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# Representation of Data

Representation of Data. Ma King Man. Reference. Text Book: Volume 2 Notes: Chapter 19. Road Map. Number System Base Conversion between binary, octal and hexadecimal system Base conversion from any base to decimal Base concersion from decimal to any base Fixed - point representation

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## Representation of Data

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1. Representation of Data Ma King Man

2. Reference • Text Book: Volume 2 • Notes: Chapter 19

3. Road Map • Number System • Base Conversion between binary, octal and hexadecimal system • Base conversion from any base to decimal • Base concersion from decimal to any base • Fixed - point representation • Floating - point number representation • Comparison between fixed - point representation and floating point representation

4. Lets start

5. Number System • Denary base • 0,1,2,3,4,5,6,7,8,9 • Binary base • 0,1 • Octal base • 0,1,2,3,4,5,6,7 • Hexadecimal base • 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F

6. Compare…

7. Base Conversion between binary, octal and hexadecimal system • Method • 3 binary digits = 1 octal digit • 4 binary digits = 1 hexadecimal digit

8. Example 1 • 1110101(ii) • =111 0101(ii) • =75(16)

9. Example 2 • 1110101(ii) • = 1 110 101(ii) • = 165(8)

10. Example 3 • 1A46(16) • = 1 1010 0100 0110(ii)

11. Example 4 • 732(8) • = 111 011 010(ii)

12. Base conversion from any base to decimal • Method: • dndn-1dn-2……d1d0d-1d-2……d-m+1d-m • dnbn+dn-1bn-1+……+d1b+d0+d-1b-1+……+d-mb-m(10)

13. Examples • E.g.1 • 1101.11(ii) • = 1x23 + 1x22 + 0x21 + 1x20 + 1x2-1 + 1x2-2 • = 13.75 • E.g.2 • 1234.56(8) • = 1x83 + 2x82 + 3x81 + 4x80 + 5x8-1 + 6x8-2 • = 668.71875 • E.g.3 • 11.8(16) • = 1x161 + 1x160 + 8x16-1 • = 17.5

14. Base conversion from decimal to any base Step 1: Convert the integral part: 234 = 11101010(ii) Step 2: Convert the fraction part 0.25 = 0.01(ii) Step 3: Therefore, 234.25 = 11101010.01(ii)

15. Lets Try…… • E.g.2 • 23.0625 • = 27.04(8) • E.g.3 • 13.125 • = 1101.001(ii)

16. Fixed – point representation • Usually the point is fixed at the right to the L.S.B. – Least significant bit – i.e. integer. • Assume 8 – bit word storage is used (8 – bit per unit) • Bit 7 is the sign bit of the number • “0” = positive • “1” = negative

17. 3 methods of fixed-point representation • Sign-and-magnitude representation • One’s complement representation (optional) • Two’s complement representation

18. Sign-and-magnitude • (i) The leftmost bit indicates the sign. • (ii) The remaining bits give the magnitude of the number. • E.g. +10101(ii) is stored as 00010101 (+21 in dec.) • And - 10101(ii) is stored as 10010101 (-21 in dec.) • Range: For 8 – bit word: • The smallest number is 11111111 (-127) • The largest number is 011111111 (+127) • Disadvantages: • 2 representations for zero: 00000000, 10000000 • Requires extra circuit to perform addition and subtraction.

19. One’s complement representation • (i) Add zeros to the left of the binary numbers until the number is fitted the given length. • (ii) For a positive integer, the binary pattern is left unchanged. • (iii) For a negative integer, 0 is replaced by 1 and 1 by 0. • E.g. +10101(ii) is stored as 00010101 • And - 10101(ii) is stored as 11101010 • Range: For 8 – bit word: • The smallest number is 10000000 (- 127) • The largest number is 01111111 (+127) • Disadvantages: • Same as sign-and-magnitude representation • Zero: 00000000, 11111111

20. Two’s complement representation (Most commonly used) • (i) and (ii) are same as 1’s complement method. • For a negative integer, add 1 to the rightmost bit of it’s 1’s complement. • E.g. +10101(ii) is stored as 00010101 • And –10101(ii) is stored as 11101011 • Range: For 8 – bit word: • The smallest number is 10000000 (-128) • The largest number is 01111111 (+127) • Advantages: • It has wider range than both sign-and-magnitude and 1’s complement, and no ambiguity for zero. • Subtraction can be done by addition, no extra circuit is required.

21. Example 1 Assume 8 – bit word and 2’s complement notation is used. E.g. 23 + 17 = 10111 + 10001 (binary no.) = 00010111 + 00010001 (internal representation) = 00101000 (internal representation) = + 101000 (binary no.) = 40

22. Example 2 E.g. 23 – 17 = 10111 – 10001 (binary no.) = 10111 + (-10001) (internal representation) = 00010111 + 11101111 (internal representation) = 00000110 (internal representation) = 110(ii) (binary no.) = + 6

23. Example 3 E.g. 17 – 23 = 10001 – 10111 = 10001 + (-10111) = 00010001 + 11101001 = 11111010 = -110(ii) = -6

24. Special cases Overflow Error Carry discard Overflow Error

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