Dr. Stamen Grigorov special : today google showing a doodle to celebrate 142th birthday of Dr Stamen Grigorov . he was the Bulgarian physician , it means from Bulgaria , he born in Studen Izvor city of Bulgaria on 27 October 1878 , in the age of 67 he died in sofia city Bulgaria .

he was a famous microbiologist also with physician skills. he was the first person in the world who identified the needy bacterium for fermentation of yogurt , that was big opportunity or success of him. he did his best contribution in development of the first tuberculosis vaccine in the world , after his contribution and other efforts of other scientists , we won the victory on tuberculosis on world level. that is why google today remembering him on his 142th birth anniversary , on this occasion of his birthday we should thankful to him for his contribution in the field of medical or science.

** Digital Electronics**

**Number System and boolean algebra**

**Number system and boolean algebra = **Any system of naming or reoresenting numbers, as the decimal system or the binary system, is called system, boolean algebra is the algebra of truth values 0 and 1 or equivalently of subsets of a given set.

A system is a combination of active and passive devices designed to perform a specific function. generally systems are classified into major categories

- Analog systems
- Digital systems

**Analog Systems**

Analog systems are those systems which process an analog signal. in an analog system, the quantities can vary over a continuous range of values.

**Example ** the amplitude of the output signal to the amplifier can have any value between zero and continuous maximum limit.

**Digital systems **

A digital system manipulates discrete quantities of information that are represented in binary form.

**Example **Digital computer, calculator, digital video and audio equipments etc.

**Advantages of Digital system Over Analog System**

- information storage is easy in digital system.
- Digital systems are generally easier to design.
- Accuracy and precision of digital systems are greater as compared to analog systems.
- These circuits are less affected by noise.
- More digital circuitry can be on IC chips.
- Digital systems represent and manipulate discrete elements of information.

**Number Systems **

A number is a collection of sysbols. in any number system, there is an ordered set of symbols known as digits. there are two types of number systems

- positional number system
- non-positional number system

in positional number system, the position of each digit of a number has some positional weight. in a non-positional number system, a digit of a number does not indicate any significance in position and weight.

the number is represented as

(S_{n – 1 }s_{n – 2 }……..s_{1}s_{o} s_{-1} s_{-2}……s – _{m})_{b}

where, b = radix or base of the number system

n = number of digits in the integer part

m = number of digits in fractional part

S_{N-1 } = Most significant bit (MSB)

S_{ – M} = Least significant bit (LSB)

On the basis of number of symbols and radix, number systems are classified as

- Decimal number system
- Binary number system
- Octal number system
- Hexadecimal number system

**Decimal Number System**

the decimal number system has 10 symbols, so the radix or base of this system is 10. the ten symbols are 0,1,2,3,4,5,6,7,8, and 9 and the radix point is known as the decimal point. the digits at the left most side of the decimal point constitute the integer part and the right side digit of the decimal point constitute the fractional part. the value of a digit is function of its position.

**Binary Number System**

In binary number System, the total number of symbols are two (0,1), so base is 2 and the radix point is known as the binary point. the symbols 0, 1 known as bit. the binary and decimal number systems are both positional notation number systems. the format used in binary number system is

………..2^{2}2^{1}2^{0} 2^{-1}2^{-2}……..

**Binary to Decimal Conversion**

**Example 1. ** convert the binary number (110101)_{2} to decimal.

**Sol. **Multiply each binary digit by the appropriate weight factor.

thus, (110101)_{2} = 1 x 2^{0} + 0 x 2^{1} + 1 x 2^{2} + 0 x 2^{3} + 1 x 2^{4} + 1 x 2^{5}

= 1 + 0 + 4 + 0 + 16 + 32 = 53

(110101)_{2} = (53)_{10}

**Do you know ?**

here, subscript 2 is used to indicate that 110101 is a base 2 number. a capital letter B can also be used i.e., 110101 B.

