Chapter 2
Number Systems + Codes
Overview
Objective: •To use positional number systems
•To convert decimals to binary integers
•To convert binary integers to decimals
•To represent binary integers in 3 forms: –Sign magnitude
–1’s complement
–2’s complement
•To reprensent number in octal (base 8) and hexadecimal numbers and convert them to decimal and binary numbers
•To understand various computer codes
Overview
Two types of representations of information:
• Numeric information-Positional number system
-Conversion between number systems
-Operation on number systems (add, substract …)
•Non numeric information-Codes- Properties of codes
Overview
Real life Data
•Numeric data
(125.5 : a price
•Nonnumeric data
(John : a name )
Representation of data in
Computer or digital circuits
000111000000 : string of 0’s and 1’s
Correspondance or Representation
Positional Number Systems (PNS)• PNS – a number is represented by a string of digits.
Each digit position is weighted by a power of the base or radix.
• Example: Decimal PNS, in base 10 digit position i is weighted by 10i
937.5210=9100 + 310 + 71 + 5.1 + 2.01 + 2.001 + 0.0001
• General Form: Base or Radix = 10D : dp-1dp-2 dp-3… d2 d1d0. d-1 d-2…d-n
D = dp-1.10p-1 + dp-2.10p-2 +…+ d1.101 + d0.100 + d-1.10-1 + d-2.10-2 + … + d-n.10-n
p-1
D = di.10i
i=-n
I. Representation of numerical information
di is a decimal digit
Positional Number Systems (PNS)
• General Form: Base or Radix = 2D : bp-1bp-2 bp-3… b2 b1b0. b-1 b-2…b-n
D = bp-1.2p-1 + bp-2.2p-2 +…+ b1.21 + b0.20 +
b-1.2-1 + b-2.2-2 + … + b-n.2-n
p-1
D = bi.2i
i=-n
bi is a binary digit
I. Representation of numerical information
Summary of PNS
• Decimal is natural (we count in base 10)- (Radix or Base 10 ) digits {0,1,…,8,9}
• Binary is used in digital system- (Radix or Base 2 ) digits {0,1}
• Octal is used for representing multibits (group of 3 bits) numbers in digital systems.- (Radix or Base 8=23) digits {0,1,…,6,7}
• Hexadecimal is used for representing multibits (group of 4 bits) numbers in digital systems- (Radix or Base 16=24) digits {0,1,…9,A,B,…,F}
I. Representation of numerical information
Summary of PNS (…) Dec Binary Oct Hex
0 0 0 (000) 0 (0000)1 01 1 (001) 1 (0001)2 10 2 (010) 2 (0010)3 11 3 (011) 3 (0011)4 100 4 (100) 4 (0100)5 101 5 ( 101) 5 (0101)6 110 6 (110) 6 (0110)7 111 7 (111) 7 (0111)8 1000 10 (001 000) 8 (1000)9 1001 11 (001 001) 9 (1001)10 1010 12 (001 010) A (1010)11 1011 13 (001 011) B (1011)12 1100 14 (001 100) C (1100)13 1101 15 (001 101) D (1101)14 1110 16 (001 110) E (1110)15 1111 17 (001 111) F (1111)16 10000 20 (010 000) 10 (0001 0000)
I. Representation of numerical information
ConversionsFrom any base r ---> Decimal
Use base 10 arythmetic to expand : p-1
D = di.ri (r is the base and di are the digits) i=0
Two solutions can be used to do this :• Polynomial expansion
D = dp-1.rp-1 + dp-2.rp-2 +…+ d1.r1 + d0.r0 Examples : 31016 = 3.162 + 1.161 + 0.160
= 3.256 + 16 + 0 = 78410
1CE816 = 1.163 + 12 .162 + 14. 161+ 8.160 = 740010
436.58 = 4.82 + 3. 81 + 6 .80 + 5. 8-1 = 286.62510
I. Representation of numerical information
ConversionsFrom any base r ---> Decimal
• Iterative multiplication (Expansion of D)
D can be written as :
((….((dp-1).r + dp-2).r +…+d2).r + d1).r + d0
Examples : 31016 = ((3).16+ 1).16 + 0)
= (49).16 + 0 = 78410
Good for programming
I. Representation of numerical information
ConversionsFrom Decimal ---> Any base r
Successive divisions of D by r yield the digits from the least to the most significant bit.
Remember that D can be written in base r as : D = dp-1.rp-1 + dp-2.rp-2 +…+ d1.r + d0.
