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Altera_Forum's avatar
Altera_Forum
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11 years ago

fixed point precision

Peace!

I want ask about the number of bits shoud I use in an fixed_point number to obtein the good precision, precision of distance eclidean.

some detail:

I want to determine an vector of foor element VECT[MOY1,MOY2,MOY3,MOY4], each element is the moyen of one quart of matrix of (64*64),

the matrix' element value varies (0-255)'''it's pixel valor'''

--> the result of division (MOY1=SUM/Nbits) has 8 bits in the integer part;

Question: how many bits of fractional part shout I use to get the good result (result of euclidean distance between two vector VECT1[MOY11,MOY12,MOY13,MOY14] and VECT[MOY21,MOY22,MOY23,MOY24]).. thank you....peace:

4 Replies

  • Altera_Forum's avatar
    Altera_Forum
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    If you sum N digital values you generally need log2(N) (approximate to the next integer) more bits to store the result. Otherwise you possibly overflow the sum or lose information.

    The same applies to the average of N values: you need log2(N) more bits if you want to have the full precision; consider, for example, that any division by 2 introduces one more decimal bit.
  • Altera_Forum's avatar
    Altera_Forum
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    thank you Cris72!

    I divide 18 bits by "10000000000" //1024//, I want to keep a big part of information

    consider that the integer part of the division result id coded in 8 bit (varies from 0 to 255),

    how many of bits shoud Allocated to the fractional part so I keep big part of information.
  • Altera_Forum's avatar
    Altera_Forum
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    your division just needs discarding 10 lsb bits from 18 so you are left with 8 bits. If you want to keep fractional bits then just don't divide as the 10 lsbs is the fractional bits. so I don't really get it.

  • Altera_Forum's avatar
    Altera_Forum
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    so 10 bits in fractional part assure the big part of information

    ..thank you Kaz !