**Example 2. ** Convert the fractional binary number 0.1011 into decimal.

**Sol.** first determine the weight of each bit and then sum the weights times the bit.

[0.1 0 1]_{2} = 1 x 0.5 + 0 x 0.25 + 1 x 0.125 + 1 x 0.0625

(0.1 0 1 1)_{2} = (0.6875)_{10}

**Another method of evaluating a binary fraction**

Considering the above example

- If we neglect the binary point, the value of 1011
_{2}is 11_{10}in decimal. - With four bits, there are 2
^{4}= 16 possible combinations. - Dividing we obtain 11/16 = 0.6875
_{10}.

**Example 3 . **Determine the decimal value 11101.011_{2}.

**Sol. ** [1 1 1 0 1. 0 1 1]_{2}

4 3 2 1 0 – 1 -2 -3

= 1 x 2^{4} + 1 x 2^{3} + 1 x 2^{2} + 0 x 2^{1} + 1 x 2^{0} + 0 x 2^{-1} + 1 x 2^{-1} + 1 x 2^{-3}

(11101.011)_{2} = (29.375)_{10}

Using the alternate method to evaluate the fraction. we have 3/8 = 0.375_{10}.

**Decimal to Binary Conversion**

**Sum of weights method ** one way to find binary equivalent is as follows

**For example **2^{3} 2^{2} 2^{1} 2^{0}

9 = 8 + 1 = 2^{3} + 2^{0} = 1 0 0 1

**Example 4. ** Convert the following decimal numbers to binary :

(a) 12

(b) 25

(c) 58

**Sol. ** (a) 12_{10} = 8 + 4 = 2^{3} + 2^{2}

2^{3}/1 2^{2}/1 2^{1}/0 2^{0}/0

(12)_{10} = (1100)_{2}

(b) (25)_{10} = 16 + 8 + 1 = 2^{4} + 2^{3} + 2^{0}

2^{4}/1 2^{3}/1 2^{2}/0 2^{1}/1 2^{0}/1

(25)_{10} = 11001)_{2}

(c) (58)_{10} = 32 + 16 + 8 + 2 =2^{5 }+ 2^{4} + 2^{3} + 2^{1}

2^{5}/1 2^{4}/1 2^{3}/1 2^{2}/1 2^{1}/1 2^{0}/1

(58)_{10} = (111010)_{2}

**Repeated division method**

A more systematic method of converting from decimal to binary is the repeated division by 2 process.

**For example**

Convert the decimal number 12 to binary.

**Convert decimal fractions to binary**

- the sum of weights method can be applied to fractional decimal number.

e.g., 0.625_{10} = 0.5 + 0.125 = 2^{-1} + 2^{-3} = 0.101_{2}

- decimal fractions can be converted to binary by repeated multiplication by 2 method.

e.g., for converting 0.3125_{10} into binary we will follow the following steps

**Step 1.** 0.3125 x 2 = 0.625

**Step 2. **0.625 x 2 = 1.25

0.25 x 2 = 0.50

0.25 x 2 = 1.00

0.0101_{2}

thus, (0.3125)_{10} = (0.0101)_{2}

**Octal Numbers**

The number system with base 8 is known as the octal numbersyste. it has eight possible digits 0, 1, 2, 3, 4, 5, 6 and 7, in octal number system, radix point is known as octal point.

it is also a positional number system: the weight of a digit is defined by its position with base 8.

**Octal to Decimal conversion**

The evaluation of an octal number in terms of its decimal equation is accomplished by multiplying each digit by its weight and summing the products.

**For example**

For converting (2154)_{8} into decimal procedure is

2154_{8} = 4 x 8^{0} + 5 x 8^{1} + 1 x 8^{3} = 1132_{10}

**Decimal to Octal conversion**

** (a) Integer part**

A method of converting a decimal number into an octal number is the repeated division by 8 method used.

**For example** Convert the given decimal number into octal number (266)_{10} = (?)_{8}

thus, (266)_{10} = (412)_{8}

**(b) Fractional part **

the fractional part of a decimal number can be converted to its octal equivalent by using the repeated multiplication method.

