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EENG 2710 Chapter 1

Number Systems and Codes

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Chapter 1 Homework

1.1c, 1.2c, 1.3c, 1.4e, 1.5e, 1.6c, 1.7e, 1.8a, 1.9a, 1.10b, 1.13a, 1.19

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Number Systems

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Binary Number System

• Uses two digits, 0 and 1.

• Represents any number using the positional notation.

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Positional Notation

• The value of a digit depends on its placement within a number.

• In base 10, the positional values are (starting to the left of the decimal) –1 (100), 10 (101), 100 (102), 1000 (103), etc.

• In base 2, the positional values are1 (20), 2 (21), 4 (22), 8 (23), etc.

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Binary Weights

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 1

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Fractional Binary Weights

2-1 2-2 2-3 2-4

½ ¼ 1/8 1/16

0.5 0.25 0.125 0.0625

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Bit

• Shorthand for binary digit, a logic 0 or 1.

• The most significant bit (MSB) is the leftmost bit of a binary number.

• The least significant bit (LSB) is the rightmost bit of a binary number.

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Binary Inputs

• Digital circuits operate by accepting logic levels (0,1) at their input(s).

• The corresponding output(s) logic level will change (0,1).

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Binary Inputs

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4-Input Digital Circuit

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Base Conversions Methods

• Series substitution method

• Sum powers of 2

• Radix method– Repeated Division

– Repeated Multiplication

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Series Substitution Method(Binary to Decimal)

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1048

1)(12)04)(18)(1

)2(1)2(0)2(1)2(1 1101 0123

(

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Sum Powers of 2(Decimal to Binary)

• Step 1:– Determine the largest power of 2 less than or

equal to the number to be converted.– Place a 1 in that positional location.

• Convert 5710 to binary

• 6410 5710 3210

32 16 8 4 2 11

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Sum Powers of 2

• Step 2:– Subtract the number found in Step 1 from the

number to be converted. • 57 – 32 = 25

– For the new number, determine if the next lowest power of 2 is less than or equal to that number.

• 25 – 16 = 9

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Sum Powers of 2

• Step 3:– If the new power of two from Step 2 is larger,

place a 0 in that positional location. – If the new value is less than or equal, place a 1

in that positional location.

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Sum Powers of 2

• Step 4:– Repeat Steps 2 and 3 until there is nothing left

to subtract. – All remaining bits are set to 0.

5710 = 1110012

32 16 8 4 2 11 1 1 0 0 1

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Radix Method (Repeated Division by 2)

1 2 5 11 23 46 22 4 10 22 46

1 0 1 1 1 0

4610 = 1011102

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Fractional Binary Numbers

• Radix point:– The generalized decimal point. The dividing

line between positive and negative powers for positional multipliers.

• Binary point:– The radix point for binary numbers.

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Fractional Binary Values

• The value immediately to the right of the binary point is 2–1 = 0.5.

• The next value to the right is 2–2 = 0.25.

• The next value to the right is 2–4 = 0.125, and so on.

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Series Substitution Method(Binary Fraction to Decimal Fraction)

7031250

45/64

1/641/161/801/2

)2(1)2(0)2(1

)2(1)2(0)2(1 10110106-5-4-

-3-2-1

.

.

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Radix Method for 0.210 to Binary (Repeated Multiplication by 2)

• Step 1:Multiply the decimal fraction by 2.• Step 2:

Integer part is 0 or 1 left of decimal point. 0.2 x 2 = 0.4 Integer part = 0

0.4 x 2 = 0.8 Integer part = 0 0.8 x 2 = 1.6 Integer part = 1 0.6 x 2 = 1.2 Integer part = 1 0.2 x 2 = 0.4 Integer part = 0 (stop

0011repeats)Read integer parts from top to bottom

Therefore, 0.210 = 0.0011 0011 0011

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Hexadecimal Numbers

• Base 16 number system.

• Primarily used as a shorthand form of binary numbers.

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Counting in Hexadecimal

• Values range from 0 to F with the letters A to F used to represent the values 10 to 15 respectively.

