2 Number Systems Floyd

8/12/2019 2 Number Systems Floyd

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2009 Pearson Education

Octal to Binary

Hexadecimal

Decimal Octal

Binary


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Octal to Binary

Technique Convert each octal digit to a 3-bit equivalent

binary representation


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Example

705 8 = ? 2

7 0 5

111 000 101

705 8 = 111000101 2


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Hexadecimal to Binary

Hexadecimal

Decimal Octal

Binary


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Hexadecimal to Binary

Technique Convert each hexadecimal digit to a 4-bit

equivalent binary representation


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Example

10AF 16 = ? 2

1 0 A F

0001 0000 1010 1111

10AF 16 = 0001000010101111 2


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Decimal to Octal

Hexadecimal

Decimal Octal

Binary


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Decimal to Octal

Technique Divide by 8 Keep track of the remainder


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Example

1234 10 = ? 8

8 1234

154 28

19 28

2 38

0 2

1234 10 = 2322 8


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Decimal to Hexadecimal

Hexadecimal

Decimal Octal

Binary


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Decimal to Hexadecimal

Technique Divide by 16 Keep track of the remainder


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Example

1234 10 = ? 16

1234 10 = 4D2 16

16 1234

77 216

4 13 = D160 4


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Binary to Octal

Hexadecimal

Decimal Octal

Binary


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Binary to Octal

Technique Group bits in threes, starting on right Convert to octal digits


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Example

10110101112

= ?8

1 011 010 111

1 3 2 7

1011010111 2 = 1327 8


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Binary to Hexadecimal

Hexadecimal

Decimal Octal

Binary


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Binary to Hexadecimal

Technique Group bits in fours, starting on right Convert to hexadecimal digits


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Example

10101110112

= ?16

10 1011 1011

2 B B

1010111011 2 = 2BB 16


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Octal to Hexadecimal

Hexadecimal

Decimal Octal

Binary


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Octal to Hexadecimal

Technique Use binary as an intermediary


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Example

10768

= ?16

1 0 7 6

001 000 111 110

2 3 E

1076 8 = 23E 16


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Hexadecimal to Octal

Hexadecimal

Decimal Octal

Binary


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Hexadecimal to Octal

Technique Use binary as an intermediary


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Example

1F0C16

= ?8

1 F 0 C

0001 1111 0000 1100

1 7 4 1 4

1F0C 16 = 17414 8


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Exercise Convert ...

Dont use a calculator!

Decimal Binary OctalHexa-

decimal

33

1110101703

1AF

Skip answer Answer

C


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Exercise Convert

Decimal Binary OctalHexa-

decimal

33 100001 41 21

117 1110101 165 75451 11100001

1703 1C3

431 110101111

657 1AF

Answer


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2009 Pearson Education, Upper Saddle River, NJ 07458. All Rights ReservedFloyd, Digital Fundamentals, 10 th ed

1s Complement

The 1s complement of a binary number is just the inverseof the digits. To form the 1s complement, change all 0s to1s and all 1s to 0s.For example, the 1s complement of 11001010 is

00110101

In digital circuits, the 1s complement is formed by usinginverters: 1 1 0 0 1 0 1 0

0 0 1 1 0 1 0 1


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2s Complement

The 2s complement of a binary number is found byadding 1 to the LSB of the 1s complement.

Recall that the 1s complement of 11001010 is00110101 (1s complement)

To form the 2s complement, add 1: +100110110 (2s complement)

Adder

Input bits

Output bits (sum)

Carryin (add 1)

1 1 0 0 1 0 1 0

0 0 1 1 0 1 0 1

1

0 0 1 1 0 1 1 0


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Signed Binary Numbers

There are several ways to represent signed binary numbers.In all cases, the MSB in a signed number is the sign bit, thattells you if the number is positive or negative.

Computers use a modified 2s complement forsigned numbers. Positive numbers are stored in true form(with a 0 for the sign bit) and negative numbers are storedin complement form (with a 1 for the sign bit).

For example, the positive number 58 is written using 8-bits as00111010 (true form).

Sign bit Magnitude bits


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Signed Binary Numbers

Assuming that the sign bit = - 128, show that 11000110 = - 58

as a 2s complement signed number:

1 1 0 0 0 1 1 0Column weights: - 128 64 32 16 8 4 2 1 .

- 128 +64 +4 +2 = - 58

Negative numbers are written as the 2s complement of thecorresponding positive number.

- 58 = 11000110 (complement form)

Sign bit Magnitude bitsAn easy way to read a signed number that uses this notation is toassign the sign bit a column weight of - 128 (for an 8-bit number).Then add the column weights for the 1s.

The negative number - 58 is written as:


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Floating Point Numbers

Express the speed of light, c, in single precision floating pointnotation. ( c = 0.2998 x 10 9)

Floating point notation is capable of representing verylarge or small numbers by using a form of scientificnotation. A 32-bit single precision number is illustrated.