**For example**

(0.256)_{10} = (?)_{8}

**Step 1. **0.256 x 8 = 2.048

0.048 x 8 = 0.384

0.384 x 8 = 3.072

0.072 x 8 = 0.576

0.576 x 8 = 4.608

(0.256)_{10} = (0.20304)_{8}

**Octal to Binary conversion**

An octal number is also known as a 3 bit binary number because each digit of an octal number is represented by three binary digits. each octal digit is represented by three bits as indicated

**Octal digit Binary**

0 000

1 001

2 010

3 011

4 100

5 101

6 110

7 111

To convert an octal number to a binary number simply replace each octal difit by appropriate three bits.

**For example**

(a) (452)_{8} = (?)_{2}

4 = 100

5 = 101

2 = 010

(452)_{8} = (100 101 010)_{2}

(b) ( 37.12)_{8} = (?)_{2}

3 7 . 1 2

011 111 . 001 010

(37.12)_{10} = (011 111 .001 010)_{2}

**Binary to Octal conversion**

**(a) Integer part**

the integer binary numbers can be converted into octal equivalent by making groups of three bits starting from LSB and moving towards MSB.

**For example** 100 011 110 = 100 011 110

4 3 6

= (436)_{8}

**(b) Fractional part **

the fractional binary number can be converted into their equivalent octal numbers by making group of three bits starting from MSB and moving towards LSB.

**For example** 0.010 101 100 = 0.010 101 100

2 5 4

(0.010 101 100)_{2} = (0.254)_{8}

**Hexadecimal Numbers**

The hexadecimal number system has a base of sixteen i.e., it is composed of 16 digits and characters.

Ten digits and six alphabetic characters make up this number system. A subscript 16 indicates a hexadecimal number.

**Binary to hexadecimal conversion**

Break the binary number into four bits group starting at the binary point and replace each group by the equivalent hexadecimal symbol.

**For example**

(a) 1110 1010 0100

E A 4

(1110 1010 0100)_{2} = (EA4)_{16}

(b) (111100110101011011)_{2} = (?)_{16}

0 0 11 1100 1101 0101 1011 = (3CD5B)_{16}

3 C D 5 B

Adding zeros

(c) (11011.111)_{2} = (?)_{16}

0011 0011 1110 = (33.E)_{16}

3 3 E

Adding two zeros

**Hexadecimal to Binary conversion**

Replace each hexadecimal symbol with the appropriate four bit.

**For example**

(a) 1 0 A E

0001 0000 1010 1110

(10AE)_{16} = (0001000010101110)_{2}

(b) 0 2 3 5

0010 0011 0101

(0.235)_{16 }= (0.001000110101)_{2}

(c) F 2 D 8 A 1

1111 0010 1101 1000 1010 0001

(F2D . 8A1)_{16} = (111100101101.100010100001)_{2}

**Hexadecimal to decimal conversion**

By multiplying each hexadecimal digit by its weight and then taking the sum of these products.

**For example**

(a) (321)_{16} = (?)_{10}

(321)_{16} = 1 x 16^{0} + 2 x 16^{1} + 3 x 16^{2}

= 1 + 32 + 768 = 801

= (801)_{10}

(b) (0.11)_{16} = (?)_{10}

(0.11)_{16} = 0.00390 + 0.0625

= (0.06640625)_{10}

(c) (A00.01)_{16} = (?)_{10}

(A00.01)_{16} = (2560.00390)_{10}

**Decimal to-hexadecimal conversion**

Repeated division of a decimal number by 16 will produce the equivalent hexadecimal number formed by the remainders of each division.

**For example**

(a) (498)_{10} = (?)_{16}

(498)_{10} = (1F2)_{16}

(b) (0.210)_{10} = (?)_{16}

0.210 x 16 = 3.36 decimal hex

3 3 MSB

0.36 x 16 = 5.76

5 5

0.76 x 16 = 12.16

12 C

0.16 x 16 = 2.56

2 2

0.56 x 16 = 8.96

8 8

0.96 x 16 = 15.36

15 F LSB

(0.210)_{10} = (35C28F)_{16}