FFT for Lift coefficient to find the frequency

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Sopo Yoon
Sopo Yoon am 17 Okt. 2022
Kommentiert: Mathieu NOE am 17 Okt. 2022
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.

Akzeptierte Antwort

Mathieu NOE
Mathieu NOE am 17 Okt. 2022
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 Kommentare
Sopo Yoon
Sopo Yoon am 17 Okt. 2022
Thank you, appreciate it.
It looks good and easy to understand.
Best
Sopo
Mathieu NOE
Mathieu NOE am 17 Okt. 2022
My pleasure !

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