Simulink and FFT function

2 次查看（过去 30 天）
marcusbarnet
marcusbarnet 2018-3-31
评论： marcusbarnet 2018-3-31
Hi to all,
I hope to be able to find a solution for my problem.
I'm using Simulink to perform data analysis on accelerations along the Z-axis coming from a accelerometer at a sampling rate of 160 Hz. I would like to calculate the FFT over 1600 samples (so, every 10 seconds) during a real-time acquisition and so I'm using this diagram:
I would like to make the stats block output the four maximum peaks from the FFT. In Matlab, the code works very well since I'm able to obtain every time four maximum peaks and their frequencies, but when I use this same code in the stats blocks I get different results.
The code I use in Matlab is:
figure(2)
N = length(raw_accelearation(start*10:stop*10));
xdft = fft(raw_accelearation(start*10:stop*10));
xdft = xdft(1:floor(N/2+1));
psdx = (1/(Fs*N)) * abs(xdft).^2;
psdx(2:end-1) = 2*psdx(2:end-1);
freq = 0:Fs/length(raw_accelearation(start*10:stop*10)):Fs/2;
L = length(psdx);
windowed_signal=psdx.*hamming(L);
plot(freq,10*log10(windowed_signal))
grid on
title('Periodogram for asphalt')
xlabel('Frequency (Hz)')
ylabel('Power/Frequency (dB/Hz)')
[maxpeaks, idx] = findpeaks(10*log10(psdx),'MinPeakDistance',500,'NPeaks',4,'MinPeakHeight',-10)
peak1 = maxpeaks(1)
freq1 = freq(idx(1))
peak2 = maxpeaks(2)
freq2 = freq(idx(2))
peak3 = maxpeaks(3)
freq3 = freq(idx(3))
peak4 = maxpeaks(4)
freq4 = freq(idx(4))
and this is the plot:
This is the code I use in the stats block in Simulink:
function [x,y, z]= stats(u)
coder.extrinsic('findpeaks');
y=10;
x=10;
z=10;
Fs=160;
raw_acceleration = u;
N = length(raw_acceleration);
xdft = fft(raw_acceleration);
xdft = xdft(1:floor(N/2+1));
psdx = (1/(Fs*N)) * abs(xdft).^2;
psdx(2:end-1) = 2*psdx(2:end-1);
freq = 0:Fs/length(raw_acceleration):Fs/2;
% plot(freq,10*log10(psdx))
% grid on
% title('Periodogram for asphalt')
% xlabel('Frequency (Hz)')
% ylabel('Power/Frequency (dB/Hz)')
[maxpeaks, idx] = findpeaks(10*log10(psdx),'MinPeakDistance',300,'NPeaks',4,'MinPeakHeight',mean(10*log10(psdx)))
peak1 = maxpeaks(1)
% freq1 = freq(idx(1))
peak2 = maxpeaks(2)
% freq2 = freq(idx(2))
% peak3 = maxpeaks(3)
% freq3 = freq(idx(3))
% peak4 = maxpeaks(4)
% freq4 = freq(idx(4))
y = peak2
x = peak1
z = length(maxpeaks)
end
The problems in Simulink are:
  • I can't output all the four peaks because the code is able to find only 3 peaks
  • If I increase the MinPeakDistance I get an error: Index exceeds matrix dimensions.
How can I solve the problems?
This is how I setup the buffer block and the window block.
  1 个评论
marcusbarnet
marcusbarnet 2018-3-31
I found out that during the running time, it finds 2 peaks and sometimes 3 peaks..

请先登录，再进行评论。

请先登录，再回答此问题。

回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Circuit Envelope Simulation 的更多信息

标签

产品

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by