Computer Systems

1. Computer Systems Data Representation andComputer Arithmetic

2. Data Representation • Binary codes used to represent numbers • Conversion between number systems • Binary  decimal • Hexadecimal  decimal • Between arbitrary number systems • Representation of positive/negative numbers • Floating point representation

3. 11001010 00001101 Bit Byte Word Bit, Byte and Word • Bit is the smallest quantity that can be handled by computer: 1 or 0. • Byte is a group of 8 bits. • Word is the basic unit of data that can be operated by computer: e.g., 16 bits. Some architectures have 8, 32, or 64-bit words

4. Content of Word • Bit Pattern • May represent many things • Actual meaning of a particular bit pattern is given by the programmer • Computer itself cannot determine the meaning of the word • Classic question: Can you tell the meaning of a word picked randomly from the memory?

5. Instruction: (op-code) A single word defines an action that is to be performed by CPU • Numeric Quantity • Character: A to Z, a to z, 0 to 9, *, -, +, etc. • ASCII Code: - Representation of a character by 7 bits. - The ASCII for W is 101 01112, or 5716

6. ASCII Code

7. Pixel (1) Pixel (Picture Element) • The smallest unit to construct a picture • 1bit/pixel for black-and-white pictures, >1 bits/pixel for gray scale or color pictures, e.g. 24bits/pixel. • A letter is represented by a group of pixels on computer screen.

8. Positional Notation • Weight is associated with the location within a number. • The 9 in 95 has weight 10 A number N in base b is represented by

9. Example:

10. Number Bases Decimal (b=10): {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} Binary (b=2): {0, 1} Octal (b=8): {0, 1, 2, 3, 4, 5, 6, 7} • Hexadecimal (b=16): {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F}

11. How many binary bits are needed to represent an n-digits decimal? The value of an m-bit binary is up to The value of an n-digit decimal is up to so OR

12. How many bits are needed to represent an n-digits octal?How many bits are needed to represent an n-digits hexadecimal? Questions

13. Conversion of Integer- Decimal to Binary Divide a decimal successively by 2, record the remainder, until the result of the division is 0. 67/2 = 33 Rem.=1 33/2 = 16 Rem.=1 16/2 = 8 Rem.=0 8/2 = 4 Rem.=0 4/2 = 2 Rem.=0 2/2 = 1 Rem.=0 1/2 = 0 Rem.=1

14. Conversion of Integer- Decimal to Hexadecimal Divide a decimal successively by 16, record the remainder, until the result is 0 432/16 = 27 Rem.=0 27/16 = 1 Rem.=B 1/16 = 0 Rem.=1

15. Conversion of integer - Binary to Decimal Conversion of integer - Hexadecimal to Decimal

16. 0 1 0 1 0 1 1 1 2 B 7 5 7 0010 1011 0111 2 5 7 0 1 0 1 0 1 1 1 1 2 010 101 111 7 Conversion of integer - Hexadecimal Binary Conversion of integer: - Octal Binary

17. 0.687510 = 0.10112 Question: Convert 0.625? 0.110? Conversion of Fraction - Decimal to Binary The fraction is repeatedly multiplied by 2, the integer part is stripped and recorded. 0.687510 * 2 =1.375 1 0.37510 * 2 = 0.75 0 0.7510 * 2 = 1.5 1 0.510 * 2 = 1.0 1 0.010 * 2 = 0.0 end

18. Convert 0.6D5? Convert 0.1100101011 Conversion of Fraction - Binary Hexadecimal 0.1010110012 = 0.1010 1100 10002 = 0.AC816 0.7BC16 = 0.0111 1011 11002 = 0.01111011112

19. Binary Coded Decimal (BCD) • A decimal digit is coded into 4 bits  1 byte can store 2 digits • Example: 1942 is encoded as 0001 1001 0100 0010 (2 bytes) • Disadvantages • Complex arithmetic • Inefficient use of storage • Advantage: • Can represent “real” number DecimalBCD 0 0000 1 0001 2 0010 3 0011 4 0100 5 0101 6 0110 7 0111 8 1000 9 1001

20. Binary Addition 0 0 1 1 0 1 1 1 + 0 1 0 1 0 1 1 0 1 1 1 1 1 Carry 1 0 0 0 1 1 0 1 Binary Subtraction ? use 2’s complementary arithmetic (later … )

21. Signed Integer Representations Sign-and-Magnitude 1’s Complement 2’s Complement

22. Problem: 00000000 = +0, 10000000 = -0 Sign-and-Magnitude Representation • Use the most significant bit (MSB) to indicate the sign of the number. • The MSB is 0 for positive, 1 for negative. • 8 bits represent -12710 to 12710 • Examples: 00101100 for +4410 10101100 for -4410

