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Digital Logic Design

Digital Logic Design. EGRE 254 Number Systems and Codes 1/12/09. Positional Number Systems. Numbers are commonly represented in the base 10 positional number system. For example. In general (see page 26).

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Digital Logic Design

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  1. Digital Logic Design EGRE 254 Number Systems and Codes 1/12/09

  2. Positional Number Systems • Numbers are commonly represented in the base 10 positional number system. • For example

  3. In general (see page 26) • A number D of radix or base r with p digits to the left of the radix point and n digits to the right of the radix point can be expressed as: di – in base 10 di – in base r

  4. Examples • 156.78 = (1x102 + 5x101 + 6x100 + 7x10-1 )8 = (1x82 + 5x81 + 6x80 + 7x8-1 )10 = (64 + 40 + 6 + .875)10 = 110.87510 • 559 = (5x101 + 5x100)9 = (5x91 + 5x90)10 = (45 + 5)10 = 5010 • 1212 = (1x121 + 2x120)10 = (12 + 2)10 = 1410 • 101100.112 =

  5. Examples • Does r10 = 10r? Prove it! • Does rb = br? Prove it! • 111b = 13310 Find b. • (41/3)b = 13b Find b. • (33/3)b = 11b Find b.

  6. Example • (5D4.A2)16 • = (5x102+Dx101+4+Ax10-1+2x10-2)16 = (5x162+13x161+4+10x16-1+2x16-2)10 = (1280+208+4+.625+.0078125)10 = (1492.6328125)10

  7. How do we convert from base 10 to base r? • We could use previous techniques, but we would have to do arithmetic in base r. • Not desirable. • Consider two cases. • Integer number. • Fractional number.

  8. Base 10 to base r (integer case)

  9. Example – Integer to Hex 1492 = (d3 d2 d1 d0)16 = d3x163+ d2x162+ d1x16+ d0 1492/16 = 93+4/16 = d3x162+ d2x16+ d1+ d0/16 d0 = 4 93 = d3x162+ d2x16+ d1 93/16 = 5 +13/16 = d3x16+ d2+ d1/16 d1 = 1310 = D16 5 = d3x16+ d2 5/16 = 0 + 5/16 = d3 + d2/16, d3 = 0, d2 = 5 149210 = 5D416

  10. In practice • You may get mixed up on direction to read the answer. • 5D416 or 4D516? • Note that 5D4 = 0…05D4 • 4D5 not same as 4D50…0

  11. Example fraction to hex (.6328125)10 = (.d-1 d-2 d-3…)16 = (d-1x16-1+d-2x16-2+d-3x16-3+…)10 How do we find d-1? Multiply by 16. 16 x .6328125 = 10.125 = d-1+d-2x16-1+d-3x16-2+… d-1 = 10 and .125 = d-2x16-1+d-3x16-2+… Multiply by 16. Then,.125 x 16 = 2.00 = d-2+d-3x16-1+… d-2 = 2 and d-n = 0 for n > 2. Therefore, (.6328125)10 = (.A2)16

  12. Example • For a number containing both an integer and a fractional part. Compute the two parts separately and combine. • 1492.632812510 = 5D4.A216

  13. Special Case converting between base r and rn • Convert 132.68 to binary. • Could convert to decimal and then binary. • Easier way.

  14. Examples • Convert (11010.11)2 to octal, and to hexadecimal. (11010.11)2 = (11,010.110)2 = 32.68 (11010.11)2 = (1,1010.1100)2 =1A.C16 • Convert (121.121)3 to base 9 = 32. • Convert (1234.567)8 to hexadecimal. • Hint 8 = 23 and 16 = 24.

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