= ((….((dp-1).r + dp-2).r +…+d2).r + d1).r + d0 D/r -----> quotient D1 = (….((dp-1).r + dp-2).r +…+d2).r + d1
remainder d0
D1/r -----> quotient D2 = (….((dp-1).r + dp-2).r +… d3).r+d2)
remainder d1
D2/r -----> quotient D3 = (….((dp-1).r + dp-2).r +…+d3)
remainder d2And so on …..
I. Representation of numerical information
ConversionsFrom Decimal ---> Any base r
• Example : conversion of 179 from Base 10 -> Base 217910 179 1 89 1 LSB 2 44 1
4 22 0 8 11 0 16 5 1 32 2 1 64 1 0 128 0 1 MSBweight Successive divisions by 2
17910 = 101100112
I. Representation of numerical information
ConversionsFrom Decimal ---> Any base r
• Example : conversion of 467 from Base 10 -> Base 8
467
3 1 58 3 LSB
16 8 7 2
448 64 0 7 MSB
467 weight successive divisions46710 = 7238
I. Representation of numerical information
Other ConversionsFrom Binary ---> Octal/Hex
• Conversion by substitution– Binary to Oct : Starting from rightmost bit, make
groups of 3 bits and convert each group to base8
1000110011102 = 100/011/001/110 = 43168
4 / 3 / 1 / 6
– Binary to Hex : Starting from rightmost bit, make groups of 4 bits and convert each group to base 16
1000110011102 = 1000/1100/1110 = 8CE16
8 / C / E
I. Representation of numerical information
Other ConversionsFrom Octal/hex ---> Binary
• Conversion by substitution– Octal to Binary : Convert each digit to a 3 bits binary
number
5768= 101 111 1102
5 7 6
– Hex to Binary : Convert each digit to a 4 bits binary number
57616 = 0101 0111 01102
5 7 6
I. Representation of numerical information
Addition of binary numbers• Example 1 (Addition) Carry 1 01111000 X 190 10111110 + Y 141 10001101 331 1 01001011 Result has more bits than binary addends
Carry equals 1 in the last bit position
• Example 2 (Addition) Carry 0 01011000 X 173 10101101 + Y 44 00101100 217 11011001 Carry equal to 0 in the last bit position
I. Representation of numerical information
Addition of binary numbers• In general, given two binary number
X = ( xn-1 … xi … x1 x0 )2 and Y = ( yn-1 … yi … y1 y0)2 the sum (X+Y) bit by bit at position i is given by (ci is the ith position’s carry)
ci-1 Addition table
+ xi ci-1 xi yi si ci
+ yi 0 0 0 0 0
ci si 0 0 1 1 0
0 1 0 1 0 0 1 1 0 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 1 1 1 1
I. Representation of numerical information
Substraction of binary numbers
• Substraction B 0 10 1
10 1
10
10
10
X 12129 1 1 1 0 0 1 0 1 - Y 46 0 0 1 0 1 1 1 0 183 1 0 1 1 0 1 1 1
B 0 1
0 1
0 11
0 10 0 10
X 2110 1 1 0 1 0 0 1 0 - Y 1 09 0 1 1 0 1 1 0 1 1 01 1 1 0 0 1 0 1
I. Representation of numerical information
Addition of hexadecimal numbers
C 1 1
X 1 9 B 9 16
+ Y C 7 E 6 16
E 1 9 F 16
I. Representation of numerical information
Addition of two hexadecimal numbers
Representation of negative numbers Signed-magnitude systems
• +99, -57, -13.3Binary uses a (sign bit, the most significant bit (MSB)) MSB = 0 (Positive) MSB = 1 (Negative)
0 10101012 = + 8510 1 10101012 = -8510
0 11111112 = + 12710 1 11111112 = -12710
0 00000002 = + 010 1 00000002 = - 010
• Problem:Too much logic needed to
- detect and compare sign bit - add or abstract magnitudes
- determine the sign of the result
I. Representation of numerical information
Representation of negative numbers Complement number system
• Difficult to change to complement
• But adding or substracting two numbers are easier once they are represented in complement number systems
I. Representation of numerical information
Representation of negative numbers Radix – Complement representation
• n digit number is substracted from rn
• Example:if r = 10 and n = 4, rn = 104 =10,000The 10’s complement of D=184910 is:
(rn – D) = 8151• Computing 10’s complement: rn – D = (rn - 1 – D) +1
(rn - 1) – D => complement of each digit with rspect to base (r-1)
In base 10 complement each digit with respect to 9 and 1 to the result
Example: The 10’s complement of 184910 is (104 - 1849)= (9999 – 1849) + 1 = 8 1 5 0 + 1 = 8151
I. Representation of numerical information
Representation of negative numbers Radix – Complement representation
• Complement of each digit :
