Data Representation and Binary Arithmetic. Lecture 2


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1 Data Representation and Binary Arithmetic Lecture 2 Computer Data Data is stored as binary; 0 s and 1 s Because twostate ( 0 & 1 ) logic elements can be manufactured easily Bit: binary digit (smallest quantity of info.) Organized in groups of bits stored in registers or memory The basic unit of data Aword: 8, 16, 32, 64 bits Usually all bits of a word are accessible simultaneously A byte is a group of 8 bits a word may be 1, 2, 4, or 8 bytes Lecture 2 2 1
2 Data Representation A computer word is whatever the programmer chooses it to represent Example: in our midterm, a student s score is recorded as in binary. What s the meaning? It might be 0 or 100! Entities which can be represented by a binary word are: an instruction or an operation code E.g. For Freescale ColdFire Family microprocessor represents STOP a numeric quantity Unsigned binary number, signed binary number Binary floating point number, complex number, etc. a character The letter A or b or TAB or a space a Graphic Character on the screen, e.g. Lecture 2 3 Binary Numbers = 1x10 2 = x10 1 = 0 + 1x10 0 = 1 + 1x101 = x102 = = 1x2 2 = 4 + 0x2 1 = 0 + 1x2 0 = 1 + 1x21 = x22 = 0.25 = Lecture 2 4 2
3 How to convert a number in base b to a decimal number? Lecture 2 5 Equivalence Table DECIMAL BINARY OCTAL HEXADECIMAL A B C D E F Lecture 2 6 3
4 Binary to Octal/Octal to Binary 1 Octal Digit corresponds to exactly 3 binary digits (8=2 3) EXAMPLES: 1) Divide into groups of 3 2) Write each digit in 3bit binary format Lecture 2 7 Binary to Hexadecimal/Hex. to Binary 1 Hexadecimal Digit corresponds to exactly 4 binary digits (16=2 4) EXAMPLES: 1) D. E 16 2) 9D0.7C 16 Write each digit in 4bit binary format Lecture 2 8 4
5 Decimal to Hex. Conversion In order to convert a decimal number into a hexadecimal number, the decimal number can first be converted into a binary number, and then into a hexadecimal number EXAMPLE: Two methods: 1. Divide by 16 or = N convert to binary (divide by 2) & group 25/16 = 1 R=9 1/16 = 0 R=1 LSB MSB 25/2 = 12 R=1 12/2 = 6 R=0 LSB 6/2 = 3 R=0 N 16 = /2 = 1 R=1 N 2 = Note: division completes until 1/2 = 0 R=1 MSB the quotient is 0 Question: Octal to Hex.? Lecture 2 9 Character Representation AMERICAN STANDARD CODE FOR INFORMATION INTERCHANGE (ASCII) A code for representing alphanumeric characters in a digital system 7 bits, 128 possible combinations 26 upper case, 26 lower case, 10 numerals, special? / } *! Control Characters Examples: CR Carriage Return LF Line Feed BS Back Space VT Vertical Tab HT Horizontal Tab ETX End of Text SOH Start of Header STX Start of Text ACK Acknowledge ENQ Inquiry Lecture
6 ASCII Character Set MSB LSB NUL DLE SP P p SOH DC1! 1 A Q a q STX DC2 2 B R b r ETX DC3 # 3 C S c s EQT DC4 $ 4 D T d t ENQ NAK % 5 E U e u ACK SYN & 6 F V f v BEL ETB 7 G W g w BS CAN ( 8 H X h x HT EM ) 9 I Y i y A 1010 LF SUB * : J Z j z Lecture B 1011 VT ESC + ; K [ k { C 1100 FF FS, < L \ l D 1101 CR GS  = M ] m } E 1110 SO RS. > N ^ n ~ F 1111 SI US /? O _ o DEL L = 4C 16 = l = 6C 16 = Example Upper and Lower Case Letters L 4C l 6C Masking Operations  to change to lower case: OR with to SET bit #5  to change to upper case: AND with to CLEAR bit #5 B6 B5 B4 B3 B2 B1 B0 B6 B5 B4 B3 B2 B1 B0 OR AND (LSB is bit #0) B6 1 B4 B3 B2 B1 B0 B6 0 B4 B3 B2 B1 B0 Lecture