23. Problem: 00000000 = +0, 11111111 = -0 1’s Complement • Just flip the bits! • The MSB is 0 for positive, 1 for negative. • 8 bits represent -12710 to 12710 • Examples: 00101100 for +4410 11010011 for -4410

24. 2’s Complement The 2’s complement for N is Example: Using 5 bits (n = 5), if N = 7, then 0 1 1 0 0 + 1 1 0 0 1 1) 0 0 1 0 1 12 -7 5 2’s complement = 1’s complement + 1 7 0 0 1 1 1 1’s complement 1 1 0 0 0 + 1 2’s complement 1 1 0 0 1 = -7

25. 2’s comp. -34? 2’s Complement (con.) • To form the two’s complement of a number, simply invert the bits (1 to0, 0 to 1), and add 1  1’s complement + 1 1310 = 0 0 0 0 1 1 0 1 -1310 = 1 1 1 1 0 0 1 0 + 1 = 1 1 1 1 0 0 1 1

26. Properties of 2’s Complement One unique 0 MSB = 0, positive number MSB = 1, negative number The range is For 5 bits, the range is -1610 (10000) to +1510 (01111) The 2’s complement of the complement of X is X itself (Prove: )

27. X -Y 310 0 1 0 0 1 + 1 1 0 1 0 1) 0 0 0 1 1 - X - Y -1510 1 0 1 1 1 + 1 1 0 1 0 1) 1 0 0 0 1 2’s Complementary Arithmetic • Subtraction is performed using addition A - B = A + (-B) X = 910 = 010012, Y = 610 = 001102 -X = -910 = 101112, -Y = -610 = 110102

28. Binary Adder/Subtractor

29. 0 1 1 0 0 + 0 1 1 0 1 1 1 0 0 1 1 0 1 0 0 + 1 0 0 1 1 1) 0 0 1 1 1 12 +13 25 -12 -13 -25 Arithmetic Overflow • The overflow happens when positive + positive  negative negative + negative  positive If are MSBs, then the overflow bit v is expressed as

30. Arithmetic Overflow In practice, • Case I: A and B are + an-1 = __ bn-1 = __ cn = __ cn-1 = __ V = _____ • Case II: A and B are – an-1 = __ bn-1 = __ cn = __ cn-1 = __ V = _____ Most significant stage of full adder

31. What if we want to represent ? We need at least m > 3.3*n = 3.3*20 = 66 bits ! Fixed-point Arithmetic 3.62510 0011.10102 00111010 + 6.510 0110.10002+ 01101000 10.12510 10100010 The fraction point is added 10100010 -> 1010.00102

32. Floating Point Numbers • Scientific Notation a is mantissa, r is radix, e is exponent is for binary

33. IEEE Floating Point Format S: Sign bit, 1 bit E: Exponent, 8 bits B: Bias, 8 bits, 12710=0111 1111 F: mantissa

34. If we have 8 bits for exponent, to represent -5, we add 12710 (0111 1111) to -5, the result is 12210=0111 10102. IEEE Floating Point - Example Normalization, e.g., Use sign and magnitude representation for signed mantissa. Use biased exponent, e.g. in excess 127

35. Represent -2345.125 in 32 bits ? 1.510 in IEEE format? IEEE Floating Point - Example -2345.125 = -100100101001.0012 1) Normalization: 2) Negative mantissa, so S = 1 3) Exponent: E = 11 + 127 = 13810 = 100010102 4) Mantissa F = 00100101001 0010…000 1 10001010 00100101001001000…0

36. IEEE Floating Point Format – Special Cases • Zero: exponent and fraction all 0s • Denormalized: exponent all 0s, fraction non-zero for single precision: • Infinity: exponents all 1s, fraction all 0s sign bit determines +infinity or -infinity • Not a Number (NaN): exponents all 1s, fraction non-zero • A good reference is available at http://research.microsoft.com/~hollasch/cgindex/coding/ieeefloat.html

37. Bit Patterns and Logical Operations AND OR NOT Exclusive OR They are bit-wise operations.

38. AND Operation • x AND y is true, if and only if both bits x and y are true. A = 1 1 0 0 1 0 1 1 B = 0 1 1 0 1 1 0 1 C=A AND B = 0 1 0 0 1 0 0 1 OR Operation • x OR y is true, if either bits x or y is true. A = 1 1 0 0 1 0 1 1 B = 0 1 1 0 1 1 0 1 C=A OR B = 1 1 1 0 1 1 1 1

39. NOT Operation • A 1 becomes 0, and a 0 becomes 1. A = 1 1 0 0 1 0 1 1 C = NOT A = 0 0 1 1 0 1 0 0 EOR (Exclusive OR) Operation • It is true, if and only if just one of inputs is true. A = 1 1 0 0 1 0 1 1 B = 0 1 1 0 1 1 0 1 C = A EOR B = 1 0 1 0 0 1 1 0