A balance has a range of 120 g and a resolution of 01mg How

A balance has a range of 120 g and a resolution of 0.1mg. How many bits are required in the number register and the number-current converter for this balance?

Solution

Number of Levels = Range/Resolution = 120g/0.1 mg = 120/0.1 * 10^3 = 1.2 * 10^(6)

Bits required in the register will be given

2^n = 1.2 * 10^(6)

taking log both sides we get

nlog(2) = log(1200000)

n = log(1200000)/log(2) = 20.194

Hence number of bits required will be 21

The converter will be given by

If a number has 1 at nth place, then the mass corresponding to that will be equal to 1mg * 2^n