Fixed Point Toolbox

I'm trying to learn the new fixed point toolbox for Matlab V7.0 R14.
I have the following Section of floating point code...

y = randn(1,2)/64;
X = zeros(2,1);
X(1) = y(1) + y(2); 
X(2) = (y(1) - y(2))/sqrt(2);

(which is a two point butterfly in a DCT if you are interested). What
do i do to convert this to fixed point? Here is my (not very good)
attempt.
*******************************************************************
% first generate 18 bit random numbers for y 
y = randn(1,2)/64;                       
y = fi(y,1,18,17,'RoundMode','floor');  

% then define fixed point representation of X
F = fimath('ProductMode','SpecifyPrecision',...
    'SumMode','SpecifyPrecision',...
    'ProductWordLength',32,...
    'ProductFractionLength',16,...
    'SumWordLength',32,...
    'SumFractionLength',16,...
    'RoundMode','floor');

% do we need to initialize X? (I used the zeros function in floating
point)
    X(1) = F.add(x(1),x(2));
    X(2) = F.sub(x(1),x(2));

% eh, no idea how to divide, because divide is defined for
numerictypes
    X(2) = F.divide(X(2),fi(sqrt(2));???
***********************************************************************
Please help?

Hi,
Your fixed-point implementation could look something like this:

% Create your fimath (and attach it to y too)
F = fimath('ProductMode','SpecifyPrecision',...
   'SumMode','SpecifyPrecision',...
    'ProductWordLength',32,...
    'ProductFractionLength',16,...
    'SumWordLength',32,...
    'SumFractionLength',16,...
    'RoundMode','floor');

% Set up y with this fimath F
yfi = fi(y,1,18,17); yfi.fimath = F;

% For X, use the fimath you created and lets say (since that info is not
here)
% that X too is a s18q17 (same as y)
N = numerictype(yfi);
X = fi(zeros(2,1),N,F);

% Then
X(1) = yfi(1) + yfi(2);
X(2) = N.divide(yfi(1)-yfi(2),fi(sqrt(2)));

% You could also compute the 1/sqrt(2) and store it in  a fi
% to get X(2) using the * operator
onebysqrt2 = fi(1/sqrt(2),1,Somewordlength,Somefractionlength);
onebysqrt2.fmath = F;
X(2) = (yfi(1)-yfi(2))*onebysqrt2;

Hope this helps,

Ashok Charry
The Mathworks

"porterboy" <porterboy76@yahoo.com> wrote in message
news:c4b57fd0.0409150441.71f4e3ed@posting.google.com...
> I'm trying to learn the new fixed point toolbox for Matlab V7.0 R14.
> I have the following Section of floating point code...
>
> y = randn(1,2)/64;
> X = zeros(2,1);
> X(1) = y(1) + y(2);
> X(2) = (y(1) - y(2))/sqrt(2);
>
> (which is a two point butterfly in a DCT if you are interested). What
> do i do to convert this to fixed point? Here is my (not very good)
> attempt.
> *******************************************************************
> % first generate 18 bit random numbers for y
> y = randn(1,2)/64;
> y = fi(y,1,18,17,'RoundMode','floor');
>
> % then define fixed point representation of X
> F = fimath('ProductMode','SpecifyPrecision',...
>     'SumMode','SpecifyPrecision',...
>     'ProductWordLength',32,...
>     'ProductFractionLength',16,...
>     'SumWordLength',32,...
>     'SumFractionLength',16,...
>     'RoundMode','floor');
>
> % do we need to initialize X? (I used the zeros function in floating
> point)
>     X(1) = F.add(x(1),x(2));
>     X(2) = F.sub(x(1),x(2));
>
> % eh, no idea how to divide, because divide is defined for
> numerictypes
>     X(2) = F.divide(X(2),fi(sqrt(2));???
> ***********************************************************************
> Please help?

Thanks Ashok, that was very useful.

> % For X, use the fimath you created and lets say (since that info is not
> here)
> % that X too is a s18q17 (same as y)
> N = numerictype(yfi);
> X = fi(zeros(2,1),N,F);

Suppose X is not the same as y. Will the divide function still work,
even though the numerictypes are different? Or do they have to be cast
to the same type?

Hi,

Yes, X can have a different numerictype and the divide would still work.
% Say X is a s16q15
Nx = numerictype(X);
Nx.divide(X,yfi) % will work and the result will be a s16q15

However, I'd rather not use divide in your fixed-point implementation. The
better practice is to
compute the 1/sqrt(2) and store that as a constant (fi) and use multiply
instead.


"porterboy" <porterboy76@yahoo.com> wrote in message
news:c4b57fd0.0409152341.41f14c2c@posting.google.com...
> Thanks Ashok, that was very useful.
>
> > % For X, use the fimath you created and lets say (since that info is not
> > here)
> > % that X too is a s18q17 (same as y)
> > N = numerictype(yfi);
> > X = fi(zeros(2,1),N,F);
>
> Suppose X is not the same as y. Will the divide function still work,
> even though the numerictypes are different? Or do they have to be cast
> to the same type?