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Introduction to Computer Architecture

Introduction to Computer Architecture. Overview. Number Systems And Binary Arithmetic. Review – Base 10 Numbers (Decimal). Base-10 The decimal number system is based on power of the base 10. For example, for the number 1259 the 9 is in the 10^0 column - 1s column

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Introduction to Computer Architecture

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  1. Introduction to Computer Architecture Fall 2009

  2. Overview Number Systems And Binary Arithmetic Fall 2009

  3. Review – Base 10 Numbers (Decimal) Base-10 The decimal number system is based on power of the base 10. For example, for the number 1259 the 9 is in the 10^0 column - 1s column the 5 is in the 10^1 column - 10s column the 2 is in the 10^2 column - 100s column the 1 is in the 10^3 column - 1000s column 1259 is 9 X 1 = 9 + 5 X 10 = 50 + 2 X 100 = 200 + 1 X 1000 = 1000 ----- 1259 Fall 2009

  4. Introducing Base 2 Number System (Binary) Base-2 (Binary) The Binary number system uses the same mechanism and concept however, the base is 2 The place values for binary are based on powers of the base 2: … 2^7 2^6 2^5 2^4 2^3 2^2 2^1 2^0 128 64 32 16 8 4 2 1 Fall 2009

  5. 5-bit Binary Number System X4 X3 X2 X1 X0 Fall 2009

  6. Different Number Systems • Base-10 (Decimal) • Example = 659 • Base-2 (Binary) • Example = 1101 • Base-16 (Hex) • Example = AE • Base-8 (Octal) • Example = 73 Fall 2009

  7. Converting Number Systems The binary number 10110011 can be converted to a decimal number 1 X 1 = 1 (right most bit or position) 1 X 2 = 2 0 X 4 = 0 0 X 8 = 0 1 X 16 = 16 1 X 32 = 32 0 X 64 = 0 1 X 128 = 128 (left most bit or position) ------ 179 in decimal Fall 2009

  8. Converting Number Systems To convert from decimal to binary requires a different method called the division/remainder method. The idea is to repeatedly divide the decimal number and resulting quotients by 2. The answer will be the remainders. Example: convert 155 to binary (Start from the top and work down) 155/2 Q = 77, R = 1 (Start) 77/2 Q = 38, R = 1 38/2 Q = 19, R = 0 19/2 Q = 9, R = 1 9/2 Q = 4, R = 1 4/2 Q = 2, R = 0 2/2 Q = 1, R = 0 1/2 Q = 0, R = 1 (Stop) Answer is 10011011. Be careful to place the digits in the correct order. Fall 2009

  9. Converting Number Systems - Review Check the answer (10011011) : 1 X 1 = 1 1 X 2 = 2 0 X 4 = 0 1 X 8 = 8 1 X 16 = 16 0 X 32 = 0 0 X 64 = 0 1 X 128 = 128 ----- 155 Fall 2009

  10. Base-16 (Hex) The hexadecimal number system is based 16, and uses the same mechanisms and conversion routines we have already examined. The place values for hexadecimal are based on powers of the base 16 The digits for 10-15 are the letters A - F (A is 10, …….., F is 15) …….. 16^3 16^2 16^1 16^0 4096 256 16 1  The binary number 10110011 can be converted to hexadecimal by grouping bits into groups of 4 bits: 1011 0011 B 3 Fall 2009

  11. Addition & Subtraction Fall 2009

  12. Division Fall 2009

  13. Binary Arithmetic Fall 2009

  14. Review – Basic Binary Fall 2009

  15. Computers can not SUBTRACT Therefore, need approaches to represent “signed” numbers Fall 2009

  16. Signed Numbers • Sign and magnitude • 1’s complement • 2’s complement The left-most bit is: • 0 for positive numbers • 1 for negative numbers Fall 2009

  17. Changing only b3 makes +/- Complementing makes +/- B V alues represented Complementing plus 1 makes +/- Sign and b b b b magnitude 1' s complement 2' s complement 3 2 1 0 + 7 + 7 + 7 0 1 1 1 + 6 + 6 + 6 0 1 1 0 + 5 + 5 + 5 0 1 0 1 + 4 + 4 + 4 0 1 0 0 + 3 + 3 + 3 0 0 1 1 + 2 + 2 + 2 0 0 1 0 + 1 + 1 + 1 0 0 0 1 + 0 + 0 + 0 0 0 0 0 - 0 - 7 - 8 1 0 0 0 - 1 - 6 - 7 1 0 0 1 - 2 - 5 - 6 1 0 1 0 - 3 - 4 - 5 1 0 1 1 - 4 - 3 - 4 1 1 0 0 - 5 - 2 - 3 1 1 0 1 - 6 - 1 - 2 1 1 1 0 - 7 - 0 - 1 1 1 1 1 More used and more efficient Figure 2.1. Binary, signed-integer representations. Fall 2009

  18. One’s Complement - Binary Addition • 1010 (neg 5) • +0010 (pos 2) • 1100 (neg 3) • 1101 (neg 2) • +0111 (pos 7) • 10100 (overflow – add the 1 back) • 0101 (pos 5) • Recall complement • 0011 Fall 2009

  19. Possibly Will NOT be able to use a Calculator for number system part of Exam 3 2ND REMINDER MUST LEARN THE CONCEPTS Fall 2009

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