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Review on Number Systems

Review on Number Systems. Decimal, Binary, and Hexadecimal. Base-N Number System. Base N N Digits: 0, 1, 2, 3, 4, 5, …, N-1 Example: 1045 N Positional Number System. Digit d o is the least significant digit (LSD). Digit d n -1 is the most significant digit (MSD).

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Review on Number Systems

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  1. Review on Number Systems Decimal, Binary, and Hexadecimal

  2. Base-N Number System • Base N • N Digits: 0, 1, 2, 3, 4, 5, …, N-1 • Example: 1045N • Positional Number System • Digit do is the least significant digit (LSD). • Digit dn-1 is the most significant digit (MSD).

  3. Decimal Number System • Base 10 • Ten Digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 • Example: 104510 • Positional Number System • Digit d0 is the least significant digit (LSD). • Digit dn-1 is the most significant digit (MSD).

  4. Binary Number System • Base 2 • Two Digits: 0, 1 • Example: 10101102 • Positional Number System • Binary Digits are called Bits • Bit bo is the least significant bit (LSB). • Bit bn-1 is the most significant bit (MSB).

  5. Definitions • nybble = 4 bits • byte = 8 bits • (short) word = 2 bytes = 16 bits • (double) word = 4 bytes = 32 bits • (long) word = 8 bytes = 64 bits • 1K (kilo or “kibi”) = 1,024 • 1M (mega or “mebi”) = (1K)*(1K) = 1,048,576 • 1G (giga or “gibi”) = (1K)*(1M) = 1,073,741,824

  6. Hexadecimal Number System • Base 16 • Sixteen Digits: 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F • Example: EF5616 • Positional Number System

  7. Binary Addition • Single Bit Addition Table 0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 = 10 Note “carry”

  8. Hex Addition • 4-bit Addition 4 + 4 = 8 4 + 8 = C 8 + 7 = F F + E = 1D Note “carry”

  9. Hex Digit Addition Table

  10. 1’s Complements • 1’s complement (or Ones’ Complement) • To calculate the 1’s complement of a binary number just “flip” each bit of the original binary number. • E.g. 0  1 , 1  0 • 01010100100  10101011011

  11. Why choose 2’s complement?

  12. 2’s Complements • 2’s complement • To calculate the 2’s complement just calculate the 1’s complement, then add 1. 01010100100  10101011011 + 1= 10101011100 • Handy Trick: Leave all of the least significant 0’s and first 1 unchanged, and then “flip” the bits for all other digits. • Eg: 01010100100 -> 10101011100

  13. Complements • Note the 2’s complement of the 2’s complement is just the original number N • EX: let N = 01010100100 • (2’s comp of N) = M = 10101011100 • (2’s comp of M) = 01010100100 = N

  14. Two’s Complement Representation for Signed Numbers • Let’s introduce a notation for negative digits: • For any digit d, define d = −d. • Notice that in binary, where d {0,1}, we have: • Two’s complement notation: • To encode a negative number, we implicitly negate the leftmost (most significant) bit: • E.g., 1000 = (−1)000 = −1·23 + 0·22 + 0·21 + 0·20 = −8

  15. Negating in Two’s Complement • Theorem: To negatea two’s complementnumber, just complement it and add 1. • Proof (for the case of 3-bit numbers XYZ):

  16. Signed Binary Numbers • Two methods: • First method: sign-magnitude • Use one bit to represent the sign • 0 = positive, 1 = negative • Remaining bits are used to represent the magnitude • Range - (2n-1 – 1) to 2n-1 - 1 where n=number of digits • Example: Let n=4: Range is –7 to 7 or • 1111 to 0111

  17. Signed Binary Numbers • Second method: Two’s-complement • Use the 2’s complement of N to represent -N • Note: MSB is 0 if positive and 1 if negative • Range - 2n-1 to 2n-1 -1 where n=number of digits • Example: Let n=4: Range is –8 to 7 Or 1000 to 0111

  18. Signed Numbers – 4-bit example Decimal 2’s comp Sign-Mag 7 0111 0111 6 0110 0110 5 0101 0101 4 0100 0100 3 0011 0011 2 0010 0010 1 0001 0001 0 0000 0000 Pos 0

  19. Signed Numbers-4 bit example Decimal 2’s comp Sign-Mag -8 1000 N/A -7 1001 1111 -6 1010 1110 -5 1011 1101 -4 1100 1100 -3 1101 1011 -2 1110 1010 -1 1111 1001 -0 0000 (= +0) 1000

  20. Signed Numbers-8 bit example

  21. Notes: • “Humans” normally use sign-magnitude representation for signed numbers • Eg: Positive numbers: +N or N • Negative numbers: -N • Computers generally use two’s-complement representation for signed numbers • First bit still indicates positive or negative. • If the number is negative, take 2’s complement to determine its magnitude • Or, just add up the values of bits at their positions, remembering that the first bit is implicitly negative.

  22. Examples • Let N=4: two’s-complement • What is the decimal equivalent of 01012 Since MSB is 0, number is positive 01012 = 4+1 = +510 • What is the decimal equivalent of 11012 = • Since MSB is one, number is negative • Must calculate its 2’s complement • 11012 = −(0010+1)= − 00112 or −310

  23. Very Important!!! – Unless otherwise stated, assume two’s-complement numbers for all problems, quizzes, HW’s, etc.The first digit will not necessarily be explicitly underlined.