r –1 – di
In base 10 : 10 –1 – di = 9 – di
Example : The complement of each digit of 345 is :
For 5 : (10 – 1) – 5 = 9 – 5 = 4
For 4 : = 9 – 4 = 5
For 3 : = 9 – 3 = 6
• In base 3 the radix complement of 121 is 101
• In base 2 the radix complement of 1110111 is 0001000
I. Representation of numerical information
Representation of negative numbers Two’s Complement
• The MSB is the sign bit
• Example:1710 = 000100012 -12710 = 100000012
11101110 01111110
+1 +1 111011112 = -1710 01111111
=12710
• The range is -(2n-1) -> (2n-1-1), For n=8010 = 000000002 -12810 = 100000002
I. Representation of numerical information
Note: The weight of the MSB is –2n-1
The range of positive numbers is 0 to 127
The range of negative numbers is –1 to -128
Representation of negative numbers Two’s Complement
• Two’s complement of 010 and -12810 010 = 000000002 -12810 = 100000002
11111111 01111111
+1 +1 1000000002 = 010 10000000 = -12810
I. Representation of numerical information
Note: Ignore carry out of the MSB position
Note: -128 does not have a positive counterpart, that is –128 is its own
Complement : this can creat problems ……
2’s Complement Addition + Substraction
+3 0011 -2 0010 -> 1101+1-> 1110
+4 +0100 + -6 0110 -> 1001+1-> + 1010
+7 0111 -8 11000
+6 0110 0110
+ -3 0011 -> 1100+1 => 1101
+3 1 0011 = 3
I. Representation of numerical information
Overflow in the last position
Overflow in the last position
Rule : Ignore any carry beyond the MSB. The result is correct if the range of the
number system Is not exceeded.
2’s Complement Addition + SubstractionI. Representation of numerical information
Addition/Substraction rule :
Assume that the signed integers are represented in 2s complement :
To compute A – B, compute the 2’s complement B’ of B and add B’ to A
Examples :
5 0101 5 0101
+2 0010 - 2 + 1110 ( = 2’)
7 0111 3 1 0011 (ignore the leftmost carry
Overflow
• Addition of 2 numbers with different signs can never overflow
• Addition of 2 numbers with same sign -3 1101 +5 0101+ -6 1010 + +6 0110 -9 1 0111 = +7 +11 1011 = -5
wrong wrongIf signs of (addends) are same and sign of sum is
different
I. Representation of numerical information
Substraction rules (2’s complement)
+4 0100 0100 - +3 -0011 1101 1 1 0001
- 3 1101 1101 - -4 1100 0011 1 1 0001
Same overflow rules apply
I. Representation of numerical information
Binary multiplication
11 1011 13 1101 33 1011 11 10000 143 11011 11011 10001111
uses a shift register and adder, and logic control• Signed multiplication + + = +
+ - = -
- - = +
I. Representation of numerical information
Binary codes for decimal numbers
Goal:
Use binary strings (sequence of 0’s and 1’s) to represent decimal information
Definition:
• A code is a set of n-bit strings. Each n-bit string represrnts a different number.
• A code word is a particular member of a code
II. Representation of non numerical information
Binary codes for decimal numbers
• A code is a set of n-bit strings. Each n-bit string represrnts a different number.
• A code word is a particular member of a code• If n bit is used to represent each code words, the total
number of possible code words is 2n
Examples:To encode a set consisting of 3 elements {A,B,C} • we need 2 bits to represent each code word. • The possible code words are {00, 01, 10, 11}
II. Representation of non numerical information
Binary codes for decimal numbers
• We need to assign a code word to each element of the original set.
Assignment 1 :
00 -----> A
01 -----> B
10 -----> C Code 11 is unused
Another assignment :
10 -----> A
01 -----> B
11 -----> C Code 00 is unused
II. Representation of non numerical information
Binary codes for decimal numbers
• To encode the 10 decimal digits {0,1,….,8,9} We need 4 bit binary code words. There are 16=24 possible code words. So 6 codes are unused.
Assignment 1 :
0 -----> 0000 5 -----> 0101
1 -----> 0001 6 -----> 0110
2 -----> 0010 7 -----> 0111
3 -----> 0011 8 -----> 1000
4 -----> 0100 9 -----> 1001
The following codes are unused: 1010, 1011, 1100, 1101, 1110, 1111
II. Representation of non numerical information