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### Slide03 Number System and Operations Part 1

• 1. Number Systems & Operations Part I 19 พฤศจิกายน 2555
• 2. Decimal Numbers (Base 10) People use decimal numbers. I hope you know this very well. However, let’s review:  Ten digits  0-9  The value of a digit is determined by its position in the number. . . . 102101100.10-110-210-3 . . .
• 3. Binary Numbers Decimal Binary 0 0 1 1 2 10 3 11 There are only 2 digits 4 100 (0 and 1) and we can 5 101 do binary counting as 6 110 shown in the table. 7 111 8 1000 9 1001 10 1010
• 4. Binary Numbers (Base 2) The weighting structure of binary numbers 2Positive power 23two 21 20 n-1 . . . of 22 .2Negative . . . two 2 power of 2 -1 -2 -n (whole number) (fractional number) 25 24 23 22 21 20 2-1 2-2 2-3 2-4 2-5 2-6 32 16 8 4 2 1 1/2 1/4 1/8 1/16 1/32 1/64 0.5 0.25 0.125 0.0625 0.03125 0.015625
• 5. Binary-to-Decimal Conversion Add the weights of all 1s in a binary number to get the decimal value. ex: convert 11011012 to decimal Weight 26 25 24 23 22 21 20 bin 1 1 0 1 1 0 1 11011012 ! = 26 + 25 + 23 + 22 + 20 ! ! ! = 64 + 32 + 8 + 4 + 1 ! ! ! = 109
• 6. Binary-to-Decimal Conversion Fractional binary example ex: convert 0.1011 to decimal Weight 2-1 2-2 2-3 24 bin 1 0 1 1 0.1011 != 2-1 + 2-3 + 2-4 ! ! ! = 0.5 + 0.125 + 0.0625 ! ! ! = 0.6875
• 7. Decimal-to-Binary Conversion Sum-of-weights method  To get the binary number for a given decimal number, ﬁnd the binary weights that add up to the decimal number. ex: convert 1210 , 2510 , 5810 , 8210 to binary ! 12 = 8+4 = 23+22 = 1100 ! 25 = 16+8+1 = 24+23+20 = 11001 ! 58 = 32+16+8+2 = 25+24+23+21 = 111010 ! 82 = 64+16+2 = 26+24+21 = 1010010
• 8. Decimal-to-Binary Conversion Repeated division-by-2 remainder method 12/2 = 6 0 LSB  To get the binary number for a given 6/2 = 3 0 decimal number, divide the decimal number by 3/2 = 1 1 2 until the quotient is 0. 1/2 = 0 1 MSB Remainders form the binary number. Stop when the whole-number quotient is 0 1210 = 11002
• 9. Decimal-to-Binary Conversion Converting decimal fractions to binary  Sum-of-weights This method can be applied to fractional decimal numbers, as shown in the following example: ! 0.625 = 0.5+0.125 = 2-1+2-3 = 0.101  Repeated multiplication by 2 Decimal fraction can be converted to binary by repeated multiplication by 2 (see details in the following slide.)
• 10. Repeated Multiplication by 2 (by example) ex: convert the decimal fraction 0.3125 to binary carry 0.3125 x 2 = 0.625 0 MSB 0.625 x 2 = 1.25 1 0.25 x 2 = 0.50 0 0.50 x 2 = 1.00 1 LSB Continue to the desired number of decimal places or stop when the fractional part is all zero 0.312510 = 0.01012
• 11. Binary Arithmetic Basic of binary arithmetic  Binary addition  Binary subtraction  Binary multiplication  Binary division
• 12. Binary Addition The four basic rules for adding digits are as follows:  0+0=0  sum of 0 with a carry of 0  0+1=1  sum of 1 with a carry of 0  1+0=1  sum of 1 with a carry of 0  1+1=10  sum of 0 with a carry of 1
• 13. Binary Addition (by example) 11 3 100 4 +11 +3 + 10 +2 110 6 110 6 111 7 110 6 + 11 +3 +100 +4 1010 10 1010 10
• 14. Binary Subtraction The four basic rules for subtracting digits are as follows:  0-0 = 0  1-1 = 0  1-0 = 1  10-1 = 1 ; 0-1 with a borrow of 1
• 15. Binary Subtraction (by example) 11 3 11 3 -01 -1 -10 -2 10 2 01 1 101 5 -011 -3 010 2
• 16. Binary Multiplication The four basic rules for multiplying digits are as follows:  0x0 = 0  0x1 = 0  1x0 = 0  1x1 = 1 Multiplication is performed with binary numbers in the same manner as with decimal numbers.  It involves forming partial products, shifting each successive partial product left one place, and then adding all the partial products.
