Lecture 8 Notes

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Unformatted text preview: per number •  x = (1+f)2^e –  f is the fracJonal part of 1+f, stored using base 2 rep. •  called the the “fracJon” or manJssa •  0 <= f < 1, 52 bits allocated for f => –  0 <= 2^52*f < 2^52 –  e is an integer exponent, stored using base 2 rep. •  11 bits allocated for e, allows for - 1022 <= e < 1024 –  s is the sign of x (just assume x is posiJve, and apply s at the end) •  s=- 1, or s=1, 1 bit allocated 7 2/20/14 “IEEE 754” •  x = (1+f)2^e –  f is the fracJonal part of 1+f, base 2 rep. –  e is an integer exponent Note: f 52 bits, base 2 rep means only certain values are possible for f •  Some Examples: What are f and e –  for x = 2? •  1x2^1 ó༏ f = 0, e = 1 –  for x = 3? •  1.5x2^1 ó༏ f = .5, e = 1 Answer Time x = (1+f)2^e –  f is the fracJonal part of 1+f, base 2 rep. –  e is an integer exponent What are f and e for x=5 and x=8? –  for x = 5 •  1.25x2^2 ó༏ f = .25, e = 2 –  for x = 8 •  1x2^3 ó༏ f = 0, e = 3 eps •  x = (1+f)2^e •  For x = 2, what is the next largest floaJng point number? x = (1.0)*2^1 => (1+2^- 52)*2^1 = (1 + eps)*2^1 = x + eps*2^1 = 2 + eps*2^1 = 2 + 2*eps is next largest floaJng point number Quiz Time x = (1+f)2^e –  f is the fracJonal part of 1+f, base 2 rep. –  e is an integer exponent What are f and e for x=5 and x=8? A)  f=.5, e=0 and f=2, e=2 B)  f=.25, e=2 and f=.5, e=4 C)  f=.25, e=2 and f=0, e=3 D)  None of the above eps •  x = (1+f)2^e •  For x = 1, what is next largest floaJng point number? x = (1.0)*2^0 => (1+2^- 52)*2^0 = 1 + (2^- 52) = 1 + eps! is next floaJng point number •  Why 2^- 52? •  2^- 52 is referred to as “machine epsilon” –  It is the difference between 1 and the next largest floaJng number that can be represented –  It is available as eps in matlab eps •  x = (1+f)2^e •  For x = 3, what is the next...
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