7 Parity Bit Often a 7bit ASCII code character is encoded as an 8 bit binary code pattern. This extra bit is traditionally placed in the most significant bit position EXAMPLE: L 4C Even Parity Parity Bit Odd Parity Even parity: the total # of 1 s in the encoded 8 bits is even Lecture 2 13 Binary Coded Decimal (BCD) A set of 4 bits is used to encode each decimal digit 0 to 9 (i.e., to ) EXAMPLE: 1) = N BCD Disadvantages of BCD encoding: BCD used 4xn bits to encode an n digit number Not as efficient to store information EXAMPLE: 1) = BCD Binary is bits needed BCD is bits needed, not as efficient Advantage: It is relatively easy to figure out what s the equivalent decimal number Lecture
8 Binary Arithmetic ADDITION SUBTRACTION = = = = 1 + borrow 1+0= = = 0 + carry 11 = 0 EXAMPLES: Addition Subtraction = = = Wrong = (overflow) = Correct For unsigned calculation: carry/borrow means overflow Signed calculation: see slides Lecture 2 15 Signed Negative Number There are two commonly used methods to represent negative numbers: 1) Sign plus magnitude 2) 2's complement All systems use the most significant bit position (MSB) as the sign bit 1  negative number 0  positive number Lecture
9 Sign and Magnitude MSB LSB Given an nbit, the range of values possible is ( 2 n11 ) to ( 2 n11 ) EXAMPLES: 1) 7 4 bit +7 = = bit +7 = = ) 65 3) = = Not possible with 8 bits / Range 127 to is fine in a 2 s complement system with 8 bits! Lecture 2 17 Ten s Complement Arithmetic The 10 s complement of a ndigit number, N, is defined as 10 n N It can be calculated by subtracting each of the digits from 9 and adding 1 to the result, e.g. Consider a 4 digit number = = 8766 Lecture
10 Ten s Complement Arithmetic The complement of the complement is the original number = = 1234 Lecture s Complement Generation of 2's complement representation of an nbit number  Subtract no. from 2 n i.e 2 n A Flip each bit of the n bits, and add Copy bits from right until 1st One, then flip rest 2 s complement n=8 sign & magnitude n ( 2 n11) ( 2 n1 ) 128 not valid 1110 Range: ( 2 n1 ) to ( 2 n11 ) ( 2 n11) to ( 2 n11 ) In 2 s complement system, A is represented by 2 s complement of A Lecture
11 2 s Complement  Examples 44 + (56) +44 = = = = and (i.e., 12) are complement to each other. So represents 12 You try the addition of two negative numbers, e.g., (44) + (56) Lecture 2 21 Overflow Assume 8bit binary numbers. Does C=0 mean V=0? Yes, Why???? real result out of range: the biggest positive number with 8bits is (09) Overflow?? Overflow?? C= V=1 No, Why??? C= =59 10 V=0 For summation of two numbers of different sign, the real result is always within range. You find the examples with C=0, V=0, and C=1, and V=1 Lecture
12 Overflow (Cont d) When the real result of an addition or subtraction operation gives too large a result, this represents an overflow condition. There is overflow in addition when positive negative + positive + negative negative positive There is no overflow in addition when positive negative positive negative + positive + negative + negative + positive positive negative There is overflow in subtraction when positive negative  negative  positive negative positive Lecture 2 23 Extension of 2 s complement numbers If we have a 4 bit 2's complement number, what is the corresponding 8bit representation? Extend the MSB! +7 = = = = Weighted values of bit positions in a 2's complement representation: n=8, 8bit a 7 a 6 a 5 a 4 a 3 a 2 a 1 a 02 n1, 2 7 = = = = 16 Lecture
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