In the examples

shown above, the partial products are first formed and the

summation is done at the end. In a computer algorithm this is not the

usual case since it is more efficient to sum the partial products as

they are formed. In special processors optimized for speed, such as

digital signal processors (DSP), dedicated hardware to do this is

called a multiply and accumulate register (MAC).

Multiplying in binary follows the same rules.

111 x 101. This can be formulated as

follows:

(111 × 1 × 22 ) + (111 × 0 × 21 ) + (111 × 1 × 20 )

With binary arithmetic, the task of multiplying by 2 is simply a shift

of one bit to the left. Similarly, dividing by 2 is a shift of one bit to

the right.

Binary Division

Division in binary follows the same concept as for long division

in decimal

(Ex)

1.4 Boolean Algebra

George Boole (1854) invented a new kind of algebra that

could be used to analyse and design digital and computer circuits.

1.4.1 Boolean laws and theorems

Certain rules and theorems are defined to facilitate the

simplification of Boolean expressions inturn making the simpler logic

circuits with reduced number of gates.

Basic laws

Commutative law:

A+B =B+A

AB = BA

Associative law:

A + (B+C) = (A+B) + C

A(BC) = (AB)C

Distributive law:

A (B+C) = AB + AC

Boolean relations about OR operations

A+0=A

A+A=A

A+1=1

Boolean relations about AND operations

A ' 1 = A

A ' A = A

A ' 0 = 0

1.4.2 Postulates

1. IdentityIdentity elements exist for each operation that leaves the result unchanged.A+0 = AA.0 = 0A+1 = 1A.1 = A

An operation between two identical variables will yield a result that is unchanged A+A = AA.A = A2. InverseFor every element, there is an inverse or complement with the following propertiesA.A’= 0A+A ‘= 1

3. The complement of a complement gives back the original quantity(A’)’ = A

1.4.3 Application:

It provides the systematic mathematical approach to study, design

and analyze the logic circuits.

Boolean expressions enables the design and simplification of logic

circuits.

1.5 Logic Gates

1.5.1 NOT operation

Fig. 2-1: Inverter symbol and Boolean notation

Ex: If A is 0 (low) ' X = NOT 0 = 1

In Boolean algebra the overbar stands for NOT operation.

1.5.2 OR operation

Fig. 2-2: OR symbol and Boolean notation

Ex: If A = 0, B = 1 ' X = A or B = 0 or 1 = l

In Boolean algebra the + sign stans for the OR operation:

X=A+B

Ex: If A = 1, B = 0 ' X = A + B = 1 + 0 = 1

1.5.3 AND operation

Fig. 2-3: AND symbol and Boolean notation

In Boolean algebra the multiplication sign stands for the AND

operation

X = AB

Ex: If A = 1, B = 0 ' X = A B = 1 ' 0 = 0

1.5.4 NOR gate

Based on the three fundamental logic operations it is possible to design additionel logic devices.

Fig. 2-6: NOR gate, symbol and truth table

1.5.5 NAND gate

Fig. 2-7: NAND grate, symbol and truth table

1.5.6 XOR gateEvents which are true only if and only if one of the motivating

events are true

Truth Table A B F 0 0 0 0 1 1 1 0 1 1 1 0

1.6 DE MORGAN’S THEOREMS

1.

This states that the inverse (i.e.)of a product [and] is equal to the

sum [or] of the complements

2.

This states that the inverse (complement) of a sum [or] is equal to

the product [and] of the complements

These theorems can be extended to cover several variables:

1.7 Canonical and Standard forms

Consider two binary variables x and y combined by an AND

operation

X’. Y’ 0 0 m0

X’. Y 0 1 m1

X . Y’ 1 0 m 2 Where X’ – Primed representing

binary ‘0’

X . Y 1 1 m 3 And X – Unprimed representing

binary ‘1’

mx – Minterm or standard

product

For n variables, there are in total 2n Minterms.

For OR operation,

X + Y 0 0 M0

X + Y’ 0 1 M1

X’+ Y 1 0 M2

X’ + Y’ 1 1 M3 Mx – Maxterm or standard sum

Each Maxterm is the complement of corresponding Minterm.

Boolean expression can be expressed in terms of Minterms and

Maxterms as follows

X Y Z f1

0 0 0 0

0 0 1 1

0 1 0 0

0 1 1 0

1 0 0 1

1 0 1 0

1 1 0 0

1 1 1 1

1.7.1 Sum of minterms:

Boolean expression for the function f1 from the given truth table is,

f1 = X’Y’Z + XY’Z’ + XYZ = m1 + m4 + m7

ie) OR ing the minterms which gives ‘1’ in the function f1

1.7.2 Product of maxterms:

Similarly,

f1 = (X+Y+Z) (X+Y’+Z) (X+Y’+Z’) (X’+Y+Z’) (X’+Y’+Z) = M0 .