• Positional multipliers are powers of 16:160 = 1, 161 = 16, 162 = 256, etc.

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Hexadecimal vs. Decimal Numbers

Decimal

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Hexadecimal

0 1 2 3 4 5 6 7 8 9 A B C D E F

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Counting In Hexadecimal

0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F

10,11,12,13,14,15,16,17,18,19,1A,1B,1C,1D,1E,1F

20,21,22,23,24,25,26,27,28,29,2A,2B,2C,2D,2E,2F

30,31,32,33,34,35,36,37,38,39,3A,3B,3C,3D,3E,3F

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Decimal-to-Hexadecimal Conversion(Repeated division by 16)

7 123 1973 31581 16112 1968 31568

7 11 5 137 B 5 D

3158110 = 7B5DH

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Hex-Decimal-Binary Table

Hex Decimal Binary Hex (Cont)

Decimal Binary

0 0 0000 8 8 1000

1 1 0001 9 9 1001

2 2 0010 A 10 1010

3 3 0011 B 11 1011

4 4 0100 C 12 1100

5 5 0101 D 13 1101

6 6 0110 E 14 1110

7 7 0111 F 15 1111

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Conversion Between Hexadecimal and Binary

• Each hexadecimal digit represents 4 binary bits.

F D 6 9

1111 1101 0110 1001

FD69H = 11111101011010012

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Signed/Unsigned Binary Numbers

• Signed Binary Number:– A binary number of fixed length whose sign

(+/–) is represented by one bit (usually MSB) and its magnitude by the remaining bits.

• Unsigned Binary Number:– A binary number of fixed length whose sign

is not specified by a bit. All bits are magnitude and the sign is assumed +.

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Unsigned Binary Arithmetic

• Sum:– Result of an Addition Operation of two (or more)

binary numbers (operands).

• Carry:– A digit (or bit) that is carried over to the next most

significant bit during an n-Bit addition operation.

• The carry bit is a 1 if the result was too large to be expressed in n bits.

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Basic Rules (Unsigned)

bit outCarry

101000001 11100

00100111 1010

10101110 10010

1111 next tocarry 1

bit outCarry

101000001 11100

00100111 1010

10101110 10010

1111 next tocarry 1

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Basic Subtraction

• Basic Subtraction of x = a – b, with a = minuend, b = subtrahend, and x = difference or result.

• Requires a Borrow Bit if a < b.

• There are other forms of subtraction such as 2’s Complement Addition used by microprocessors (such as in a PC).

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Binary Subtraction with Borrow Examples

1 0101

1 10 101 -

LSB to ripplesBorrow 0111(10) 10000

1 010

1 100 1001 -

StageBorrow 110(10) 1110

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Signed Binary Numbers

• Sign Bit:– A bit (usually the MSB) that indicates whether

a number is positive (= 0) or negative (= 1).

• Magnitude Bits:– The bits of a signed binary number that tell how

large it is in value.

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Signed Binary Numbers

• True-Magnitude Form:– A form of signed binary whose magnitude bits are the

TRUE binary form (not complements).

• 1’s Complement:– A form of signed binary in which negative numbers are

created by complementing all bits.

• 2’s Complement:– A form of signed binary in which the negative numbers

are created by complementing all the bits and adding a 1 (1’s Complement + 1).

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True-Magnitude Form

• 5-Bit Numbers Negative Sign (S = 1)• +25 = 011001 (Note sign bit (MSB) Sign = 0)• –25 = 111001 (Same as +25 with sign = 1)• +12 = 001100• –12 = 101100

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1’s Complement Form

• 8-Bit 1’s Complement Negative (S = 1)

• +57 = 00111001

• –57 = 11000110 (All Bits Inverted)

• +72 = 01001000

• –72 = 10110111

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2’s Complement Form

57 = 0011 1001

-57 = 1100 0110

+ 1

1100 0111

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Signed Binary Addition (8-Bit)

Signed Addition Positive (S = 0) +30 = 00011110 +75 = 01001011 105 01101001

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Subtraction with 1’s Complement• Add the 1’s Complement and then Carry.

