Binary Numbers 1 The Decimal Number System cont

Binary Numbers 1

The Decimal Number System (con’t) • The decimal number system is also known as base 10. The values of the positions are calculated by taking 10 to some power. • Why is the base 10 for decimal numbers? o Because we use 10 digits, the digits 0 through 9. 2

The Decimal Number System - base 10 • The decimal number system is a positional number system with a base 10. • Example: 1011 1000 1 x 23 • 000 0 x 22 10 1 1 x 21 1 x 20 10112 = 1000 + 10 + 1 = 1 x 23 + 0 x 22 + 1 x 21 + 1 x 20 = 1110 3

The Binary Number System • The binary number system is also known as base 2. The values of the positions are calculated by taking 2 to some power. • Why is the base 2 for binary numbers? o Because we use 2 digits, the digits 0 and 1. 4

The Binary Number System – base 2 • The decimal number system is a positional number system with a base 10. • Example: 5623 5000 5 x 103 • 600 6 x 102 20 3 2 x 101 3 x 100 5623 = 5000 + 600 + 20 + 3 = 5 x 103 + 6 x 102 + 2 x 101 + 3 x 100 5

Why Bits (Binary Digits)? • • • Computers are built using digital circuits – Inputs and outputs can have only two values – True (high voltage) or false (low voltage) – Represented as 1 and 0 Can represent many kinds of information – Boolean (true or false) – Numbers (23, 79, …) – Characters (‘a’, ‘z’, …) – Pixels – Sound Can manipulate in many ways – Read and write – Logical operations – Arithmetic – … 6

Base 10 and Base 2 • Base 10 – Each digit represents a power of 10 – 5417310 = 5 x 104 + 4 x 103 + 1 x 102 + 7 x 101 + 3 x 100 • Base 2 – Each bit represents a power of 2 – 101012= 1 x 24 + 0 x 23 + 1 x 22 + 0 x 21 + 1 x 20 = 2110 7

The Binary Number System (con’t) • The binary number system is also a positional numbering system. • Instead of using ten digits, 0 - 9, the binary system uses only two digits, 0 and 1. • Example of a binary number and the values of the positions: 1 0 0 1 1 0 1 2 6 25 24 23 22 21 20 8

Converting from Binary to Decimal 1 0 0 1 1 0 1 26 25 24 23 22 21 20 1 X 20 = 1 0 X 21 = 0 1 X 22 = 4 1 X 23 = 8 0 X 24 = 0 0 X 25 = 0 1 X 26 = 64 7710 20 = 1 21 = 2 22 = 4 23 = 8 24 = 16 25 = 32 26 = 64 9

Converting from Binary to Decimal (con’t) Practice conversions: Binary 11101 1010101 100111 10 Decimal

Converting From Decimal to Binary (con’t) • • Make a list of the binary place values up to the number being converted. Perform successive divisions by 2, placing the remainder of 0 or 1 in each of the positions from right to left. Continue until the quotient is zero. Example: 4210 2 5 24 23 22 21 20 32 16 8 4 2 1 1 0 1 0 42/2 = 21 21/2 = 10 10/2 = 5 5/2 = 2 2/2 = 1 1/2 = 0 and R = 1 and R = 0 and R = 1 11

Example 1210 We repeatedly divide the decimal number by 2 and keep remainders – – 12/2 = 6 6/2 = 3 3/2 = 1 1/2 = 0 and and R=0 R=1 The binary number representing 12 is 1100

Converting From Decimal to Binary (con’t) Practice conversions: Decimal 59 82 175 13 Binary

Exercises • Find the binary the decimal number represented by the following binary sequences: – 110101 – 10111010 • Represent the number 135 in base 2.

Fractional Numbers Examples: 456. 7810 = 4 x 102 + 5 x 101 + 6 x 100 + 7 x 10 -1+8 x 10 -2 1011. 112 = 1 x 23 + 0 x 22 + 1 x 21 + 1 x 20 + 1 x 2 -1 + 1 x 2 -2 = 8 + 0 + 2 + 1/2 + ¼ = 11 + 0. 5 + 0. 25 = 11. 7510 • Conversion from binary number system to decimal system Examples: 111. 112 = 1 x 22 + 1 x 21 + 1 x 20 + 1 x 2 -1 + 1 x 2 -2 = 4 + 2 + 1/2 + ¼ = 7. 7510 2 1 0 -1 -2 Examples: 11. 0112 22 4 21 2 20 1 2 -1 ½ 2 -2 ¼ -3 2 -3 1/8 15