M2 .M3.M5. M6

ie) AND ing the maxterms which gives ‘0’ in the function f1

1.8 Karnaugh Map

It is a systematic method to simplify Boolean expression using

a map. Map is a diagram made up of cells where each cell gives the

output value for the corresponding input combination. Totally there

are 2n cells for ‘n’ input variables. The map presents a visual

diagram of all possible ways a function may be expressed in a

standard form.

1.8.1 Two- and Three- variable maps

A Two-variable map is shown below.

There are 4 cells for 2 variables. Each cell for a minterm. Figure (a)

shows the minterms and (b) the relationship between the cells and

variables.

Example:

F1 = XY is represented by this mapping method as,

Since XY is equal to m3, a ‘1’ is placed inside the cell

that belongs to m3.

Similarly, F2 = X + Y as,

A Three-variable map is shown below.

Here 8 cells are not arranged in a binary sequence, but in a

sequence similar to Gray code. The characteristic of this sequence is

that only 1 bit changes from 0 to 1 or 1 to 0 in the listing sequence.

ie) the minterms in the adjacent cells are differing by a single bit.

This is the basic principle behind K-Map and can be clarified by

taking a function having m5 and m7 as,

m5 + m7 = XY’Z + XYZ = XZ(Y’+Y) = XZ.

Two minterms of 3 literals is solved into a single minterm of 2

literals.

The two cells are considered to be adjacent even though they do not

touch each other like m0 and m2.

Example:

#1) Simplify the Boolean function F(X, Y, Z) = ∑ (2, 3, 4, 5)

Mark 1 in each minterm that represents the function.

Find the adjacent cells (shown in two rectangles)

Upper rectangle area is enclosed by the variables X’ and

Y

Similarly lower rectangle by variables X and Y’

So the resultant function is F(X, Y, Z) = ∑ (2, 3, 4, 5) = X’Y + XY’

1.8.2 Four- variable map

A Four-variable map is shown below.

The minterm corresponding to each cell can be obtained by

the concatenation of the respective row and column. For instance,

third row (11) and second column (01) gives 1101, the binary

equivalent of decimal 13 representing m13.

The map minimization is the similar procedure as that of the

three variable type.

No other combination of squares can simplify the function.

1.8.3 Five- variable map

Maps for more than 4 variables are not as simple to use. It

needs 32 cells. A Five-variable map is shown below

It consists of 2 four-variable maps, where A distinguishes between

two maps. Minterms 0 throgh 15 belong with A = 0 and Minterms 16

throgh 31 belong with A = 1. Each map retains adjacency when

taken separately. Each cell in the A = 0 map is adjacent to the

corresponding cell in the A = 1 map like 4 and 20, 15 and 31.

Simply, for 2k adjacent squares, for k = 0,1,2..n, in an n-

variable map will represent an area that gives a term of n-k literals.

When n=k, the entire areas of the is combined to give the identity

function.

1.8.4 Product of Sum simplification

With minor modification, POS form is obtained as the result of

simplification from the map. Simplified expression for the

complement of the function,F’, is obtained if the cells with 0’s are

combined into valid squares. The complement of F’ inturn gives us F

in POS form.(By Demorgan’s Theorem)

Ex:

#1) Simplify the given Boolean expression in SOP and POS form.

F(A,B,C,D) = ∑(0,1,2,5,8,9,10)

Combining cells with 1’s gives the SOP form,

F = B’D’ + B’C’ + A’C’D

Combining cells with 0’s gives the form,

F = AB + CD + BD’

Applyind Demorgan’s theorem, we get the simplified POS form

F = (A’ + B’) (C’+ D’) (B’ + D)

Gate implementation of the function

F = B’D’ + B’C’ + A’C’D F = (A’ + B’) (C’+ D’) (B’ + D)

1.8.4 NAND and NOR Implementation

NAND and NOR gates are easier to fabricate and are the basic

gates used in all IC digital logic families. So Boolean functions

interms of NAND and NOR is necessory.

The invert OR symbol is followed from the Demorgan’s thoerem.

Similarly for NOR gate,

A one-input NAND or NOR behaves like an inverter

1.8.4.1 NAND implementation

Boolean expression in sum of product form is needed for NAND

implementation.

Consider for example, F = AB+CD+E is implemented in three ways

as shown below.

