浮点转换-二进制->十进制

Question

Here's the number I'm working on

1 01110 001 = ____
1 sign bit, 5 exp bits, 3 fraction bits
bias = 15

Here's my current process, hopefully you can tell me where I'm missing something

1. Convert binary exponent to decimal
   01110 = 14
2. Subtract bias
   14 - 15 = -1
3. Multiply fraction bits by result
   0.001 * 2^-1 = 0.0001
4. Convert to decimal
   .0001 = 1/16

The sign bit is 1 so my result is -1/16, however the given answer is -9/16. Would anyone mind explaining where the extra 8 in the fraction is coming from?

Answer 1

You seem to have the correct concept, including an understanding of the excess-N representation, but you're missing a crucial point.

The 3 bits used to encode the fractional part of the magnitude are 001 , but there is an implicit 1. preceding the fraction bits, so the full magnitude is actually 1.001 , which can be represented as an improper fraction as 1+1/8 => 9/8 .

2^(-1) is the same as 1/(2^1) , or 1/2 .

9/8 * 1/2 = 9/16 . Take the sign bit into account, and you arrive at the answer -9/16 .

Answer 2

For normalized floating point representation, the Mantissa (fractional bits) = 1 + f. This is sometimes called an implied leading 1 representation. This is a trick for getting an additional bit of precision for free since we can always adjust the exponent E so that significant M is in the range 1<=M < 2 ...

You are almost correct but must take into consideration the implied 1. If it is denormalized (meaning the exponent bits are all 0s) you do not add an implied 1.

I would solve this problem as such...

1  01110  001

bias = 2^(k-1) -1 =        14

Exponent = e - bias       

14 - 15 = -1

1. Take the fractional bits ->> 001
2. Add the implied 1 ->> 1.001
3. Shift it by the exponent, which is -1. Becomes .1001
4. Count up the values, 1(1/2) + 0(1/4) + 0(1/8) + 1(1/16) = 9/16
5. With the a negative sign bit it becomes -9/16

hope that helps!