• Uses an End around carry addition method.

1111 0000

1

1110 0000 1

65) Comp s(1' 1110 1011 65)( 0001 0100 65 -

0000 0101 80)( 0000 0101 80

(80 – 65)

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2’s Complement Subtraction

• Add 2’s Complement to Minuend.

Discord Carry Bit From Results

(80 – 65)

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2’s Complement Subtraction20010 – 510 = 19510

(Use 16 bit word)

20010 = 00000000110010002

510 = 00000000000001012

-510 = 11111111111110102 = 1’s complement

+ 1

11111111111110112

00000000110010002

+ 11111111111110112

100000000110000112 = 00000000110000112 = 19510

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Negative Results

• If the True-Magnitude Form is used for subtraction, the results are incorrect.

• If the result is from 1’s or 2’s Complement and the result is negative (S = 1), the magnitude is found by taking the complement of the result.

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Negative Result Example

Thus, = -1510

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More Binary Addition

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More Binary Subtraction

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Binary Multiplication

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Binary Division

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Range of Signed Numbers

• Range of Positive Numbers is 0 to 2n – 1 for a number with n magnitude bits.

• Range of Negative Numbers is –1 to –2n for a number with n magnitude bits.

• 8-Bit Example:8-Bit Number Range is –2n x +2n – 1

or –128 to +127

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Sign Bit Overflow

• Overflow:– An erroneous carry into the sign bit of a signed

binary number

– Results from a sum or difference that is larger than can be represented by the magnitude bits.

• Results in a False Positive or False Negative Number.

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False Negative Overflow

• Addition of two 8-Bit Positive Numbers:

• Two positive numbers added with a result greater than the range of +127 for 8-bit numbers causes an overflow.

(False) Negative is Result 1011 1010

0000 0110 96

1011 0100 75

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False Positive Overflow

• Addition of two 8-Bit Negative Numbers:

• Two Negative numbers were added to produce a False Positive Result due to overflowing the negative range of 8-bit numbers (0 to –128).

(False) Positive is Result 1111 0110

1111 1011 65

0000 1011 80

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Hexadecimal Addition

• Similar to decimal addition with a range of digits of 0 to 9 and A to F.

• Examples:

F + 1 = 10

F + F = 1E

F + F + 1 = 1F

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414FH Hex

4)(15) ( 1) 4)( (

9)(12) 1)(10)( (

(11)(3) 6) 2)( (

1 1 Carry

5)(16)(20)(1 (3)

9)(12) 1)(10)( ( 1A9CH

3) (11)( 6) 2)( ( 26B3H

Equivalent Decimal Hex

Hexadecimal Addition

• For sums greater than 15, subtract 16 and carry 1 to the next position.

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Hexadecimal Subtraction

C17H Hex

7) ( (1) 0)(12)(

(12) (9) (10) (1) -

3) (10)(16 6) (1)(16

1 1 Borrow

position. previous the from

)(16 10Hborrow digit, tsignifican least the subtract To

(12)(1)(10)(9) 1A9CH

(3)(2)(6)(11) 26B3H

Equivalent Decimal Hex

10

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Hexadecimal Subtraction

2F00H – 4000H

Convert 4000H to hexadecimal equivalent of 2’s complement:

FFFF

4000

BFFF

+ 1

C000

Add numbers: 2F00

+ C000

EF00H

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Hexadecimal Subtraction

2F00H – 4000H

a. Converting 4000 to binary = 0100 0000 0000 0000

b. Take 1’s complement = 1011 1111 1111 1111

c. Take 2’s complement = +1 +1 +1 +1

1100 0000 0000 0000

d. Change to Hexadecimal = C000H

e. Add numbers: 2F00

+ C000

EF00H

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More Hexadecimal Addition

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More Hexadecimal Subtraction

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Hexadecimal Multiplication

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Hexadecimal Division

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Octal Addition

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Octal Subtraction

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Octal Multiplication

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Octal Division

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Excess 8 code

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Floating-point Number Formats

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Developing a Floating Point number

If n = 1101.0101, convert n to a floating point number. n = (0.11010101) x 24

The mantissa M = 0.110101012sm The exponent E = 011 + 10000 = 100112

E = (1, 0011)exess 16

Combining M & E:

N = 0, 1, 0011, 11010101)fp

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BCD Codes

• BCD Code (Binary-Coded Decimal): A code used to represent each decimal digit of a number by a 4-Bit Binary Value.

