How to use extended word length and fraction in HDLFFT?

Hi

When I extend word and fraction length by using sfi(ddc_out,50,30) from sfi(ddc_out,32,24).

But I've got the below error messages during run the below code

[Yf(loop),validOut(loop)] = HDLFFT8(complex(y_fixed(i)),(loop <= N));

What am I supposed to do to resolve this problem?

close all

N = 8;
y_fixed = sfi(ddc_out,50,30);

Yf = zeros(1,3*N);
validOut = false(1,3*N);

for loop = 1:1:3*N 
    if (mod(loop, N) == 0) 
        i = N; 
    else
        i = mod(loop, N); 
    end
      [Yf(loop),validOut(loop)] = HDLFFT8(complex(y_fixed(i)),(loop <= N));

end

%%code
function [yOut,validOut] = HDLFFT8(yIn,validIn)
%HDLFFT8 
% Processes one sample of FFT data using the dsp.HDLFFT System object(TM)
% yIn is a fixed-point scalar or column vector. 
% validIn is a logical scalar value.
% You can generate HDL code from this function.
    persistent fft8;
    if isempty(fft8)
      fft8 = dsp.HDLFFT('FFTLength',8);
    end    
    [yOut,validOut] = fft8(yIn,validIn);
  end

*Error using dsp.HDLFFT/parenReference
Changing the data type on input 1 is not allowed without first calling the release() method.

Error in HDLFFT8 (line 12)
    [yOut,validOut] = fft8(yIn,validIn);

Error in AD9467_Cal_171219 (line 151)
      [Yf(loop),validOut(loop)] = HDLFFT8(complex(y_fixed(i)),(loop <= N));

>>

Answer 1

[Bharath Venkataraman]

Satish,

I am not able to reproduce the error. Here is my code.

close all
N = 8;
ddc_out = randi([0 1],8,1);
y_fixed = sfi(ddc_out,50,30);
Yf = zeros(1,3*N);
validOut = false(1,3*N);
for loop = 1:1:3*N 
    if (mod(loop, N) == 0) 
        i = N; 
    else
        i = mod(loop, N); 
    end
      [Yf(loop),validOut(loop)] = HDLFFT8(complex(y_fixed(i)),(loop <= N));
end

disp('Output of FFT: ');
Yf(validOut)

HDLFFT8.m:
function [yOut,validOut] = HDLFFT8(yIn,validIn)
%HDLFFT8 
% Processes one sample of FFT data using the dsp.HDLFFT System object(TM)
% yIn is a fixed-point scalar or column vector. 
% validIn is a logical scalar value.
% You can generate HDL code from this function.
    persistent fft8;
    if isempty(fft8)
      fft8 = dsp.HDLFFT('FFTLength',8);
    end    
    [yOut,validOut] = fft8(complex(yIn),validIn);
  end

Bharath

[satish guruma]

Dear Bharath,

Would you please help me how to get the verification values correctly?

I've succeeded with as your recommends, I came across all zero value as the below. but when I use hdlcoder with HDLFFT8.m .

ans =

    1×0 empty double row vector

close all
clear
clc
N = 8;  

ddc_out = [43.1071670298783 + 6.93920051078942i,196.370439792336 + 72.8360842474480i,45.0743298749615 + 68.5707875747732i,-253.063222261192 + 131.949254363931i,3287.75721314405 + 496.424894456540i,10796.7475951656 + 4062.11599418717i,7668.00136364171 + 7105.42396362013i,5142.92229218845 + 9422.14012629434i];
y_fixed = sfi(ddc_out,50,30);
Yf = zeros(1,3*N);
validOut = false(1,3*N);
ddcout = zeros(1,3*N);

for loop = 1:1:3*N 
    if (mod(loop, N) == 0) 
        i = N; 
    else
        i = mod(loop, N); 
    end

      [Yf(loop),validOut(loop)] = HDLFFT8(complex(y_fixed(i)),(loop <= N));

           disp('test1 : Output of FFT: ');
           Yf(loop)

end
disp('test2 : Output of FFT: ');
Yf(validOut)

[Walter Roberson]

By the way, you can simplify

      if (mod(loop, N) == 0) 
        i = N; 
    else
        i = mod(loop, N); 
    end

as

i = mod(loop-1, N) + 1;

[Bharath Venkataraman]

Since you are already using fixed-point types, in Step 1 of Fixed point conversion, I suggest that you skip the conversion and use the datatypes as is.