Fractional numbers Examples: 7. 7510 = (? )2 1. Conversion of the integer part: same as before – repeated division by 2 7 / 2 = 3 (Q), 1 (R) 3 / 2 = 1 (Q), 1 (R) 1 / 2 = 0 (Q), 1 (R) 710 = 1112 2. Conversion of the fractional part: perform a repeated multiplication by 2 and extract the integer part of the result 0. 75 x 2 =1. 50 extract 1 0. 5 x 2 = 1. 0 extract 1 0. 7510 = 0. 112 0. 0 stop write in the same order Combine the results from integer and fractional part, 7. 7510 = 111. 112 How about choose some of 4 2 Examples: try 5. 625 1 1/2 1/4 =0. 5 =0. 25 1/8 =0. 125 16

Fractional Numbers (cont. ) Exercise 1: Convert (0. 625)10 to its binary form Solution: 0. 625 x 2 = 1. 25 extract 1 0. 25 x 2 = 0. 5 extract 0 0. 5 x 2 = 1. 0 extract 1 stop 0. 0 Solution : (0. 625)10 = (0. 101)2 Exercise 2: Convert (0. 6)10 to its binary form 0. 6 x 2 = 1. 2 extract 1 0. 2 x 2 = 0. 4 extract 0 0. 4 x 2 = 0. 8 extract 0 0. 8 x 2 = 1. 6 extract 1 0. 6 x 2 = (0. 6)10 = (0. 1001 …)2 17

Fractional Numbers (cont. ) Exercise 3: Convert (0. 8125)10 to its binary form Solution: 1 extract 1 0. 8125 x 2 = 1. 625 0. 625 x 2 = 1. 25 extract 0. 25 x 2 = 0. 5 extract 0 0. 5 x 2 = 1. 0 extract 1 0. 0 stop (0. 8125)10 = (0. 1101)2 18

Fractional Numbers (cont. ) • Errors – One source of error in the computations is due to back and forth conversions between decimal and binary formats Example: (0. 6)10 + (0. 6)10 = 1. 210 Since (0. 6)10 = (0. 1001 …)2 Lets assume a 8 -bit representation: (0. 6)10 = (0. 1001)2 , therefore 0. 6 0. 1001 + 0. 6 + 0. 1001 1. 00110010 Lets reconvert to decimal system: (1. 00110010)b= 1 x 20 + 0 x 2 -1 + 0 x 2 -2 + 1 x 2 -3 + 1 x 2 -4 + 0 x 2 -5 + 0 x 2 -6 + 1 x 2 -7 + 0 x 2 -8 = 1 + 1/8 + 1/16 + 1/128 = 1. 1953125 Error = 1. 2 – 1. 1953125 = 0. 0046875 19

Bits, Bytes, and Words • • • A bit is a single binary digit (a 1 or 0). A byte is 8 bits A word is 32 bits or 4 bytes Long word = 8 bytes = 64 bits Quad word = 16 bytes = 128 bits Programming languages use these standard number of bits when organizing data storage and access. 20

Adding Two Integers: Base 10 • From right to left, we add each pair of digits • We write the sum, and add the carry to the next column 0 1 1 + 0 0 1 2 Sum 1 0 0 1 Carry 0 1 1 1 9 8 + 2 6 4 Sum 4 6 Carry 0 1 21

Example 10011110 1101111 + + 111 ----------= 101 0 0 101 1101 ---------= 1111100

Fractional Numbers (cont. ) • Errors – One source of error in the computations is due to back and forth conversions between decimal and binary formats Example: (0. 6)10 + (0. 6)10 = 1. 210 Since (0. 6)10 = (0. 1001 …)2 Lets assume a 8 -bit representation: (0. 6)10 = (0. 1001)2 , therefore 0. 6 0. 1001 + 0. 6 + 0. 1001 1. 00110010 Lets reconvert to decimal system: (1. 00110010)b= 1 x 20 + 0 x 2 -1 + 0 x 2 -2 + 1 x 2 -3 + 1 x 2 -4 + 0 x 2 -5 + 0 x 2 -6 + 1 x 2 -7 + 0 x 2 -8 = 1 + 1/8 + 1/16 + 1/128 = 1. 1953125 Error = 1. 2 – 1. 1953125 = 0. 0046875 23

Binary subtraction

Binary subtraction (Cont)

Binary subtraction (Cont)

Exercise • 10010101 - 11011 =? • 10000001 - 111 = ?