  24. Arithmetic Subtraction • Borrow Method • This is the technique you learned in grade school • For binary numbers, we have 0 - 0 = 0 1 - 0 = 1 1 - 1 = 0 1 0 - 1 = 1 with a “borrow”

  25. Binary Subtraction • Note: • A – (+B) = A + (-B) • A – (-B) = A + (-(-B))= A + (+B) • In other words, we can “subtract” B from A by “adding” –B to A. • However, -B is just the 2’s complement of B, so to perform subtraction, we • 1. Calculate the 2’s complement of B • 2. Add A + (-B)

  26. Binary Subtraction - Example • Let n=4, A=01002 (410), and B=00102 (210) • Let’s find A+B, A-B and B-A 0 1 0 0 + 0 0 1 0  (4)10 A+B  (2)10 0 11 0 6

  27. Binary Subtraction - Example 0 1 0 0 - 0 0 1 0  (4)10 A-B  (2)10 0 1 0 0 + 1 1 1 0  (4)10 A+ (-B)  (-2)10 10 0 1 0 2 “Throw this bit” away since n=4

  28. Binary Subtraction - Example 0 0 1 0 - 0 1 0 0  (2)10 B-A  (4)10 0 0 1 0 + 1 1 0 0  (2)10 B + (-A)  (-4)10 1 1 1 0 -2 1 1 1 02 = - 0 0 1 02 = -210

  29. “16’s Complement” method • The 16’s complement of a 16 bit Hexadecimal number is just: =1000016 – N16 • Q: What is the decimal equivalent of B2CE16 ?

  30. 16’s Complement • Since sign bit is one, number is negative. Must calculate the 16’s complement to find magnitude. 1000016 – B2CE16 = ? • We have 10000 - B2CE

  31. 16’s Complement FFF10 - B2CE 4 D 3 2

  32. 16’s Complement • So, 1000016 – B2CE16 = 4D3216 = 4×4,096 + 13×256 + 3×16 + 2 = 19,76210 • Thus, B2CE16 (in signed-magnitude)represents -19,76210.

  33. Why does 2’s complement work?

  34. Sign Extension

  35. Sign Extension • Assume a signed binary system • Let A = 0101 (4 bits) and B = 010 (3 bits) • What is A+B? • To add these two values we need A and B to be of the same bit width. • Do we truncate A to 3 bits or add an additional bit to B?

  36. Sign Extension • A = 0101 and B=010 • Can’t truncate A! Why? • A: 0101 -> 101 • But 0101 <> 101 in a signed system • 0101 = +5 • 101 = -3

  37. Sign Extension • Must “sign extend” B, • so B becomes 010 -> 0010 • Note: Value of B remains the same So 0101 (5) +0010 (2) -------- 0111 (7) Sign bit is extended

  38. Sign Extension • What about negative numbers? • Let A=0101 and B=100 • Now B = 100  1100 Sign bit is extended 0101 (5) +1100 (-4) ------- 10001 (1) Throw away

  39. Why does sign extension work? • Note that: (−1) = 1 = 11 = 111 = 1111 = 111…1 • Thus, any number of leading 1’s is equivalent, so long as the leftmost one of them is implicitly negative. • Proof:111…1 = −(111…1) = = −(100…0 − 11…1) = −(1) • So, the combined value of any sequence of leading ones is always just −1 times the position value of the rightmost 1 in the sequence. 111…100…0 = (−1)·2n n

  40. Number Conversions

  41. Decimal to Binary Conversion Method I: Use repeated subtraction. Subtract largest power of 2, then next largest, etc. Powers of 2: 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2n Exponent: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 , n 20 21 22 23 24 25 26 27 28 29 210 2n

  42. Decimal to Binary Conversion Suppose x = 156410 Subtract 1024: 1564-1024 (210) = 540  n=10 or 1 in the (210)’s position Subtract 512: 540-512 (29) = 28  n=9 or 1 in the (29)’s position 28=256, 27=128, 26=64, 25=32 > 28, so we have 0 in all of these positions Subtract 16: 28-16 (24) = 12  n=4 or 1 in (24)’s position Subtract 8: 12 – 8 (23) = 4  n=3 or 1 in (23)’s position Subtract 4: 4 – 4 (22) = 0  n=2 or 1 in (22)’s position Thus: 156410 = (1 1 0 0 0 0 1 1 1 0 0)2

  43. Decimal to Binary Conversion Method II: Use repeated division by radix. 2 | 1564 782 R = 0 2|__24_ 12 R = 0 2|_____ 391 R = 0 2|_____ 6 R = 0  2|_____ 195 R = 1 2|_____ 3 R = 0 2|_____ 97 R = 1 2|_____ 1 R = 1 2|_____ 48 R = 1 2|_____ 0 R = 1 2|_____ 24 R = 0 Collect remainders in reverse order 1 1 0 0 0 0 1 1 1 0 0

  44. Binary to Hex Conversion • Divide binary number into 4-bit groups 0 1 1 0 0 0 0 1 1 1 0 0 Pad with 0’s If unsigned number 2. Substitute hex digit for each group Pad with sign bit if signed number 61C16

  45. Hexadecimal to Binary ConversionExample • Convert each hex digit to equivalent binary (1 E 9 C)16 (0001 1110 1001 1100)2

  46. Decimal to Hex Conversion Method II: Use repeated division by radix. 16 | 1564 97 R = 12 = C 16|_____ 6 R = 1  16|_____ 0 R = 6 N = 61C 16

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