Perform FFT from imported excel data. Help please
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Hi, I have the attached time in first column and current in second column. i am trying to perform FFT and plot magnitude vs Frequency. My code is showing an empty plot.
could you please tell me where i am going wrong and kindly provide correction. I will be grateful. Thanks.
Sampling frequency is 20KHZ
[D,S,R] = xlsread('test data.csv');
RD = str2double(R);
Signal = RD(:,2); % Sampling Interval (milliseconds)
Ts = 0.00005; % Sampling Interval (seconds)
Fs = 1/Ts; % Sampling Frequency (Hz)
Fn = Fs/2; % Nyquist Frequency (Hz)
N = length(Signal);
meanSignal = mean(Signal); % ‘Signal’ Mean
FTSignal = fft(Signal-meanSignal)/N; % Normalised Fourier Transform Of Baseline-Corrected ‘Signal’
Fv = linspace(0, 1, fix(numel(FTSignal)/2)+1)*Fn; % Frequency Vector
Iv = 1:numel(Fv); % Index Vector
[pks,locs] = findpeaks(abs(FTSignal(Iv))*2, 'MinPeakHeight',0.044);
figure
plot(Fv, abs(FTSignal(Iv))*2)
grid
xlabel('Frequency (Hz)')
ylabel('Amplitude')
plotIdx = 1:Iv(max(locs));
figure
plot(Fv(plotIdx), abs(FTSignal(Iv(plotIdx)))*2)
hold on
plot(Fv(plotIdx(locs)), pks, '^r', 'MarkerFaceColor','r')
hold off
xlabel('Frequency (Hz)')
ylabel('Amplitude')
text(0.005, 0.08, 'Choose a peak and frequency')
2 个评论
Walter Roberson
2020-2-25
"Sampling frequency is 20KHZ"
No it is not. If you examine the times in the first column, the difference is 0.0005 which is 2kHz. Your Ts is not correct for the data file.
回答(1 个)
Ridwan Alam
2020-2-26
编辑:Ridwan Alam
2020-2-26
Signal = D(:,2);
RD = str2double(R) is making a matrix of NaNs. So, you can use D instead of R.
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