Figure (a) and (c) looks similar but NAND implementation needs one

more NAND gate for complementing E. By Demorgan’s theorem,

F = [ (AB)’ . (CD)’ . E’ ]’ = AB + CD + E.

Thus the two level implementation is possible with the NAND gates

for the given Boolean expression.

1.8.4.2 NOR implementation

NOR function is the dual of the NAND function. So all the

procedures and rules for NOR logic are the dual of the NAND logic.

Boolean expression in product of sum form is needed for NOR

implementation.

For the same example given above, F = (A+B) . (C+D) . E

To obtain product of sum from a map, it is necessary to

combine cells with 0’s and then complement the function.

1.8.4.3 Don’t-Care conditions

The logical sun of minterms associated with the Boolean

function specifies the condition under which the function is equal to

1. The function is equal to 0 for the rest of the minterms. Ie) all the

combinations of variables are valid. In some practical applications

the function is not specified for certain combinations of variables.

For example four bit codes for BCD greater than 9 has six invalid

combinations. These unspecified minterms are known as don’t care

conditions and used for further simplification. To distinguish don’t

care condition from 0 and 1, it is marked with X. Don’t care cell is

assumed to be 1 or 0 when choosing adjacent squares.

Ex. #1) Simplify F(w,x,y,z) = ∑ (1,3,7,11,15)

That has the don’t care conditions

d(w,x,y,z) = ∑ (0,2,5)

All the five 1’s should be included to form the sum of products. But

X may or may not be included.

By including 0 and 2 don’t care items, F = yz+w’x’

By including 5 don’t care item, F = yz+w’z

Expression in POS form is by grouping 0’s and include

minterms 0 and 2 to get it in the simplified version.

F’ = z’ + wy’

Complementing F’ yields

F(w,x,y,z) = z(w’+y) = yz+w’z

For this case minterms 0 and 2 are included with 0’s and minterm 5

with the 1’s.

1.9 TABULATION METHOD

If the number of variables exceeds five or six, excessive

number of cells make the grouping difficult whereby the tabulation

method is a step-by-step procedure.This method was first tabulated

by Quine and later improved by Mccluskey. Hence it is known as

Quine- Mccluskey method.

It consists of two parts. First is to find the Prime implicants.

Second to choose among the prime implicants those that give an

expression with least number of literals.

Example 1) Find minimal SOP for f = ∑ (1,2,3,4,5,7,8,9,10,11,14,15)

Solution :

i) write the given minterms in binary form

minterm

sA B C D

1 0 0 0 1

2 0 0 1 0

3 0 0 1 1

4 0 1 0 0

5 0 1 0 1

7 0 1 1 1

8 1 0 0 0

9 1 0 0 1

10 1 0 1 0

11 1 0 1 1

14 1 1 1 0

15 1 1 1 1

ii) Arrange the numbers in increasing number of 1’s

No. Of 1’s minterms A B C D

1

1 ^ 0 0 0 1

4 ^ 0 1 0 0

8 ^ 1 0 0 0

2

3 ^ 0 0 1 1

5 ^ 0 1 0 1

9 ^ 1 0 0 1

10 ^ 1 0 1 0

3

7 ^ 0 1 1 1

11 ^ 1 0 1 1

14 ^ 1 1 1 0

4 15 ^ 1 1 1 1

iii) Compare adjacent groups to find numbers which differ by 1

variable.

^denotes the numbers undergone the comparision

Combinati

onA B C D

(1,3) ^ 0 0 _ 1

(1,9)^ _ 0 0 1

(2,3)^ 0 0 1 _

(2,10)^ _ 0 1 0

(8,9)^ 1 0 0 _

(8,10) ^ 1 0 _ 0

(3,7)^ 0 _ 1 1

(3,11)^ _ 0 1 1

(9,11)^ 1 0 _ 1

(10,11)^ 1 0 1 _

(10,14) ^ 1 _ 1 0

(7,15) ^ _ 1 1 1

(11,15)^ 1 _ 1 1

(14,15)^ 1 1 1 _

_ denotes the differing variable position

iv) Form 4 cell combination,by having adjacent group

comparison, with 1 bit varying and _ in same position

Combinati

onA B C D

(1,3,9,11) _ 0 _ 1

(2,3,10,11) _ 0 1 _

(8,9,10,11) 1 0 _ _

(3,7,11,15) _ _ 1 1

(10,11,14,15

)1 _ 1 _

From the above table no more combination is possible.

v) Prime Implicants table

Prime

Implicants1 2 3

4 5 6 7 8 9 1

0

1

1

1

2

1

3

1

4

1

5

(1,3,9,11)* X X X X

(2,3,10,11)* X X X X

(8,9,10,11)* X X X X

(3,7,11,15)* X X X X

(10,11,14,1

5)*

X X X X

Essential

Prime

^ ^ ^ ^ ^

Implicants

Essential Prime Implicants is the one which has only one X in the

column of minterms

* mark is placed on the Prime Implicants which has Essential

Prime Implicants inside.

