Chapter%201:%20Binary%20Systems

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Chapter 1 Binary Systems

• Topics in this Chapter
• Digital Systems
• Numbering Systems
• Number Base Conversions
• Complements
• Arithmetic Operations
• Binary Coded Decimal (BCD)
• Binary Storage and Registers
• Binary Logic

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Digital Systems

• Discrete quantities weekly salaries, income
  taxes, letters or alphabet, digits
• Digital systems digital telephones, digital
  cameras, digital computers and etc.
• Digital systems have the ability to manipulate
  discrete quantities.
• Some digital systems can operate with extreme
  reliability by using error-correcting code. For
  example DVD

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Numbering Systems

• Decimal numbering systems
• A decimal digit can have the values of
  0,1,2,3,4,5,6,7,8 and 9. The number of
  combination of 2 decimal digits is 102 . In
  general for any numbering system the number of
  combination of n digits can be determined by
• Bn
• where B is the base of numbering system and n is
  the number of digits to be combined.
• For example for two decimal digits the largest
  number is 99 so the two decimal digits end at 99.
  Using the formula of Bn the total number of
  combinations for two decimal digits would be 102
  100

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Numbering Systems

• A decimal number such as 7,392.42 is equal to
• 7 103 3 102 9 101 2 100 4
  10-1 2 10-2
• In general for any numbering system
• an an-1 an-2 a1 a0. a-1 a-2. a-m
• the equal decimal number is
• an rn an-1 rn-1 .a1 r a0 a-1
  r-1 am rm
• where aj is the coefficient
• and
• the number expressed in a base-r system

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Octal Numbering System

• The Octal numbering system is a base-eight
  numbering system with eight digits of
• 0,1,2,3,4,5,6,7
• Decimal Octal
• ------------ ---------
• 0 0
• 1 1
• 7 7
• 8 10
• 9 11

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Octal Numbering System

• For example to count in octal the digits combine
  after reaching a count of 7
• 1,..7,10,11,12,,15,16,17,20,21,,75,76,77,100
• for two octal digits the largest number is 77
  so the two octal digits end after at 77. Using
  the formula of Bn the total number of
  combinations for two octal digits would be 82
  64
• To find the decimal number equal to an octal
  number
• (127.4)8 1 82 2 81 7 80 4 8-1
  (87.5)10

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The Hexadecimal Numbering System

• It is a base-sixteen numbering system. That is
  there are 16 digits in this system
• Hexadecimal Decimal
• ------------------ -----------
• 0 0
• 1 1
• 2 2
• ..
• 9 9
• A 10
• B 11
• C 12
• D 13
• E 14
• F 15

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The Hexadecimal Numbering System

• For example to count in hexadecimal
• F,10,11,12,19,1A,1B,1C,1D,1E,1F,20,
• 21,.99,9A,,9F,A0,A1..,FE,FF,100
• for two hexadecimal digits the largest number
  is FF so the two hexadecimal digits end after at
  FF. Using the formula of Bn the total number of
  combinations for two hexadecimal digits would be
  162 256.
• To find the decimal number equal to a hexadecimal
  number
• (B65F)16 11 163 6 162 5 161 5 160
  (46,687)10

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The Binary Numbering System
Only two digits required 1,0
21 20 Base 10 Equivalent
0 0
1 1
1 0 2
1 1 3
10
The Binary Numbering System

• For four binary digits the largest number is 1111
  so the four binary digits end at 1111. Using the
  formula of Bn the total number of combinations
  for four binary digits would be 24 16.
• To find the decimal number equal to a binary
  number
• (110101)2 32 16 4 1 (53)10
• (0.1101)2 (1.5 1.25 0.125 1.0625 )10
• (0.8125) 10

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Adding Binary Numbers

• Same rule as decimal Example

0 1
0 0 1
1 1 10
1 1 1
1 1 1
1 1 1 0
carry
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Multiplication of Binary numbers

• Multiplication Table Example

0 1
0 0 0
1 0 1
1 1 1
1 1 1
1 1 1
1 1 1
1 1 1
1 1 0 0 0 1
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Number Base Conversions

• For converting decimal number to binary number
  REPETETIVE DIVISION is used as follows
• The following algorithm generates the binary
  digits.
•  
• Q decimal number
• While Q is not equal to 0 do the following
• Binary digit Q mod 2
• Q Q / 2 (quotient)
• End While

