As per my understanding, when the code file is executed, it does not return the required FFT “yOut” variable value.
I was able to verify the issue and have provided a modified script that executes FFT function with the required FFT “yOut” output.
The issue was arising because the function defined in the main code was “HDLFFT” and it was being called as “HDLFFT128_final” in the testbench.
% Modified script
function [yOut, validOut] = HDLFFT(yIn, validIn) %function name
persistent fftObj;
if isempty(fftObj)
fftObj = dsphdl.FFT('FFTLength', 128);
end
[yOut, validOut] = fftObj(yIn, validIn);
end
N = 1024;
Fs = 800000;
number_of_samples = 0:N-1;
time = (0:N-1) / Fs;
signal = cos(2*pi*200000*time);
sampled_signal = cos(2*pi*8*number_of_samples/Fs);
signal_fixed = fi(sampled_signal, 0, 32, 24);
signal_zeros = zeros(1, N);
validOut = false(1, N);
for loop = 1:1:3*N
if (mod(loop, N) == 0)
i = N;
else
i = mod(loop, N);
end
[signal_zeros(loop), validOut(loop)] = HDLFFT(signal_fixed(i), (loop <= N)); % function being called
end
signal_zeros = signal_zeros(validOut);
fft_reverse = bitrevorder(signal_zeros);
disp(fft_reverse);
figure(1)
stem(0:length(fft_reverse)-1, fft_reverse)
title('FFT Output')
Here are some documentation links which might be helpful:
- Compute fast Fourier transform (FFT) - MATLAB - MathWorks India – FFT using DSP HDL Toolbox
- Fast Fourier transform - MATLAB fft - MathWorks India – Basic understanding of FFT
I hope this is useful.