Binary Multiplication 1 0 0 0 two = 8 ten 1 0 0 1 two = 9 ten ______ multiplicand multiplier 1000 0000 1000 ______ 1 0 0 0 two = 72 ten Spring 2007, Jan. 17 partial products ELEC 2200 (Agrawal) 28

Binary Division 13 11/147 11 37 33 4 Spring 2007, Jan. 17 Quotient Divisor / Dividend Partial remainder Remainder 00001101 1011/10010011 1011 001110 1011 001111 100 ELEC 2200 (Agrawal) 29

Bitwise Operators: Shift Left/Right • Shift left (<<): Multiply by powers of 2 – Shift some # of bits to the left, filling the blanks with 0 53 0 0 1 1 0 0 53<<2 1 1 0 0 0 0 • Shift right (>>): Divide by powers of 2 – Shift some # of bits to the right • For unsigned integer, fill in blanks with 0 • What about signed integers? Varies across machines… – Can vary from one machine to another! 53 0 0 1 1 0 0 53>>2 0 0 1 1 0 1 30

Boolean Algebra to Logic Gates • Logic circuits are built from components called logic gates. • The logic gates correspond to Boolean operations +, *, ’. OR + AND * NOT ’ • Binary operations have two inputs, unary has one

AND Logic Gate: A A*B B Truth Table: A 0 0 1 B 0 1 0 A*B 0 0 0 1 1 1

OR Logic Gate: A A+B B Truth Table: A 0 0 1 B 0 1 0 A+B 0 1 1 1

NOT Logic Gate: A (also called an inverter) A’ or A Truth Table: a 0 1 A 1 0

n-input Gates • Because + and * are binary operations, they can be cascaded together to OR or AND multiple inputs. A B C A+B+C A B C ABC

• NAND and NOR Gates NAND and NOR gates can greatly simplify circuit diagrams. As we will see, can you use these gates wherever you could use AND, OR, and NOT. NAND NOR A B 0 0 1 1 1 0 A B 0 0 1 0 1 0 0 1 1 0

XOR and XNOR Gates • XOR is used to choose between two mutually exclusive inputs. Unlike OR, XOR is true only when one input or the other is true, not both. XOR XNOR A B 0 0 1 1 1 0 A B 0 0 1 0 1 0 0 1 1 1

Binary Sums and Carries a 0 0 1 1 b 0 1 Sum 0 1 1 0 a 0 0 1 1 XOR b 0 1 Carry 0 0 0 1 AND 0100 0101 69 + 0110 0111 103 1010 1100 172 38

Design Hardware Bit by Bit • Adding two bits: a 0 0 1 1 b 0 1 half_sum 0 1 1 0 carry_out 0 a 0 b 0 1 • Half-adder circuit Spring 2007, Jan. 17 ELEC 2200 (Agrawal) XOR half_sum AND carry_out 39

Half Adder (1 -bit) A B Half Adder S C A B S(um) C(arry) 0 0 0 1 1 0 1 0 1

Half Adder (1 -bit) A B Sum Carry A B S(um) C(arry) 0 0 0 1 1 0 1 0 1

Full Adder A Carry In (Cin) B Full Adder S Cout Cin A B S(um) Cout 0 0 0 0 1 1 0 0 1 0 1 1 1 0 0 1 1 1

Full Adder A B H. A. S Cin Cout

Full Adder A A S A Half Adder B Cin B S S Half Adder C B C Cout

4 -bit Ripple Adder using Full Adder Carry A 3 B 3 A 2 B 2 A 1 B 1 A 0 B 0 A B A B Full Cin Adder Cout Full Cin Adder Cout S S S 3 S 2 S 1 S 0 S A B C Half Adder S Cout A H. A. B Cin Full Adder

Working with Large Numbers 0101000010100111 = ? • Humans can’t work well with binary numbers; there are too many digits to deal with. • Memory addresses and other data can be quite large. Therefore, we sometimes use the hexadecimal number system. 46

The Hexadecimal Number System • The hexadecimal number system is also known as base 16. The values of the positions are calculated by taking 16 to some power. • Why is the base 16 for hexadecimal numbers ? – Because we use 16 symbols, the digits 0 and 1 and the letters A through F. 47

The Hexadecimal Number System (con’t) Binary Decimal Hexadecimal 0 1 10 11 100 101 110 111 1000 1001 0 1 2 3 4 5 6 7 8 9 48 0 1 2 3 4 5 6 7 8 9 Binary Decimal Hexadecimal 1010 1011 1100 1101 1110 1111 10 11 12 13 14 15 A B C D E F

The Hexadecimal Number System (con’t) • Example of a hexadecimal number and the values of the positions: 3 C 8 B 0 5 1 166 165 164 163 162 161 160 49

Example of Equivalent Numbers Binary: 1 0 0 0 0 1 1 12 Decimal: 2064710 Hexadecimal: 50 A 716 Notice how the number of digits gets smaller as the base increases. 50

Summary • • • Convert binary to decimal Decimal to binary Binary operation Logic gates Use of logic gates to perform binary operations – Half adder – Full adder • The need of Hexadecimal

Next lecture (Data representation) • Put this all together – negative and position integer representation – unsigned – Signed – Excess – Tow’s complement • Floating point representation – Single and double precision • Character, colour and sound representation