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Conversion Example

• Convert (58)10 to (?)2
• 58 mod 2 0 29
• 29 mod 2 1 14
• 14 mod 2 0 7
• 7 mod 2 1 3
• 3 mod 2 1 1
• 1 mod 2 1 1 Ans (111010)2

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Conversion Example

• To convert a fraction, it multiples by 2 to give
  an integer and a fraction and only the fraction
  again multiplies by 2. This process continues
  until the faction becomes zero or until the
  number of digits have sufficient accuracy
• (0.6875) 10 ( ? ) 2
• 0.6875 2 1 0.3750
• 0.3750 2 0 0.7500
• 0.7500 2 1 0.5000
• 0.5000 2 1 0.0000
• ? result is (0.1011)2

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Hex to Binary

• Converting from Hex to Binary is easy
• Every hex digit becomes 4 binary digits
• (1AF5) 16
• (0001 1010 1111 0101) 2
• (1101011110101)2
• (306.D)16
• ( 0011 0000 0110. 1101 )2

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Binary to Hex

• Just as simple, reverse to process
• (111001010101011101) 2
• (0011 1001 0101 0101 1101) 2
• (3955D) 16

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Octal to Binary

• Converting from Octal to Binary is trivial
• Every octal digit becomes 3 binary digits
• (175 ) 8
• (001 111 101) 2
• (1111101) 2

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Binary to Octal

• Just as simple, reverse to process
• (11001010101011101) 2
• (011 001 010 101 011 101) 2
• (312535) 8
• Using the hex and octal equivalent instead of
  binary numbers are more convenient and less prone
  to errors.

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Complements

• One of the use of complements is the simplifying
  subtraction operation. We want to be able to
  perform subtraction through the addition
  operation.
• There are two types of complements for each
  base-r system
• rs complement ( radix complement )
• (r-1)s complement ( diminished radix complement)

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(r-1)s Complements

• (r-1)s complement for a number N with n digits
  in base-r numbering system is defined as
• (rn -1 ) N
• For example for decimal numbers since r is 10,
• (r-1)s complement of a number N is
• ( 10n 1) N
• For example 9s complement of 12389 is
• (105 -1 ) - 12389 99999 -12389 87610
• Or 9s complement of 2389 is
• (104 -1) 2389 9999 -2389 7610

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(r-1)s Complements

• 1s complement of a number N, is defined as (2n
  1 ) N
• For example for 1011001 it is equal to
• ( 27 1 ) 1011001 0 100110
• Therefore, 1s complement of binary numbers is
  formed by changing 1s to 0s and 0s to 1s
• For example 1s complement of 0001111 is 1110000

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(rs complement)

• (rs) complement for number N in base r with n
  digits
• is defined as rn N
• also it is clear rs complement ( r-1)s
  complement 1
• Therefore we can use (r-1)s complements of the
  previous examples to calculate their rs
  complement
• 10s complement of 2389 7610 1 7611
• 2s complement 101100 010011 1s
  complement
• 
  1
• 
  --------------
• 
  010100
• 2s complement of 101100 by using rn N
  definition generates same result
• 2n N 1000000
• - 0101100
• ----------------------
• 0010100

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Subtraction with complements (Unsigned Numbers)

• For doing the subtraction on two n digits
  unsigned numbers M and N the in base r the
  following algorithm can be used
• 1- Add M to rs complement of N
• 2- If MgtN subtract rn form the result
• 3- if MltN place a negative sign - in front of
  rs complement of the result

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Subtraction with complements (Unsigned Numbers)

• For example 3250 -72532
• M
  03250
• 10s complement of N 27468
• 
  ------------
• 
  30718
• Since N is gt M the result is
• - (10s complement 30718) -69282

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Subtraction with complements (Unsigned Numbers)

