Unable to see fft output during co-simulation.

Question

Jaykishan Solanki 2024-7-9

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2135651-unable-to-see-fft-output-during-co-simulation

回答： Sahas 2024-7-11

I am converting the following code to VHDL but after cosimulation I am only seeing validOut, how can I see the fft output as well?

function [yOut, validOut] = HDLFFT(yIn, validIn)

persistent fftObj;

if isempty(fftObj)

fftObj = dsphdl.FFT(FFTLength=128)

end

[yOut, validOut] = fftObj(yIn, validIn);

end

the main function

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)] = HDLFFT128_final((signal_fixed(i)),(loop <= N));

end

signal_zeros = signal_zeros(validOut == 1);

fft_reverse = bitrevorder(signal_zeros);

disp(fft_reverse);

figure(1)

stem(0:N-1,fft_reverse)

title('FFT')

the above is testbench

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Sahas 2024-7-11

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2135651-unable-to-see-fft-output-during-co-simulation#answer_1484183

在 MATLAB Online 中打开

Hi @Jaykishan Solanki,

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')