• 17. Binary Multiplication (by example) 11 3 101 5 x11 x3 x111 x7 11 9 101 35 +11 101 1001 +101 100011
• 18. Binary Division Division in binary follows the same procedure as division in decimal. 10 2 11 3 11 110 3 6 10 110 2 6 11 6 10 6 000 0 10 0 10 00
• 19. 1’s and 2’s Complements They are important since they permit the presentation of negative numbers. The method of 2’s complement arithmetic is commonly used in computers to handle negative numbers.
• 20. Finding the 1’s complement Very simple: change each bit in a number to get the 1’s complement ex: ﬁnd 1’s complement of 111001012 Binary 1 1 1 0 0 1 0 1 1’s complement 0 0 0 1 1 0 1 0
• 21. Finding the 2’s Complement Add 1 to the 1’s complement to get the 2’s complement. +1 ex: 10110010  01001101  01001110 1’s complement 2’s complement An alternative method:  Start at the right with the LSB and write the bits as they are up to and including the ﬁrst 1.  Take the 1’s complement of the remaining bits. 10110010!! 10111000!binary ! ! ! 01001110!! 01001000!2’s comp
• 22. Signed Numbers Digital systems, such as computer, must be able to handle both positive and negative numbers. A signed binary number consists of both sign and magnitude information.  The sign indicates whether a number is positive or negative.  The magnitude is the value of the number.
• 23. Signed Numbers There are 3 forms in which signed integer numbers can be represented in binary:  Sign-magnitude (least used)  1’s complement  2’s complement (most important) Non-integer and very large or small numbers can be expressed in ﬂoating-point format.
• 24. The Sign Bit The left-most bit in a signed binary number is the sign bit.  It tells you whether the number is positive (sign bit = 0) or negative (sign bit = 1).
• 25. Sign-Magnitude Form The left-most bit is the sign bit and the remaining bits are the magnitude bits.  The magnitude bits are in true binary for both positive and negative numbers. ex: the decimal number +25 is expressed as an 8-bit signed binary number as: 00011001 While the decimal number -25 is expressed as 10011001
• 26. Sign-Magnitude Form “ In the sign-magnitude form, a negative number has the same magnitude bits as the corresponding positive number but the sign bit is a 1 rather than a 0. “
• 27. 1’s Complement Form Positive numbers in 1’s complement form are represented the same way as the positive sign-magnitude. Negative numbers are the 1’s complements of the corresponding positive numbers. ex: the decimal number +25 is expressed as: 00011001 While the decimal number -25 is expressed as 11100110
• 28. 1’s Complement Form “ In the 1’s complement form, a negative number is the 1’s complement of the corresponding positive number. “
• 29. 2’s Complement Form Positive numbers in 2’s complement form are represented the same way as the positive sign-magnitude and 1’s complement form. Negative numbers are the 2’s complements of the corresponding positive numbers. ex: the decimal number +25 is expressed as: 00011001 While the decimal number -25 is expressed as 11100111
• 30. 2’s Complement Form “ In the 2’s complement form, a negative number is the 2’s complement of the corresponding positive number. “
• 31. Decimal Value of Signed Numbers Sign-magnitude:  Both positive and negative numbers are determined by summing the weights in all the magnitude bit positions where these are 1s and ignoring those positions where there are 0s.  The sign is determined by examination of the sign bit.
• 32. Decimal Value of Signed Numbers Sign-magnitude (by example) ex: decimal values of these numbers (expressed in sign-magnitude) 1) 10010101 2) 01110111 1) 10010101 2) 01110111 magnitude magnitude 26 25 24 23 22 21 20 26 25 24 23 22 21 20 ! 0 0 1 0 1 0 1 1 1 1 0 1 1 1 ! = 16+4+1 = 21 = 64+32+16+4+2+1 = 119 ! sign sign ! = 1  negative = 0  positive ! Hence: 10010101 = -21 Hence: 01110111 = 119
• 33. Decimal Value of Signed Numbers 1’s complement:  Positive – determined by summing the weights in all bit positions where there are 1s and ignoring those positions where there are 0s.  Negative – determined by assigning a negative value to the weight of the sign bit, summing all the weights where there are 1’s, and adding 1 to the result.
• 34. Decimal Value of Signed Numbers 1’s complement (by example) ex: decimal values of these numbers (expressed in 1’s complement) 1) 00010111 2) 11101000 1) 00010111 2) 11101000 -27 26 25 24 23 22 21 20 -27 26 25 24 23 22 21 20 ! 0 0 0 1 0 1 1 1 1 1 1 0 1 0 0 0 ! ! = 16+4+2+1 = +23 = (-128)+64+32+8 = -24 ! +1 ! Hence: 00010111 = +23 Hence: 11101000 = -23
• 35. Decimal Value of Signed Numbers 2’s complement:  Positive – determined by summing the weights in all bit positions where there are 1s and ignoring those positions where there are 0s.  Negative – the weight of the sign bit in a negative number is given a negative value.
