Goal_1 Numbering System and Digital Codes

8/10/2019 Goal_1 Numbering System and Digital Codes

1/25

ELEC 224 J. Altiti1

Goal 1

Digital & Analog Quantities

Number Systems


2/25

ELEC 224 J. Altiti2

Digital & Analog Quantities

Analog quantity have continuous values. The figure below shows a graph of air temperature vs time

of day.

The air temperature changes over a continues range of values. During the day the temperature

does not go from say 70oto 75oinstantaneously. It takes on all the infinite values in between.

another examples of analog quantity is the sine wave shown below. Other examples are distance,

sound, time and pressure

Example of an analog electronic system is shown below


3/25

ELEC 224 J. Altiti3

Digital quantity have discrete values. Varies in discrete (separate) stepsDigital technology is widely used. Examples:

o Computerso Manufacturing systemso Medical Scienceo Transportationo Entertainmento Telecommunications

The figure below shows a graph of a digital signal .

The digital sequence of the above signal is 1010101010 in reality 5v 0v 5v 0v 5v 0v 5v 0v 5v 0v

another example of a digital signal is shown below. Note that each level represent

The digital sequence of the above signal is 11010100 in reality 5v 5v 0v 5v 0v 5v 0v 0v

Example of a system that uses digital and analog electronics is shown below


4/25

ELEC 224 J. Altiti4

Advantages and disadvantages of both systems

Number SystemsUnderstanding digital systems requires an understanding of the decimal,binary, octal, and

hexadecimalnumbering systems. The most familiar is the decimal number system (conventional

number system) that uses ten digits: 0,1,2,3,4,5,6,7,8, and 9.

Decimal Number System:Base: 10

Decimal Digits: 0 1 2 3 4 5 6 7 8 9

Weights: ----- 10000 1000 100 10 1 . 0.1 0.01 0.001 0.0001 ------

----- 104 103 102 101 100 . 10-1 10-2 10-3 10-4.

Example: The number 2745.21410.

The digit 2is the Most Significant Digit (MSD)

The digit 4is the Least Significant Digit (LSD)

The decimal number 2745.214 consist of sevendecimal digits

The formula below is used to calculate the largest decimal number that can be represented with n

digits

NumberDecimalLargest1Basen

n: Number of digits

Base: Number system base (10)

With 5 decimal digits (n = 5) the largest decimal number is

999991105

Analog Digital Susceptible to noise Very Immune to noise

Less efficient in transmission More efficient in transmission

Less bandwidth Greater bandwidth


5/25

ELEC 224 J. Altiti5

Example: 598.74210= 0598.742010= 00598.7420010= 000598.74200010

100 10 1 . 0.1 0.01 0.001 0.0001

102

101

100

. 10-1

10-2

10-3

10-4

5 9 8 . 7 4 2 0

(5x102) + (9x101) + (8x100) + (7x10-1) + (4x10-2) + (2x10-3) + (0x 10-4)

(5x100) + (9x10) + (8x1) + (7x0.1) + (4x 0.01) + (2x0.001) + (0 x 0.0001)

500 + 90 + 8 + 0.7 + 0.04 + 0.002 + 0

= 598.74210

Binary Number System:

Base: 2

Binary Digits (Bit): 0 1

Weights: ----- 16 8 4 2 1 . 0.5 0.25 0.125 0.0625 ------

----- 24 23 22 21 20 . 2-1 2-2 2-3 2-4.

Example: 1010102

The 1is the Most Significant Bit (MSB)

The 0 is the Least Significant Bit (LSB)

The binary number 101010 consist of 6 Bits (Binary Digits)

What is the decimal number that is equal to the binary number 101010 ?

