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?

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,
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?

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?