• For Binary numbers the algorithm is same. See the
  examples in the book
• Subtraction also can be done by (r-1)s
  complement. In that case if MgtN, after
  discarding the end carry one should be added to
  result
• For example for X 1010100 -Y 1000011
• 
  2s compl. 1s compl.
• X
  1010100 1010100
• Y
  0111101 0111100
• 
  ----------------- ---------------
• 
  10010001 10010000
• -
  10000000 - 10000000
• 
  1
• 
  -------------------- ----------------
• 
  0010001 0010001
• The problem is how we can simplify the
  subtraction for signed numbers? For example how
  about 2- (-1)?
• For doing that first we should represent a signed
  number and then devise an algorithm for the
  arithmetic operations of the signed numbers. See
  the next slides

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Signed Binary Numbers

• Positive integer numbers can be represented by
  unsigned numbers. However we need a notation for
  negative numbers
• A negative binary number assumed to be signed or
  unsigned. For example 01001 can be considered
• unsigned binary 9 or
• signed binary 9
• Or 11001 can be considered
• unsigned binary 25 or
• signed binary -9
• If the representation of the signed numbers uses
  0 for and 1 for - , this system is called
  Signed-magnitude convention

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Signed Binary Numbers

• Another system to represent negative numbers is
  signed complement system
• For example
• 9 is 00001001 in two systems
• but -9 is
• 10001001 in signed-magnitude
• 11110110 in signed-1s-complement
  (complementing all bits include sign bit or
  subtracting from 2n -1)
• 11110111 in signed-2s-complement
• ( 2complement of all bits or subtracting
  from 2n )

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Signed Binary Numbers

• Representation method only matters when we are
  talking about negative numbers
• All negative numbers have 1 in leftmost bit
• The signed magnitude is mostly used in ordinary
  arithmetic
• The 1s complement is mostly used in logical
  operations
• The 2s complement is mostly used in computer
  arithmetic

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• Decimal signed-2s signed-1s
  signed
• complement complement
  magnitude
• ---------- -----------
  ------------- -----------
• 7 0111 0111
  0111
• 2 0010 0010
  0010
• 1 0001 0001
  0001
• 0 0000 0000
  0000
• -0 --- 1111
  1000
• -1 1111 1110
  1001
• -2 1110 1101
  1010
• -7 1001 1000
  1111
• -8 1000 -------
  ------

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Arithmetic Addition (Signed Numbers)

• The addition of binary numbers in
  signed-magnitude system follows the same rules as
  ordinary arithmetic. (sign of larger number)
• For adding the numbers in 2s complement form the
  result can be obtained by adding the two numbers
  including their sign bits. The carry of the sign
  bits is discarded and negative results are
  automatically in 2s complement form. It means in
  order to get the negative result we must find the
  2s complement form of the result including the
  sign bit.
• When adding 2 numbers of same sign (pos or neg)
  in 2s complement form, if the result cannot be
  shown in the available bits then the overflow
  occurs that indicates the result is wrong.
  Generally when the sign of the result shown in
  available bits is different from the sign of the
  numbers, overflow has been occurred. See the next
  slide.

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Arithmetic Addition (2s Comp)

• Two positive numbers
• 0 1001 0 1001
• 0 0100 0 1000
• ----------------
  -----------
• 0 1101 1 0001
• Neg
  result
• 
  wrong, overflow

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Arithmetic Addition (2s Comp.)

• For example
• 9 0 1001
• -4 1 1100
• --------------------------------
• 1 0 0101 (5)
• Discarded final carry. Note that sign bit also
  participates in the process

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Arithmetic Addition (2s Comp.)

• For example
• 6 0 0000110
• -13 1 1110011
• --------------------------------
• 1 1111001 gt 0000111 (2s
• 
  comp.)
• it is equal to 7 so the result is -7

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Arithmetic Addition (2s Comp.)

• For example
• -9 1 0111
• -4 1 1100
• --------------------------------
• 11 0011 gt 01101 (2s comp.)
• it is equal to 13 so the result is -13

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Arithmetic Subtraction

• In 2s complement format it is very simple
• Take the 2s complement of sabtrahend (the
  second number) including the sign bit and add it
  to minuend (the first number) including the sign
  bit and discard a carry out of sign bit
• By taking 2s complement of the subtrahend its
  sign can be changed. Thus, we can change
  subtraction to addition operation.