32 16 8 4 2 1

25 24 23 22 21 20

1 0 1 0 1 0

(1x25) + (0x24) + (1x23) + (0x22) + (1x21) + (0x20)

(1x32) + (0x16) + (1x8) + (0x4) + (1x2) + (0x1)

32 + 0 + 8 + 0 + 2 + 0

= 4210. The binary number 1010102= 4210in decimal

The formula below is used to calculate the largest decimal number that can

be represented in binary with n bits

NumberDecimalLargest1Basen

n: number of bits

Base: number system base (2)

With 2 bits (binary digits) the largest decimal number that can be represented in

binary is

3122

10

That is with 2 bits, a range of decimal numbers from 0 to 3 can be represented. That

is a total of 4 binary combinations where each binary combination represent a

decimal number.


6/25

ELEC 224 J. Altiti6

The formula below is used to calculate the range of decimal numbers (binary

combination) that can be represented in binary with n bits.n

Base n: number of bits

Base: number system base (2)

Example: How many decimal numbers can we represent in binary with 2 bits

422

2 1

b1 b0 Decimal

0 0 0

0 1 1

1 0 2

1 1 3

Example: How many bits are required to represent the decimal number 3710

With 5 bits (binary digits) the largest decimal number is

10

5 3112

and 32 (25) decimal numbers (from 0 to 31) can be represented

With 6 bits (binary digits) the largest decimal number is

10

6 6312

and 64 (26

) decimal numbers (from 0 to 63) can be represented

From the above it is obvious that 6 bits are needed to represent the decimal

number 37

Weights ----- 32 16 8 4 2 1

----- b5 b4 b3 b2 b1 b0

1 0 0 1 0 1

Example: The table below is for 4 bits (n =4).

b3 b2 b1 b0


7/25

ELEC 224 J. Altiti7

Example: 1011.1012= 01011.1012= 001011.101002= 0001011.1010002

8 4 2 1 . 0.5 0.25 0.125

23 22 21 20 . 2-1 2-2 2-3

1 0 1 1 . 1 0 1

(1x23) + (0x22) + (1x21) + (1x20) + (1x2-1) + (0x 2-2) + (1x 2-3)

(1x8) + (0x4) + (1x2) + (1x1) + (1x0.5) + (0x0.25) + (1x0.125)

8 + 0 + 2 + 1 + 0.5 + 0 + 0.125

= 11.62510. The binary number 1011.1012= 11.62510in decimal

and the decimal number 11.62510= 1011.1012in binary

Octal Number System:Base: 8

Octal Digits: 0 1 2 3 4 5 6 7

Weights: ----- 4096 512 64 8 1 . 0.125 0.015625 ------

----- 84 83 82 81 80 . 8-1 8-2

Octal numbers were used in old computer systems. Now it is rarely used.

Example: 50738

The 5is the Most Significant Digit (MSD)

The 3 is the Least Significant Digit (LSD)

The octal number 5073 consist of 4 octal digits

What is the decimal number that is equal to the octal number 5073 ?

512 64 8 1

83 8

2 8

1 8

0

5 0 7 3

(5x83) + (0x82) + (7x81) + (3x80)

(5x512) + (0x64) + (7x8) + (3x1)

2560 + 0 + 56 + 3

= 261910. The octal number 50738= 261910in decimal


8/25

ELEC 224 J. Altiti8

Hexadecimal Number System:Base: 16

Hexadecimal Digits: 0 1 2 3 4 5 6 7 8 9 A B C D E F(10) (11) (12) (13) (14) (15)

Weights: ----- 4096 256 16 1 . 0.0625 0.00390625 ------

----- 163 162 161 160 . 16-1 16-2 -------.

Example: 3A91C16 or 3A91Ch

The 3is the Most Significant Digit (MSB)

The C is the Least Significant Digit (LSB)

The Hex number 3A91C consist of 5 hexadecimal digits

What is the decimal number that is equal to the hex number 5C3A ?