The resultant function consisting of Essential Prime Implicants and

the Prime Implicants covering the remaining minterms. Here all the

Prime Implicants are Essential Prime Implicants.

Therefore Y = B’D + B’C + AB’ + CD + AC

1.10 Summary

First unit provides us the basics of digital number representation

and manipulation.Also the coding methodologies used in various

digital devices. It enables the one to simplify the boolean expression

to arrive at a simpler logical circuit.

Self Test

1. A NOT gate has... a. Two inputs and one output b. One input and one

outputc. One input and two outputs d. none of above

2. A NAND gate is equivalent to an AND gate followed by.... gate. a. NOR b. NOTc. XOR d. none

3. Numbers are stored and transmitted inside a computer ina. binary form b. ASCII code form c. decimal form d. alphanumeric form

4. A gigabyte represents a.1 billion bytes b. 1000 kilobytes

c. bytes d. 1024 bytes

5. .A parity bit is a. used to indicate uppercase letters b. used to detect

errors c. is the first bit in a byte d. is the last bit in a

byte

6. 110012 -100012=

a.10000 b.01000c.00100 d.00001

7. Which is correct expression? a. A.A’=0 b. A+1=Ac. A+A’=A d. A.A'=1

8. 23.6 10=……….2a.11111.10011 b.10111.10011c.00111.101 d.10111.1

9. Which of these are universal gates?a. only NOR b. only NANDc. both NOR and NAND d. NOT,AND,OR

10. Boolean algebra obeysa. commutative law b. associative lawc. distributive law d. All the above

11. Which of the following is called as Demorgan’s theorem?a. A.A’=0 b. A’’=Ac. (A+B)’=A’.B’ d. A+A = A

12. In which of the following function, each term is known as minterm

a. SOP b. POSc. Hybrid d. both SOP and POS

13. For the logic circuit shown, the required input condition (A,B,C) to make the output x= 1 is

a. 1,0,1 b. 0,0,1c. 1,1,1 d. 0,1,1

14. 2’s complement representation of 17 is a. 01110 b. 01111c. 11110 d. 10001

15. Decimal 43 in hexadecimal and BCD number system is respectively

a. B2, 0100 0011 b. 2B, 0100 0011c. 2B, 0011 0100 d. B2, 0100 0100

16. x = 01110 and y = 11001 are two 5-bit binary numbers represented in two’s complement format. The sum of x and y represented in two’s complement format using 6 bits is:

a. 100111 b.001000c. 000111 d.101001

17. The output of Logic gate in the figure is

a. 0 b. 1c. A d. F

Review Questions

Part-A

1. Find the hexadecimal equivalent of the decimal number 256.

2. Find the octal equivalent of the decimal number 64.

3. What is meant by weighted and non-weighted coding?

4. Convert A3BH and 2F3H into binary and octal respectively.

5. Find the decimal equivalent of (123)9.

6. Find the octal equivalent of the hexadecimal number AB.CD.

7. Encode the ten decimal digits in the 2 out of 5 code.

8. Show that the Excess – 3 code is self –complementing.

Answers for self test

1. b 2. b 3. a 4. c 5.b 6. b 7.a 8. b 9. c 10. d11. c12.a

13. d14. b15. b16. c17. c

9. Find the hexadecimal equivalent of the octal number 153.4.

10. Find the decimal equivalent of (346)7.

11. A hexadecimal counter capable of counting up to at

least (10,000)10 is to be constructed. What is the minimum

number of hexadecimal digits that the counter must have?

12. Convert the decimal number 214 to hexadecimal.

13. Convert 231.3 4 to base 7.

14. Give an example of a switching function that contains

only cyclic prime implicant.

15. Give an example of a switching function that for which

the MSP from is not unique.

16. Express x+yz as the sum of minterms.

17. What is prime implicant?

18. Find the value of X = A B C (A+D) if A=0; B=1; C=1 and

D=1.

19. What are ‘minterms’ and ‘maxterms’?

20. State and prove Demorgan’s theorem.

21. Find the complement of x+yz.

22. Define the following : Minterm and Maxterm

23. State and prove Consensus theorem.

24. What theorem is used when two terms in adjacent

squares of K map are combined?

