Get max, min and precision of a fi

What is the easiest way to find the maximum or minimum value that a fi can take?

What is the easiest way to find the precision of a fi?

Answer 1

[Jan]

a  = fi(pi, true, 16, 12);
xi = intmax(a)

This is code taken from https://www.mathworks.com/help/fixedpoint/ref/intmax.html , which I found as first link by asking an internet search engine for "Matlab fi maximum value".

[EDITED] Maybe you want https://www.mathworks.com/help/fixedpoint/ref/realmax.html :

a  = fi(pi, true, 16, 12);
xd = realmax(a)

[Justin]

But the maximum that a fi(#, true, 16, 12) can take is:

(2^(16-1) - 1) / (2^12) = 7.9998.

The code above gives x = 32767

i was hoping for something like:

>> a = fi(pi, true, 16, 12);
>> x = a.get_max()
x = 7.9998

[Jan]

I cannot test this, because I do not have the FI toolbox. But what about realmax?

[Justin]

Yes, that looks good. found everything I need here:https://uk.mathworks.com/help/fixedpoint/data-type-operators-and-tools.html

and here: https://uk.mathworks.com/help/fixedpoint/data-analysis.html

upperbound and lowerbound (or range) and realmin get me what I want. Thanks for your help navigating the documentation!

Answer 2

[Justin]

I have answered my own question by creating the following function:

function info = fiInfo(inFi)
    if inFi.Signed 
        info.max = (2^(inFi.WordLength - 1) - 1) / 2^inFi.FractionLength;
        info.min = -2^((inFi.WordLength - 1) - inFi.FractionLength);
        info.precision = 2^-inFi.FractionLength;
    else
        info.max = (2^inFi.WordLength - 1) / 2^inFi.FractionLength;
        info.min = 0;
        info.precision = 2^-inFi.FractionLength;
    end
end

resulting in

>> a = fi(pi, true, 16, 12);
>> fiInfo(a)

ans = 

  struct with fields:

          max: 7.9998
          min: -8
    precision: 2.4414e-04

or

>> a = fi(pi, false, 16, 12);
>> fiInfo(a)

ans = 

  struct with fields:

          max: 15.9998
          min: 0
    precision: 2.4414e-04

This seems to be a lot of faffing to get at some basic info. Surely there is a better way?

Answer 3

[Harry]

% A fixed-point object with no value, 18-bit word length, and 16-bit fraction length

a = fi([],1,18, 16)

a = 

[]

          DataTypeMode: Fixed-point: binary point scaling
            Signedness: Signed
            WordLength: 18
        FractionLength: 16

% The easiest way to find the maximum or minimum value that a fi can represent is

range(a)

ans = 
   -2.0000    2.0000

          DataTypeMode: Fixed-point: binary point scaling
            Signedness: Signed
            WordLength: 18
        FractionLength: 16

% The easiest way to find the precision of a fi is

eps(a)

ans = 
   1.5259e-05

          DataTypeMode: Fixed-point: binary point scaling
            Signedness: Signed
            WordLength: 18
        FractionLength: 16