4096 256 16 1163 162 161 160

5 C 3 A

(5x163) + (Cx162) + (3x161) + (Ax160)

(5x4096) + (12x256) + (3x16) + (10x1)

20480 + 03072 + 48 + 10

= 2361010. The Hexadecimal number 5C3A16= 2361010in decimal

Binary Coded Decimal (BCD) Number SystemBinary Coded Decimal (BCD) is another way to present decimal numbers in binary form. BCD iswidely used and combines features of both decimal and binary systems. Each decimal digit is

converted to a 4 bit binary equivalent.

Decimal Digit BCD Equivalent

0 00001 00012 00103 00114 0100

5 01016 01107 01118 10009 1001

Example: What is the decimal number that is equal to the 100001110100BCD

1000 0111 0100

8 7 4

So the BCD 100001110100 is equal to the decimal number 874

Example: What is the BCD equivalent to the decimal number 1095 ?1 0 9 5

0001 0000 1001 0101

So the decimal number 1095 has a BCD equivalent of 0001000010010101


9/25

ELEC 224 J. Altiti9

Gray Codehas the property that only one bit changes from one number to the next in the sequence. The gray

code is used in applications where numbers change rapidly.

The table shows the representations of decimal numbers in Binary, Octal, Hexadecimal, BCD

and Gray code.

Decimal Binary Gray

0 0000 0000

1 0001 0001

2 0010 0011

3 0011 0010

4 0100 0110

5 0101 0111

6 0110 0101

7 0111 0100

8 1000 1100

9 1001 1101

10 1010 1111

11 1011 1110

12 1100 1010

13 1101 1011

14 1110 1001

15 1111 1000

Decimal Binary Octal Hexadecimal BCD Gray

0 0000 0 0 0000 000

1 0001 1 1 0001 0001

2 0010 2 2 0010 0011

3 0011 3 3 0011 0010

4 0100 4 4 0100 0110

5 0101 5 5 0101 01116 0110 6 6 0110 0101

7 0111 7 7 0111 0100

8 1000 10 8 1000 1100

9 1001 11 9 1001 1101

10 1010 12 A 0001 0000 1111

11 1011 13 B 0001 0001 1110

12 1100 14 C 0001 0010 1010

13 1101 15 D 0001 0011 1011

14 1110 16 E 0001 0100 1001

15 1111 17 F 0001 0101 1000


10/25

ELEC 224 J. Altiti10

ASCII CodeASCIIAmerican Standard Code for Information Interchange. Each code represents a character

or function found on a computer keyboard. ASCII code is used to transfer information between

computers

and printers, and for internal storage.

The ASCII code is a seven bit code. With 7 bits there are 128 possible combinations.

27= 128 possible codes (from 0 to 127)

The Extended ASCII code (80h to FFh or 12810to 25510) represent non-English alphabetic

characters such as:

- Currency symbols

- Greek letters

- Math symbols

- Drawing characters

- Bar graphing characters

- Shading characters

The following web site has the complete ASCII tablehttp://www.lookuptables.com/
http://www.lookuptables.com/http://www.lookuptables.com/http://www.lookuptables.com/http://www.lookuptables.com/


11/25


Conversion between Number Systems

The Figure below demonstrate the various possible conversion between number systems

g 3 bits : Group every 3 bits g 4 bits : Group every 4 bits

f 3 bits : Form 3 bits for every octal digit f 4 bits : Form 4 bits for each hexadecimal digit

The following terms are used in digital systems Bit, Nibble, Byte, KiloByte, MegaByte and GigaByte

-Bit: a single binary digit.

Example: 1 or 0

- Nibble: a group of 4 binary digits (4 bits). 1510is the maximum decimal number that can be

represented with one nibble.

Example: 1101

- Byte: a group of 8 binary digits (8 bits). 25510is the maximum decimal number that can be

represented with one byte.Example: 10110101

- Word: a group of 16 binary digits (16 bits). 65535 is the maximum decimal number that can be

represented with one word.

Example: 1011101101110011 = ------------------10

- KiloBit: is 1024 binary digits (1024 bits).

- KiloByte: is 1024 Byte = 1024 x 8 bits = 8192 bits.

