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I am working with a library that needs input in a Fixed Point notation. I’d like to figure out a way to convert the floating point results into fixed point representation.

The fixed point length is 16 bit. Numbers can be represented with a variable number of bits and I can specify the format of the bits. For instance, I can represent a number as Signed[3.13], upto 3 bits are used to represent the integer portion, and 13 bits are used for the fraction (resolution of 2^-13).

Q1. What is the best method to convert Mathematica floating point output into a 16 bit number. Ideally, I’d like a function like:

f[x_, signbits_, integerbits_, fractionbits_]

Q2. Given a set of numbers, what is the best method to determine the ideal number of integer bits and fractional bits used for the representation in order to minimize truncation errors

data = {0.0000618365, 0.0000701533, -0.0000747471, 0.0000595436, 0.0000705533, \
0.0000728675, 0.0000711056, 0.0000684559, 0.0000753624, -0.0000557638}
J. M.'s missing motivation
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Pam
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  • have you had a look at ComputerNumber? – Stefan Feb 27 '14 at 12:22
  • I’ve actually figured this out and was going to post a detailed post when I find some time this weekend. – Pam Feb 27 '14 at 15:28
  • Pam, I notice that you never posted the answer you intended to. I have trouble returning to such things myself, but in case you merely forgot here's a reminder. :-) – Mr.Wizard Jul 03 '14 at 19:25
  • Thanks. Am away for the long weekend... Will post next week! – Pam Jul 03 '14 at 23:40

1 Answers1

6

Answering my own question from a while ago.

Turns out the easiest method is using MMA’s built-in Computer Arithmetic package

<< ComputerArithmetic`

(*Set Math Parameters*)

SetArithmetic[6, 10, ExponentRange -> {-20, 20}];

fpConvert[x_, integerbits_, fractionbits_] := ComputerNumber[IntegerPart[x] + Round[FractionalPart[x], 2^-fractionbits]];

Let’s test the output:

N[fpConvert[Pi, 3, 2], {8, 8}]
N[fpConvert[Pi, 3, 4], {8, 8}]
N[fpConvert[Pi, 3, 8], {8, 8}]
N[fpConvert[Pi, 3, 13], {8, 8}]

Works fine.

Mr.Wizard
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Pam
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