Floating Arithmetic

Question

a. Consider a fixed-point representation using decimal digits, in which the implied radix point can be in any position (e.g. to the right of the least significant digit, to the right of the most significant digit, and so on). How many decimal digits are needed to represent the approximations of both Planck’s constant and Avogadro’s number the implied radix point must be in the same position for both numbers?

b. Now consider a decimal floating-point format with the exponent stored in a biased representation with a bias of 50. A normalized representation is assumed. How many decimal digits are needed to represent these constants in this floating- point format?

Can you please check this?

I got answer A as 53 bits

While for answer B

Since they have given Biased for decimal = 50

Then Planck Constant which is 6.63*10^-27 for this Biased Exponent Field= -27+50= 23

Similarly for Avogadro numbers which is 6.02*10²³

For this Biased Exponent Field= 23+50 = 73

Now, they asked us to find number of digit in decimal only to represent this number

For Planck Constant

Sign bit	Exponent Field	Significand
0	23	63

Similarly for Avogadro

Sign bit	Exponent Field	Significand
0	73	02

Here 6 is a hidden bit in both cases

Therefore to represent this number both we required 5 digit

Is it right?

Answer 1

a. Scaling factor needed is 10^-29 to represent Plank constant. Now Plank constant is 663 and Avogadro number is 602..0 (50 0'2). So, we need 50 + 3 = 53 decimal digits. 

b. "6 is a hidden bit in both cases" Because of normalization? In binary normalized representation 1 on the left of decimal point is taken as implied. But in decimal this "digit" can be any from 1-9. So, we can't just imply it. 

Normalization just means that the leading significant digit is non-zero. 1 being taken implicitly in binary is something done in addition to normalized representation because '1' is the only possibility there. 

Comments

okay got thank you :D It mean when i would have number in its required format . we cant again decode into its real value , because for a hidden bit there would be choice of (0-9) right ? So since power of planck is raised to -23 so we choose 0.(  something )and for avogadro it is raised to 23 (positive value ) so can be express in 1 .(something ) right ?

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Actually not 0-9 but 1-9 because of normalization. Still when decoded we have no idea which digit from 1-9, and that is the problem.

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Yeah that only 1-9 sorry . I mean that only but :P