- MegaByte: 1024 kiloByte = 1024 x 8192 bits = 8,388,608 bits.

- GigaByte: 1024 MegaByte = 1024 x 8,388,608 bits = 8,589,934,592 bits .


12/25


Decimal to Binary Conversion

The integer portion of the decimal number will be divided by 2 repeatedly till the quotient is equal 0. Thefraction part will be multiplied by 2 till the fraction is equal 0.

Example: Convert the decimal number 25.8125 to binary.

Divison Quotient Reminder Bit

25/2 = 12 1 (LSB)

12/2 = 6 0 0

6/2 = 3 0 0

3/2 = 1 1

1/2 = 0 1 (MSB)

Multiply Integer Fraction Bit

0.8125 x 2 = 1 0.625 1 (MSB)

0.625 x 2 = 1 0.250 1

0.250 x 2 = 0 0.5 0

0.5 x 2 = 1 0 1 (LSB)

Finally 25.812510= 11001.11012


13/25


Decimal to Octal Conversion

The integer portion of the decimal number will be divided by 8 repeatedly till the quotient is equal 0. Thefraction part will be multiplied by 8 till the fraction is equal 0.

Example: Convert the decimal number 25.8125 to octal.

Divison Quotient Reminder Digit

25/8 = 3 1/ 8 1 (LSD)

3/8 = 0 3/ 8 3 (MSD)

Multiply Integer Fraction Bit

0.8125 x 8 = 6 0.5 6 (MSD)

0.5 x 8 = 4 0 4 (LSD)

Finally 25.812510= 31.648

Decimal to Hexadecimal Conversion

The integer portion of the decimal number will be divided by 16 repeatedly till the quotient is equal 0. The

fraction part will be multiplied by 16 till the fraction is equal 0.

Example: Convert the decimal number 25.8125 to octal.

Divison Quotient Reminder Digit

25/16 = 1 9/ 16 9 (LSD)1/16 = 0 1/ 16 1 (MSD)

Multiply Integer Fraction Digit

0.8125 x 16 = 13 0 D (MSD)

Finally 25.812510= 19.D16

Binary to Hexasecimal Conversion

1. Starting from the Right of the binary number group every 4-bits2. Replace each group with the hexadecimal digit equivalent

Example: Find the hexadecimal and decimal equivalent of the following binary number 100111110010 ?

Hexadecimal

1. 1001 1111 0010

2. 9 F 2

Decimal

2048 1024 512 256 128 64 32 16 8 4 2 1

211 210 29 28 27 26 25 24 23 22 21 20

1 0 0 1 1 1 1 1 0 0 1 0= 2048 + 256 + 128 + 64 + 32 + 16 + 2 = 2,54610

Finally 1001111100102= 9F216= 2,54610


14/25


Binary to Octal Conversion1. Starting from the Right of the binary number group every 3-bits

2. Replace each group with the octal digit equivalent

Example: Find the octal equivalent of the following binary number 100111110010 ?

Hexadecimal

1. 100 111 110 010

2. 4 7 6 2

Finally 1001111100102= 47628= 2,54610

Hexadecimal to Binary ConversionReplace each hexadecimal digit with its 4-bit binary equivalent (form 4-bits for each hexadecimal digit)

Example: Find the binary equivalent of the following hexadecimal number F7C9?

F 7 C 9

1111 0111 1100 1001

Finally F7C916= 11110111110010012= 63,43310

Octal to Binary Conversion

Replace each octal digit with its 3-bit binary equivalent (form 3-bits for each octal digit)

Example: Find the binary equivalent of the following octal number 3741?

3 7 4 1

011 111 100 001

Finally 37418= 0111111000012= 2,01710

Hexadecimal to Octal Conversion

1. Convert each hexadecimal digit to its 4-bit binary equivalent.2. follow the procedure for converting binary to octal.

Example: Find the octal equivalent of the following hexadecimal number D94A?