25. How will you use a 4 input NAND gate as a 2 input NAND

gate?

26. How will you use a 4 input NOR gate as a 2 input NOR

gate?

27. Show that the NAND connection is not associative.

28. What happens when all the gates is a two level AND-OR

gate network are replaced by NOR gates?

29. What is meant by multilevel gates networks?

30. Show that the NAND gate is a universal building block.

31. Show that a positive logic NAND gate is the same as a

negative logic NOT gate.

32. Distinguish between positive logic and negative logic.

33. What is the exact number of bytes in a system that

contains (a) 32K byte, (b) 64M bytes, and (c) 6.4G byte?

34. List the truth table of the function:

1. F = x y + x y’ + y ’z

Part-B

1. (a) Explain how you will construct an (n+1) bit Gray code from an

n bit

Gray code.

(b) Show that the Excess – 3 code is self –complementing.

2. (a) Prove that (x1+x2)(x1’x3’+x3) (x2’ + x1x3) =x1’x2.

(b) Simplify using K-map to obtain a minimum POS expression:

(A’ + B’+C+D) (A+B’+C+D) (A+B+C+D’) (A+B+C’+D’)

(A’+B+C’+D’)

(A+B+C’+D)

3. Reduce the following equation using Quine McClucky method of

minimization F (A,B,C,D) = ∑m(0,1,3,4,5,7,10,13,14,15)

4. (a) State and Prove idempotent laws of Boolean algebra.

(b) using a K-Map ,Find the MSP from of F=∑(0,4,8,12,3,7,11,15)

+d(5)

5. (a) With the help of a suitable example, explain the meaning of

an redundant prime implicant.

(b) Using a K-Map, Find the MSP form of F= ∑ (0-3, 12-15) +∑d

(7, 11)

6. (a) Simplify the following using the Quine – McClusky

minimization technique D = f(a,b,c,d) = ∑

(0,1,2,3,6,7,8,9,14,15).Does Quine –McClusky take care of don’t

care conditions? In the above problem, will you consider any don’t

care conditions? Justify your answer

(b) List also the prime implicants and essential prime

implicants for the above case.

7. (a) Determine the MSP and MPS focus of F= ∑ (0, 2, 6, 8, 10, 12,

14, 15)

(b) State and Prove Demorgan’s theorem.

8. Determine the MSP form of the Switching function

F = ∑ ( 0,1,4,5,6,11,14,15,16,17,20-

22,30,32,33,36,37,48,49,52,53,56,63)

9. (a) Determine the MSP form of the Switching function

F( a,b,c,d) =∑(0,2,4,6,8) + ∑d(10,11,12,13,14,15)

(b) Find the Minterm expansion of f(a,b,c,d) = a’(b’+d) + acd’

10. Simplify the following Boolean function by using the Tabulation

Method

F= ∑ (0, 1, 2, 8, 10, 11, 14, 15)

11. State and prove the postulates of Boolean algebra

12. (a) Find a Min SOP and Min POS for f = b’c’d + bcd + acd’ +

a’b’c + a’bc’d

13. Find an expression for the following function usingQuine

McCluscky method F= ∑ (0, 2, 3,5,7,9,11,13,14,16,18,24,26,28,30).

14. State and prove the theorems of Boolean algebra with

illustration.

15. (a) Find the MSP representation for

F(A,B,C,D,E) = ∑m(1,4,6,10,20,22,24,26) + ∑d

(0,11,16,27)

using K-Map method.

(b) Why does a good logic designer minimize the use of NOT gates?

16. Simplify the Boolean function F(A,B,C,D) = ∑ m (1,3,7,11,15) +

∑d (0,2,5) .if don’t care conditions are not taken care, What is the

simplified Boolean function .What are your comments on it?

Implement both circuits.

17. (a) F3 = f(a,b,c,d) = ∑ (2,4,5,6) F2 = f(a,b,c,d) = ∑ (2,3,,6,7)

F1 = f(a,b,c,d) = ∑ (2,5,6,7) .Implement the above Boolean

functions

(i) When each is treated separately and

(ii)When sharing common term

(b) Convert a NOR with an equivalent AND gate

18. What are the advantages of using tabulation method?

Determine the prime implicants of the following function using

tabulation method.

F( W,X,Y,Z) = ∑ (1,4,6,7,8,9,10,11,15)

23 (a) Explain about common postulates used to formulates various

algebraic structures.

(b) Given the following Boolean function F= A”C + A’B + AB’C +

BC

Express it in sum of minterms & Find the minimal SOP expression.