S E (8 bits) F (23 bits)

Sign bit Magnitude with MSB droppedBiased exponent (+127)

In scientific notation, c = 1 .001 1101 1110 1001 0101 1100 0000 x 228.

0 10011011 001 1101 1110 1001 0101 1100

In binary, c = 0001 0001 1101 1110 1001 0101 1100 0000 2.

S = 0 because the number is positive. E = 28 + 127 = 155 10 = 1001 1011 2.F is the next 23 bits after the first 1 is dropped. In floating point notation, c =


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Arithmetic Operations with Signed Numbers

Using the signed number notation with negativenumbers in 2s complement form simplifies additionand subtraction of signed numbers.

Rules for addition : Add the two signed numbers. Discardany final carries. The result is in signed form.Examples:

00011110 = +30 00001111 = +15

00101101 = +45

00001110 = +14 11101111 = - 17

11111101 =-

3

11111111 = - 1 11111000 = - 8

11110111 =-

91Discard carry


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01000000 = +128 01000001 = +129

10000001 =-

126

10000001 = - 127 10000001 = - 127

100000010 = +2

Note that if the number of bits required for the answer isexceeded, overflow will occur. This occurs only if bothnumbers have the same sign. The overflow will beindicated by an incorrect sign bit.

Two examples are:

Wrong! The answer is incorrectand the sign bit has changed.

Discard carry


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Rules for subtraction : 2s complement the subtrahend andadd the numbers. Discard any final carries. The result is insigned form.

00001111 = +151

Discard carry

2s complement subtrahend and add: 00011110 = +30

11110001 = - 15

Repeat the examples done previously, but subtract:0001111000001111-

0000111011101111

1111111111111000- -

00011111 = +31

00001110 = +14

00010001 = +1700000111 = +71

Discard carry

11111111 = - 1

00001000 = + 8

(+30) (+15)

(+14) (- 17)

(- 1) (- 8)


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Hexadecimal Numbers

Hexadecimal uses sixteen characters torepresent numbers: the numbers 0through 9 and the alphabetic charactersA through F.

0

12345

6789101112131415

0

12345

6789ABCDEF

0000

00010010001101000101

0110011110001001101010111100110111101111

Decimal Hexadecimal Binary

Large binary number can easily be converted to hexadecimal bygrouping bits 4 at a time and writingthe equivalent hexadecimal character.

Express 1001 0110 0000 1110 2 inhexadecimal:

Group the binary number by 4-bitsstarting from the right. Thus, 960E


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Hexadecimal Numbers

Hexadecimal is a weighted numbersystem. The column weights are

powers of 16, which increase fromright to left.

.

1 A 2 F 16

6703 10

Column weights 163 16 2 16 1 16 0

4096 256 16 1 .{

Express 1A2F 16 in decimal.

Start by writing the column weights:4096 256 16 1

1(4096) + 10(256) +2(16) +15(1) =

0

12345

6789101112131415

0

12345

6789ABCDEF

0000

00010010001101000101

0110011110001001101010111100110111101111

Decimal Hexadecimal Binary


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Octal Numbers

Octal uses eight characters the numbers0 through 7 to represent numbers.There is no 8 or 9 character in octal.

0

12345

6789101112131415

0

12345

671011121314151617

0000

00010010001101000101

0110011110001001101010111100110111101111

Decimal Octal Binary

Binary number can easily beconverted to octal by grouping bits 3 ata time and writing the equivalent octalcharacter for each group.

Express 1 001 011 000 001 110 2 in

octal:Group the binary number by 3-bitsstarting from the right. Thus, 113016 8


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Octal Numbers

Octal is also a weighted numbersystem. The column weights are

powers of 8, which increase from rightto left.

.

3 7 0 2 8

1986 10

Column weights 83 82 81 80

512 64 8 1 .{

Express 3702 8 in decimal.

Start by writing the column weights:512 64 8 1

3(512) + 7(64) +0(8) +2(1) =

0

12345

6789101112131415

0

12345

671011121314151617

0000

00010010001101000101

0110011110001001101010111100110111101111

Decimal Octal Binary


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BCD

Binary coded decimal (BCD) is aweighted code that is commonlyused in digital systems when it isnecessary to show decimalnumbers such as in clock displays.

0

12345

6789101112131415

0000

00010010001101000101

0110011110001001101010111100110111101111

Decimal Binary BCD

000100010001000100010001

0000

00010010001101000101

0110011110001001000000010010001101000101

The table illustrates thedifference between straight binary andBCD. BCD represents each decimal

digit with a 4-bit code. Notice that thecodes 1010 through 1111 are not used inBCD.


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BCD

You can think of BCD in terms of column weights ingroups of four bits. For an 8-bit BCD number, the columnweights are: 80 40 20 10 8 4 2 1.