1. D 9 4 A

1101 1001 0100 1010

D94Ah = 11011001010010102

2. 001 101 100 101 001 010

1 3 4 5 1 2

Finally D94A16= 11011001010010102= 1345128


15/25


Octal to Hexadecimal Conversion1. Convert each octal digit to its 3-bit binary equivalent.

2. follow the procedure for converting binary to hexadecimal.

Example: Find the hex equivalent of the following octal number 647?

1. 6 4 7

110 100 111

6478= 1101001112

2. 0001 1010 0111

1 A 7

Finally 6478= 1101001112= 1A716

Binary to Gray Code Conversion1. The MSB in Gray code is the same as the corresponding MSB in the binary number.

2. Going from left to right, add each adjacent pair of binary code bits to get the next gray code bit.

Discard Carries

Binary addition

1

0+

0

0

0+

1

1

1+

0

1+

0

1

1

1

Example: Convert 101102to Gary code.

1 + 0 + 1 + 1 + 0 Binary

1 1 1 0 1 Gray

Finally 101102= Gray code 11101

Gray to Binary Code Conversion1. The MSB in the binary number is the same as the corresponding bit in the Gray code.

2. Add each binary bit generated to the Gray code bit in the next adjacent position. Discard carries.

Example: Convert the Gary code 11011 to binary .

1 0+

1 1

0 1

0 1

++0

1

+

Gray

Binary

Finally Gray code 11011 = 100102= 1810


16/25


Signed Numbers

Since it is only possible to show magnitude with a binary number, the sign (+ or -) is shown by adding an

extra sign bit.

A sign bit of 0 indicates a positive number.

A sign bit of 1 indicates a negative number.

There are 2 methodes to represent signed numbers

Sign-magnitude System

2s complement system

Sign-magnitude SystemThe last bit is the sign bit and does has no weight. It is used indicate the sign (+ or -). The remaining bits

corresponds to the number value.

Example:

Although the sign-magnitude system is straight forwad, computers and calculators do not normally use it,

because the circuit implementation is complex.

2s complement System

The 2s complement system is the most commonly used way to represent signed numbers.

To change the sign of a binary number perform the 2s complementas follows

Invert each bit. 1 to 0 and 0 to 1. This process is called the 1s complement

Add 1 to the 1s complement.

A short cut to performing 2s complement Starting from the right of the binary number, leave all the bits unchanged till the first 1 is

encountered

Invert all the bits after the first encountered 1

A number is negated when converted to the opposite sign. A binary number can be negated by taking the 2s

complement of it.

.


17/25


Example: Find the negative of the decimal number 45 in binary using 7-bits.

4510= 0 1 0 1 1 0 12

Perform 1s complement1 0 1 0 0 1 02

Add 1 to 1s complement

1 0 1 0 0 1 0

+

1

1 0 1 0 0 1 1 = -4510

Finally + 4510= 0 1 0 1 1 0 12 and - 4510 = 1 0 1 0 0 11

Range of Values with 2s complement System

The formula below is used to calculate the minimum (negative) and the maximum (positoive)

numbers can be calculated as follows

minimum (negative) number =(2n1)

maximum (positive) number = + (2

n1

1)n: number of bits

Example: what are the range of signed and unsigned decimal numbers that can be represented with

4-bits ?

Signed

Minimum (negative) decimal number =(241) =(23) =8

Maximum (positive) decimal number = + (2n11) = (231) = 7

Unsigned

Maximum unsignd decimal number = 241 = 15

Finally

All signed decimal number from8 to 7 can be represented with 4-bits.

All unsigned decimal numbers from 0 to 15 can be represented with 4-bits.


18/25


The figure below demonstrate the signed and unsigned decimal numbers represented with

4-bits


19/25


Arithmetic Operations with Signed numbers

The basic arithmetic operations are Addition and Subtraction In this sections only Binary, Hexadecimal and

BCD arithmetic operations will be discussed.

