フィルターのクリア

FFT for Lift coefficient to find the frequency

3 ビュー (過去 30 日間)
Sopo Yoon
Sopo Yoon 2022 年 10 月 17 日
コメント済み: Mathieu NOE 2022 年 10 月 17 日
Hello, everyone!
I have different data files(.txt) includes simulation time, lift coefficient and drag coefficient.
I want to find the some frequencies which make periodic motion of lift.
The code is follwed :
load CdCl5.txt; a=CdCl5; Time5=a(1:end,2); Lift5=a(1:end,4);
load CdCl7.txt; a=CdCl7; Time7=a(1:end,2); Lift7=a(1:end,4);
figure(1) % Graph of the life coeffi.
subplot(211)
plot(Time5, Lift5)
title('Scheme 1')
xlabel('Time')
ylabel('C_L')
subplot(212)
plot(Time7, Lift7)
title('Scheme 2')
xlabel('Time')
ylabel('C_L')
% Parameter
N = length(Time5);
T = Time5(end); % Time length
dt = T/N;
df = 1/T; % frequency interval
f = df*(0:N-1); % frequency sequence
X5 = fft(Lift5)/N;
X7 = fft(Lift7)/N;
figure(2)
subplot(211)
plot(f,abs(X5))
title('Scheme 1')
ylabel('|X5(f)|')
xlabel('Frequncy[hz]')
subplot(212)
plot(f,abs(X7))
title('Scheme 2')
ylabel('|X7(f)|')
xlabel('Frequncy[hz]')
Do I analyze the figure(2) graphs to find the frequencies?
Could someone help me?
Thank you so much!
Sopo.

採用された回答

Mathieu NOE
Mathieu NOE 2022 年 10 月 17 日
hello
I prefer to put the repetitive fft lines in a subfunction , so the code can be cleaner and more compact
for the fft result , we are interested only in the positive half frequency points
you can use max or findpeaks to get the peaks value
a = readmatrix('CdCl5.txt');
Time5=a(1:end,2); Lift5=a(1:end,4);
a = readmatrix('CdCl7.txt');
Time7=a(1:end,2); Lift7=a(1:end,4);
figure(1) % Graph of the life coeffi.
subplot(211)
plot(Time5, Lift5)
title('Scheme 1')
xlabel('Time')
ylabel('C_L')
subplot(212)
plot(Time7, Lift7)
title('Scheme 2')
xlabel('Time')
ylabel('C_L')
% FFT plot
[f1,fft_spectrum1] = do_fft(Time5,Lift5);
[f2,fft_spectrum2] = do_fft(Time7,Lift7);
figure(2)
[PKS1,LOCS1] = max(fft_spectrum1);
f1_peak = f1(LOCS1)
subplot(211)
semilogx(f1,fft_spectrum1,f1_peak,PKS1,'dr')
text(f1_peak*1.25,PKS1 , (['f1 = ' num2str(f1_peak) ' Hz']));
title('Scheme 1')
ylabel('|X5(f)|')
xlabel('Frequency[hz]')
subplot(212)
[PKS2,LOCS2] = max(fft_spectrum2);
f2_peak = f2(LOCS2)
semilogx(f2,fft_spectrum2,f2_peak,PKS2,'dr')
text(f2_peak*1.25,PKS2 , (['f2 = ' num2str(f2_peak) ' Hz']));
title('Scheme 2')
ylabel('|X7(f)|')
xlabel('Frequency[hz]')
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function [freq_vector,fft_spectrum] = do_fft(time,data)
dt = mean(diff(time));
Fs = 1/dt;
nfft = length(data); % maximise freq resolution => nfft equals signal length
fft_spectrum = abs(fft(data))/nfft;
% one sidded fft spectrum % Select first half
if rem(nfft,2) % nfft odd
select = (1:(nfft+1)/2)';
else
select = (1:nfft/2+1)';
end
fft_spectrum = fft_spectrum(select,:);
freq_vector = (select - 1)*Fs/nfft;
end
  2 件のコメント
Sopo Yoon
Sopo Yoon 2022 年 10 月 17 日
Thank you, appreciate it.
It looks good and easy to understand.
Best
Sopo
Mathieu NOE
Mathieu NOE 2022 年 10 月 17 日
My pleasure !

サインインしてコメントする。

その他の回答 (0 件)

カテゴリ

Help Center および File ExchangeFourier Analysis and Filtering についてさらに検索

Community Treasure Hunt

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

Start Hunting!

Translated by