What are the column weights for the BCD number1000 0011 0101 1001 ?

8000 4000 2000 1000 800 400 200 100 80 40 20 10 8 4 2 1

Note that you could add the column weights where there isa 1 to obtain the decimal number. For this case:

8000 + 200 +100 + 40 + 10 + 8 +1 = 835910


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BCDA lab experiment in which BCDis converted to decimal is shown.


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Gray code

Gray code is an unweighted codethat has a single bit change betweenone code word and the next in asequence. Gray code is used toavoid problems in systems where anerror can occur if more than one bitchanges at a time.

0

12345

6789101112131415

0000

00010010001101000101

0110011110001001101010111100110111101111

Decimal Binary Gray code

0000

00010011001001100111

0101010011001101111111101010101110011000


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Gray codeA shaft encoder is a typical application. Three IRemitter/detectors are used to encode the position of the shaft.The encoder on the left uses binary and can have three bitschange together, creating a potential error. The encoder on theright uses gray code and only 1-bit changes, eliminating

potential errors.

Binary sequence Gray code sequence


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ASCII

ASCII is a code for alphanumeric characters and controlcharacters. In its original form, ASCII encoded 128characters and symbols using 7-bits. The first 32 charactersare control characters, that are based on obsolete teletyperequirements, so these characters are generally assigned toother functions in modern usage.

In 1981, IBM introduced extended ASCII, which is an 8-

bit code and increased the character set to 256. Otherextended sets (such as Unicode) have been introduced tohandle characters in languages other than English.


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2009 Pearson Education, Upper Saddle River, NJ 07458. All Rights ReservedFloyd, Digital Fundamentals, 10th

ed

Parity Method

The parity method is a method of error detection forsimple transmission errors involving one bit (or an oddnumber of bits). A parity bit is an extra bit attached toa group of bits to force the number of 1s to be either

even (even parity) or odd (odd parity).The ASCII character for a is 1100001 and for A is1000001. What is the correct bit to append to make both ofthese have odd parity?

The ASCII a has an odd number of bits that are equal to 1;therefore the parity bit is 0. The ASCII A has an evennumber of bits that are equal to 1; therefore the parity bit is 1.


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2009 Pearson Education, Upper Saddle River, NJ 07458. All Rights ReservedFloyd, Digital Fundamentals, 10th

ed

Cyclic Redundancy Check

The cyclic redundancy check (CRC) is an error detection methodthat can detect multiple errors in larger blocks of data. At thesending end, a checksum is appended to a block of data. At thereceiving end, the check sum is generated and compared to the sentchecksum. If the check sums are the same, no error is detected.

Selected Key Terms


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Selected Key Terms

Byte

Floating-pointnumber

Hexadecimal

Octal

BCD

A group of eight bits

A number representation based on scientificnotation in which the number consists of anexponent and a mantissa.

A number system with a base of 16.

A number system with a base of 8.

Binary coded decimal; a digital code in which eachof the decimal digits, 0 through 9, is represented bya group of four bits.

Selected Key Terms


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Selected Key Terms

Alphanumeric

ASCII

Parity

Cyclicredundancy

check (CRC)

Consisting of numerals, letters, and othercharacters

American Standard Code for InformationInterchange; the most widely used alphanumericcode.

In relation to binary codes, the condition ofevenness or oddness in the number of 1s in a codegroup.

A type of error detection code.


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1. For the binary number 1000, the weight of the columnwith the 1 is

a. 4

b. 6

c. 8

d. 10



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2. The 2s complement of 1000 is

a. 0111

b. 1000c. 1001

d. 1010



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3. The fractional binary number 0.11 has a decimal value of

a.

b. c.

d. none of the above



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4. The hexadecimal number 2C has a decimal equivalentvalue of

a. 14

b. 44

c. 64




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5. Assume that a floating point number is represented in binary. If the sign bit is 1, the

a. number is negative

b. number is positive

c. exponent is negative

d. exponent is positive



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6. When two positive signed numbers are added, the resultmay be larger that the size of the original numbers, creatingoverflow. This condition is indicated by

a. a change in the sign bit

b. a carry out of the sign position

c. a zero result

d. smoke



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7. The number 1010 in BCD is

a. equal to decimal eight

b. equal to decimal tenc. equal to decimal twelve

d. invalid



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8. An example of an unweighted code is

a. binary

b. decimalc. BCD

d. Gray code



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9. An example of an alphanumeric code is

a. hexadecimal

b. ASCIIc. BCD

d. CRC



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10. An example of an error detection method fortransmitted data is the

a. parity check

b. CRC

c. both of the above




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Answers:

1. c

2. b3. c

4. b

5. a

6. a

7. d8. d

9. b

10. c

2 Number Systems Floyd

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