Binary Addition

The parts of any addition function are:

1. Augend2. Addend3. Sum

Four conditions for adding numbers:

1. Both numbers are positive.2. A positive number that is larger than a negative number.3. A negative number that is larger than a positive number.4. Both numbers are negative.

Signs for Addition

1. When both numbers are positive, the sum is positive.2. When the larger number is positive and the smaller is negative, the sum is positive. The carry is

discarded.


20/25


3. When the larger number is negative and the smaller is positive, the sum is negative (2scomplement form).

4. When both numbers are negative, the sum is negative (2s complement form). The carry bit isdiscarded.

Example: Add 910to 410in binary .

910= 0010112

410= 0001002

Finally 910 + 410= 1310= 011012

Example: Add 910to -410in binary.

910= 010012

-410= 111002

Finally 910 + (-410) = 510= 01012


21/25


Example: Add -910to 410in binary.

-910= 101112

410= 001002

Finally -910 + 410= -510= 110112

Example: Add -910 and -410in binary.

-910= 101112

-410= 111002

Finally -910 + (-410) = -1310= 100112

Example: Add -910 and 910in binary.

-910= 101112

910= 010012

Hexadecimal Addition

1. Add the hex digits in decimal.2. If the sum is 15 or less express it directly in hex digits.

3. If the sum is greater than 15, subtract 16 and carry 1 to the next position.4. When the MSD in a hex number is 8 or greater, the number is negative. When the MSD is 7 or

less, the number is positive.


22/25


Example: Add 2510to 3010in hexadecimal.

2510= 1916

3010= 1E16

Finally 2510+ 3010= 5510= 3716Example: Add A73486516to 1259D616.

Example: Add 34A87BC16to 12DF45816.

BCD Addition

1. Convert each BCD number to its 4-bit binary equivalent.2. Add the two BCD numbers using the binary addition.3. If the sum of two BCD numbers is less than or equal to 9, the sum is a valid BCD number.4. If the sum of two BCD numbers is greayer than 9, a binary 6 is added. This will always cause a

carry.

Example: Add the 8 to 5 using BCD.

8 1000

+5 +0101

13 1101 is 13 ( > 9)

Note that the result is MORE THAN 9, so add 6.


23/25


8 1 0 0 0

+5 + 0 1 0 1

13 1 1 0 1 is 13 (> 9)

1 1 1 0 1 13

+ 0 1 1 0 add 6

1 0 0 1 1 0001= 1 and 0011 = 3

0001| 0011 Final answer (two digits) = 13

Finally adding 8BCDto 5BCD= 13BCD

Example: Add 1897BCDto 2905BCD

Finally adding 1897BCDto 2905BCD= 4802BCD


24/25


Binary Subtraction

The parts of any subtraction function are:

1. Minuend2. Subtrahend3. Difference

Subtraction is addition with the sign of the subtrahend changed.

1. The number subtracted (subtrahend) is negatedby taking the 2s complement.2. The result is added to the minuend.3. The answer represents the difference.4. Discard any final carry bit.

Example: Compute 1310- 510in binary.

1310510 = 1310+ (-510)

1310 = 011012

510 = 001012 -510= 110112

Finally 1310+ (-510) = 810= 011012+ 110112= 010002

Example: Compute 5101210.

5101210 = 510+ (-1210)

510 = 001012

1210 = 011002 -1210= 101002

Finally 510+ (-1210) = -710= 001012+ 101002= 110012


25/25

Hexadecimal Subtraction

1. Subtract the hex digits in decimal.

2. If the sum is 15 or less express it directly in hex digits.3. If the difference is negative, borrow 1 from next hex digit and add 16 to the differenc.

Example: Compute 84162A16

Finally 84162A16= 5A16

Example: Compute 4787C141612DF45816

Finally 4787C141612DF45816= 34A87BC16

Goal_1 Numbering System and Digital Codes

Documents