• 36. Decimal Value of Signed Numbers 2’s complement (by example) ex: decimal values of these numbers (expressed in 2’s complement) 1) 01010110 2) 10101010 1) 01010110 2) 10101010 -27 26 25 24 23 22 21 20 -27 26 25 24 23 22 21 20 ! 0 1 0 1 0 1 1 0 1 0 1 0 1 0 1 0 ! = 64+16+4+2 = +86 = (-128)+32+8+2 = -86 ! Hence: 01010110 = +86 Hence: 10101010 = -86
• 37. Range of Signed Integer Numbers The range of magnitude of a binary number depends on the number of bits (n). Total combinations = 2n  8 bits = 256 different numbers  16 bits = 65,536 different numbers  32 bits = 4,294,967,296 different numbers
• 38. Range of Signed Integer Numbers For 2’s complement signed numbers:  Range = -(2n-1) to +(2n-1-1)  where there is one sign bit and n-1 magnitude ex: Negative Positive Boundary Boundary 4 bits -(23) = -8 (23-1) = +7 8 bits -(27) = -128 (27-1) = +127 16 bits -(215) = -32,768 (215-1) = +32767
• 39. Floating-point numbers How many bits do we need to represent very large number? Floating-point number consists of two parts plus a sign.  Mantissa – represents the magnitude of the number.  Exponent – represents the number of places that the decimal point (or binary point) is to be moved.  Decimal number example: 241,506,800 Mantissa = 0.2415068 Exponent = 109 Can be written as FP as 0.2415068 x 109
• 40. Binary FP Numbers The format deﬁned by ANSI/IEEE Standard 754-1985  Single-precision  Double-precision  Extended-precision Same basic formats except for the number of bits.  Single-precision = 32 bits  Double-precision = 64 bits  (Double) Extended-precision = 80 bits
• 41. Single-Precision Floating-Point Binary Numbers Standard format:  Sign bit (S) – 1 bit  Exponent (E) – 8 bits  Mantissa or fraction (F) – 23 bits S(1) E(8) F(23) Single-precision FP Binary Number Format
• 42. Single-Precision Floating-Point Binary Numbers Mantissa  The binary point is understood to be to the left of the 23 bits.  Effectively, there are 24 bits in the mantissa because in any binary number the left most bit is always 1. (say 001101100 is 1101100.)  Therefore, this 1 is understood to be there although it does not occupy an actual bit position. S(1) E(8) F(23) Single-precision FP Binary Number Format
• 43. Single-Precision Floating-Point Binary Numbers Exponent  The eight bits represent a biased exponent which is obtained by adding 127.  The purpose of the bias is to allow very large or very small numbers without requiring a separate sign bit for the exponents.  The biased exp allows a range of actual exp values from -126 (000000012) to +128 (111111102) S(1) E(8) F(23) Single-precision FP Binary Number Format
• 44. Single-Precision Floating-Point Binary Numbers Not easy, is it? Let’s see an example. ex: 10110100100012 (assumption: positive number) It can be expressed as 1 plus a fractional binary number. Hence: 1011010010001 = 1.011010010001 x 212 The exponent,12, is expressed as a biased exponent as followed: 12+127 = 139 = 10001011 Therefore, we get: 0 10001011 01101001000100000000000
• 45. Single-Precision Floating-Point Binary Numbers Let’s do the opposite way:  To evaluate a binary number in FP format.  General formula: ! Number = (-1)S(1+F)(2E-127) ex: 1 10010001 10001110001000000000000 Number = (-1)(1.10001110001)(2145-127) ! ! ! = (-1)(1.10001110001)(218) ! ! ! = -11000111000100000002
• 46. Single-Precision Floating-Point Binary Numbers Let’s review:  The exponent can be any number between -126 to +128; that means extremely large and small numbers can be expressed.  Say, a 32-bit FP number can replace a binary integer number having 129 bits.  Distinctive point: Because the exponent determines the position of the binary point, numbers containing both integer and fractional parts can be represented.
• 47. Single-Precision Floating-Point Binary Numbers There are 2 exceptions to the format for FP numbers:  The number 0.0 is represented by all 0s. x 00000000 00000000000000000000000  Infinity is represented by all 1s in the exponent and all 0s in the mantissa. x 11111111 00000000000000000000000
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