# how to implement this function in fixed point and generate the hdl code for the function?

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i have the function below and i want to generate hdl code for it and to convert it to fixed point for virsualization in simulink, i get the error that (exp and spdiags) are not supported for hdl code generation. i also have a simple testbench for calling this function.

f

unction [c, count]=sfft(x)

% Input: x, can be a vector (row or column) or matrix. When x is

% matrix, the Fourier transform is performed columnwise.

% The length of x is expected to be a power of 2. If not,

% zeros will be padded to the end of x to increase its length

% the next power of 2. Multidimensional array inputs are not

% supported.

%

% Output: c, the complex Fourier coefficients. Its preserves the

% dimensions of the x when zero-padding is not applied to x.

% If zero-padding is applied, c will be longer or taller

% than x.

%

global count % for counting the number of multiplications.

%% Prepare for the recursive call.

count=0; zeropad=false;

[siz1 siz2]=size(x);

if siz1==1 % x is a row vector

n=siz2;

ell=log2(n);

if mod(ell, 1)~=0,

ell=ceil(ell); n=2^ell; x(end+1:n)=0; zeropad=true;

end

x=x(:); % recur_sfft requires a column vector.

c=recur_sfft(x);

c=c.';

else % x is a column vector or matrix

n=siz1;

ell=log2(n);

if mod(ell, 1)~=0,ell=ceil(ell); n=2^ell; x(end+1:n,:)=0; end

c=recur_sfft(x);

end

if zeropad

warning('The input has been zero-padded to increase its length to the next power of 2.')

end

function c=recur_sfft(x)

global count

n=size(x,1);

if n>2

c_even=recur_sfft(x(1:2:n,:)); % (0:2:n-1)+1

c_odd=recur_sfft(x(2:2:n,:)); % (1:2:n-1)+1

% even means (0:2:n-1),and odd means 1:2:n-1 but they need to be added by 1 to meet Matlab's 1-based indices.

k=1:n/2;

w=exp(-pi*2i/n);

D=spdiags(w.^(k-1).',0, k(end), k(end));

c_odd=D*c_odd;

count=count+k(end)*size(c_odd,2);

% counting number of multiplications

% fill in the second half of c first

c(k+n/2,:)=c_even-c_odd;

% which has also pre-allocated RAM for the first half of c

% now do the first half of X.

c(k,:)=c_even+c_odd;

else

% the bottom is reached, 2-point of x.

c=[x(1,:)+x(2,:)

x(1,:)-x(2,:)];

end

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### Answers (1)

Bharath Venkataraman
on 3 Aug 2022

Edited: Bharath Venkataraman
on 5 Aug 2022

##### 0 Comments

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