• Valid Digits for 0 to 9 are 0000 to 1001.– The binary codes 1010 to 1111 are invalid

• Called an 8421 Code due to the decimal weight of each bit position.

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BCD Examples

Each digit is a 4-Bit Binary group:

(84)10 = 1000 0100

(4987)10 = 0100 1001 1000 0111BCD

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Gray Code• A binary code that progresses so that only one

bit changes between two successive codes. 000 001

011 010 110 111 101 100

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How to build a 4-bit Gray Code Table(A binary code that progresses so that only one bit changes between two

successive codes.)

3-bit Gray code0 0 00 0 10 1 10 1 01 1 01 1 11 0 11 0 0

4-bit Gray code0 0 0 00 0 0 10 0 1 10 0 1 00 1 1 00 1 1 10 1 0 10 1 0 01 1 0 01 1 0 11 1 1 11 1 1 01 0 1 01 0 1 11 0 0 11 0 0 0

2-bit Gray code0 00 11 11 0

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4-bit Gray Code Table 4-bit Gray Code Sequence

Q3 Q2 Q1 Q00 0 0 00 0 0 10 0 1 10 0 1 00 1 1 00 1 1 10 1 0 10 1 0 01 1 0 01 1 0 11 1 1 11 1 1 01 0 1 01 0 1 11 0 0 11 0 0 0

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A Gray Code Disk

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ASCII Code

• American Standard Code for Information Interchange.

• A seven-bit alphanumeric code used to represent text letters, numerals, punctuation, and special controls.

• An expanded 8-bit form is becoming more widespread.

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ASCII Code

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Error Detection and Correction Codes

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Error Detection and Correction Codes

• Definitions

• Parity Codes– Odd parity– Even parity

• Hamming Codes– Code 1– Code 2

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Error Detection and Correction Codes

Definitions• Error – an incorrect value of one or more binary bits.• Single Error - an incorrect value of one binary bits.• Multiple Error - incorrect values of many binary bits.• Distance between code words – d

– Code word I = 01101100– Code word J = 11000100 – d (I,J) =3

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Definitions Continued

• dmin – minimum distance between two code word – If code provides t error correction plus detection of s

additional errors, then 2t + s + 1 dmin

– At least dmin errors are needed to transform one code word to another

– Less than dmin errors, then a detectable non-code word results. Thus, if the non-code word is closer to a valid code word, then the original code word can be deduced and the error can be corrected.

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Parity Basics

• Parity: A digital system that checks for errors in a n-Bit Binary Number or Code.

• Even Parity: A parity system that requires the binary number and the parity bit to have an even # of 1s.

• Odd Parity: A parity system that requires the binary number and the parity bit to have an Odd # of 1s.

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Parity Basics

• Parity Bit: A bit appended on the end of a binary number or code to make the # of 1s odd or even depending on the type of parity in the system.

• Parity is used in transmitting and receiving data by devices in a PC called UARTs, that are on the COM Port.

• UART = Universal asynchronous

Receiver/Transmitter

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Parity Basics

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Parity Calculation

N1 = 0110110:

– It has four 1s (an even number).– If Parity is ODD, the Parity Bit = 1 to make it

an odd number (5). N = 1110110– If Parity is EVEN, the Parity Bit = 0 to keep it

an even number (4). N = 0110110

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Hamming Code Table

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Hamming Code ProblemError Word = 1100110, Use Hamming Code 1 to determine the code word.

Code word