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Arithmetic Subtraction (2s Comp)

• For example
• (9) 0 1001
• -(4) 0 0100 is changed
  to
• (9) 0 1001
• (-4) 1 1100
• ----------------
• 1 0 0101 (5)
  discard carry

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Arithmetic Subtraction (2s Comp)

• For example
• (-4) 1 1100
• -(9) 0 1001 is changed
  to
• (-4) 1 1100
• 1 0111
• ----------------
• 1 1 0 011 (2s
  comp 01101 means magnitude is 13) discard carry
  gt (-13)

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Arithmetic Subtraction

• For example
• -6 11111010
• -13 11110011 gt is changed to
• -6 11111010
• 13 00001101
• -----------------
• 100000111 gt discard carry
• ? 00000111 is 7

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Binary Coded Decimal (BCD)

• 4 bits used to encode one decimal digit
• For example (4321)10 0100 0011 0010 0001
• The problem is BCD can not be used for conversion
  from decimal to binary and from binary to
  decimal. Because there is no decimal digits for
  1010, 1011,1100,1101,1110,1111
• For example (0111 0010 1100)2 (72?)
• there is no decimal digits (0..9) for 1100

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BCD Arithmetic

• Add 2 BCD numbers using regular binary addition
• Check each nibble ( 4-bit) if result is greater
  than 9 add 6 to it
• If there is carry between 2 nibble or coming
  from 2th nibble add 6

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BCD Arithmetic

• Example 27 0010 0111
• 34 0011 0100
• -----------------
• 0101 1011 gt 9
• 0110
• ----------------
• 01100001

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BCD Arithmetic

• Example 1 carry
• 59 0101 1001
• 39 0011 1001
• ----------------
• 1001 0010
• 0110
• ----------------
• 1001 1000 (98)

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BCD Arithmetic

• 1 1 carry
• 98 1001 1000
• 89 1000 1001
• ---------------
• 1 0010 0001
• 0110 0110
• ---------------
• 11000 0111 (187)
  

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Character Representation

• ASCII American Standard Code for Information
  Interchange
• 128 characters (7 bits required)
• Contains
• Control characters (non-printing)
• Printing characters (letters, digits, punctuation)

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ASCII Characters
Hex Equiv. Binary Character
00 00000000 NULL
07 00000111 Bell
09 00001001 Horizontal tab
0A 00001010 Line feed
0D 00001110 Carriage return
20 00100000 Space (blank)
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ASCII Characters
Hex Equiv. Binary Character
30 00110000 0
31 00110001 1
39 00111001 9
41 01000001 A
42 01000010 B
61 01100001 a
62 01100010 b
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Error-Detecting Code

• To detect the error in data communication, an
  eighth bit is added to ASCII character to
  indicate its parity.
• A parity bit is an extra bit included with a
  message to make the total number of 1s either
  even or odd
• The 8-bit characters included parity bits (with
  even parity) are transmitted to their
  destination. If the parity of received character
  is not even it means at least one bits has been
  changed.

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Error-Detecting Code

• Example even parity odd
  parity
• ASCII A 1000001 01000001 11000001
• ASCII T 1010100 11010100 01010100
• The method of error checking with even parity
  detects any odd combinations of errors in each
  character that is transmitted. An even
  combination of errors is undetected with this
  method.

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Binary Storage and Registers

• Register is a group if binary cells that are
  responsible for storing and holding the binary
  information.
• Register transfer operation is transferring
  binary operation from one set of registers to
  another set of registers.
• Digital logic circuits process the binary
  information stored in the registers.

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Binary Logic

• Binary logic or Boolean algebra deals with
  variables that take on two discrete values.
• Binary logic consists of binary variables (e.g.
  A,B,C, x,y,z and etc.) that can be 1 or 0 and
  logical operations such as
• AND x.yz or xyz (see AND truth table)
• OR x y z (see OR truth table)
• NOT x z (not x is equal to z)
• Note that binary logic is different from binary
  arithmetic. For example in binary logic 1 1 1
  (1OR1) but in binary arithmetic 1 1 10 (1
  PLUS 1)

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Logic Gates

• Logic gates are electronic circuits that operate
  on one or more input signals to produce an output
  signal.
• Electrical signal can be voltage.
• Voltage-operated circuits respond to two separate
  levels equal to logic 1 or 0.
• Logical gates can be considered as a block of
  hardware that produced the equivalent of logic 1
  or logic 0 output signals if input logical
  requirement are satisfied

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