Information Representation Computer use a binary systems Why
Information Representation Computer use a binary systems Why binary? Electronic bi-stable environment on/off, high/low voltage Bit: each bit can be either 0 or 1 1 0 1 0 -3 mv 0 mv Reliability Digital v. s, Analog With only 2 values, can be widely separated, therefore clearly differentiated “drift” causes less error Example: 1 0 010101 Lecture 4: Information Representation 10 000001
Binary Representation in Computer System All information of diverse type is represented within computers in the form of bit patterns. e. g. , text, numerical data, sound, and images One important aspect of computer design is to decide how information is converted ultimately to a bit pattern Writing software also frequently requires understanding how information is represented along with accuracies Lecture 4: Information Representation
Number Systems Decimal Number System Base is 10 or ‘D’ or ‘Dec’ Ten symbols: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 Each place is weighted by the power of 10 Example: 123410 or 1234 D = 1 x 103 + 2 x 102 + 3 x 101 + 4 x 100 = 1000 + 200 + 30 + 4 $1, 000 $10 Lecture 4: Information Representation $1
Binary Number System Base is 2 or ‘b’ or ‘Bin’ Two symbols: 0 and 1 Each place is weighted by the power of 2 Example: 10112 or 1011 B = 1 x 23 + 0 x 22 + 1 x 21 + 1 x 20 = 8 + 0 + 2 +1 = 1110 11 in decimal number system is 1011 in binary number system Lecture 4: Information Representation
Conversion between Decimal and Binary Conversion from decimal number system to binary system Question: represent 3410 in the binary number system Answer: using the divide-by-2 technique repeatedly 34 17 div-by-2 8 div-by-2 0 4 div-by-2 1 2 div-by-2 0 1 Remainder If we write the remainder from right to left : 3410 1 x 25 + 0 x 24 + 0 x 23 + 0 x 22+ 1 x 21 1000102 Lecture 4: Information Representation + 0 x 20
Converting decimal to binary Example: 162: 162 / 2 81 / 2 40 / 2 20 / 2 10 / 2 5/2 2/2 1/2 = 81 = 40 = 20 = 10 =5 =2 =1 =0 rem rem 0 1 0 0 0 1 So, 16210 = 101000102 Lecture 4: Information Representation
Some Exercises 13 D = (? ) B 1101 B 23 D = (? ) B 72 D = (? ) B Lecture 4: Information Representation
Conversion from binary number system to decimal system Example: check if 1000102 is 3410 using the : weights” appropriately 1000102 1 x 25 + 0 x 24 + 0 x 23 + 0 x 22+ 1 x 21 + 0 x 20 32 + 0 + 3410 Lecture 4: Information Representation 0 + 2 + 0
Some Exercises Ex: 0101 B ( ? ) D Bit 4 23 = 8 Bit 3 22 = 4 Bit 2 21 = 2 Bit 1 20 = 1 0 1 Ex: 1100 B ( ? ) D 4 + 1 1 1 0 0 8 4 2 1 = 12 D Ex: 0101 1100 B ( ? ) D 0 1 1 1 0 0 128 64 32 16 8 4 2 1 Lecture 4: Information Representation = 92 D = 5 D
Binary Arithmetic on Integers Addition a 0 0 1 1 b 0 1 a+b 0 1 1 10 Carry bit Example: find binary number of a + b • If a = 13 D , b = 5 D 13 D + 5 D 18 D • If a = 15 D, b = 10 D 15 D + 10 D 25 D Lecture 4: Information Representation 1 1 + 1 1 1 0 1 b 1 0 0 1 0 b 1 1 1 1 b + 1 0 b 1 1 0 0 1 b
Binary Arithmetic on Integers (cont. ) Multiplication a 0 0 1 1 b 0 1 axb 0 0 0 1 Example: if a = 100001 b , b = 101 b , find a x b x 1 0 0 1 b 1 0 1 b 33 D 5 D 100001 0000000 10000100 1 0 0 1 b 165 D Lecture 4: Information Representation
Hexadecimal Number System Base = 16 or ‘H’ or ‘Hex’ 16 symbols: { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A(=10), B(=11), C(=12), D(=13), E(=14), F(=15)} Hexadecimal to Decimal (an-1 an-2…a 1 a 0)16 = (an-1 x 16 n-1 + an-2 x 16 n-2 + …+ a 1 x 161 + a 0 x 160 )D Example: (1 C 7)16 = (1 x 162 + 12 x 161 + 7 x 160 )10 = (256 + 192 + 7)10 = (455)10 Decimal to Hexadecimal Repeated division by 16 Similar in principle to generating binary codes Example: (829)10 = (? )16 Divide-by-16 Quotient 829 / 16 51 / 16 3 / 16 Hence, (829)10 = (3313)16 Lecture 4: Information Representation 51 3 0 Remainder 13 3 3 (33 D)16 Stop, since quotient = 0
Hexadecimal Conversions Hexadecimal to Binary Expand each hexadecimal digit to 4 binary bits. Example: (E 29)16 = (1110 | 0010 | 1001)2 Binary to Hexadecimal Combine every 4 bits into one hexadecimal digit Example: (0101 | 1111 | 1010 | 0110)2 = (5 FA 6)16 Lecture 4: Information Representation
Octal Number System Base = 8 or ‘o’ or ‘Oct’ 8 symbols: { 0, 1, 2, 3, 4, 5, 6, 7} Octal to Decimal (an-1 an-2…a 1 a 0)8 = (an-1 x 8 n-1 + an-2 x 8 n-2 + …+ a 1 x 81 + a 0 x 80 )10 Example: (127)8 = (1 x 82 + 2 x 81 + 7 x 80 )10 = (64 + 16 + 7)10 = (87)10 Decimal to Octal Repeated division by 8 (similar in principle to generating binary codes) Example: (213)10 = (? )8 Divide-by -8 Quotient Remainder 213 / 8 26 / 8 3/8 26 3 0 5 2 3 Lecture 4: Information Representation Octal digit Lower digit = 5 Second digit =2 Third digit =3 Stop, since quotient = 0 Hence, (213)10 = (325)8
Octal to Binary Expand each octal digit to 3 binary bits. Example: (725)8 = (111 | 010 | 101)2 Binary to Octal Combine every 3 bits into one octal digit Example: (110 | 011)2 = (623)8 Lecture 4: Information Representation
Mini Homework 1) Convert the following binary numbers to decimal numbers: (a) 0011 B (b) 0101 B (c) 0001 0110 B (d) 0101 0011 B 2) Convert the following decimal numbers to binary: (a) 21 D (b) 731 D (c) 1, 023 D Lecture 4: Information Representation
3) Convert the following binary numbers to hexadecimal numbers: (a) (b) (c) (d) (a) (b) (c) 0011 B 0101 B 0001 0110 B 0101 0011 B 21 D 731 D 1, 023 D 4. ) Perform the following binary additions and subtractions. Show your work without using decimal numbers during conversion. (a) (b) 111 B + 101 B 1001 B + 11 B Lecture 4: Information Representation
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