Frequency analysis using FFT

18 次查看（过去 30 天）
Thomas Leiw
Thomas Leiw 2015-9-11
评论： Star Strider 2015-9-13
Dear colleagues, my question is How to analyse the frequencies and its corresponding magnitude within an array by using FFT function in MatLab?
I am certain that the values in the array have the following information:
  • Already sampled at 1kHz
  • Contains 30001 samples
  • The sampled waveform is superimposed with multiple frequencies (e.g. 15Hz, 34Hz, etc)
I would like to analyse the magnitudes of each frequency band that is contained in the samples.
Thank you very much for any help!
Thomas

请先登录，再回答此问题。

采纳的回答

Star Strider
Star Strider 2015-9-12
The fft documentation has everything you need in the code between the first two plot images. Your Nyquist frequency is 500 Hz.
To get the peak values and frequencies, use the Signal Processing Toolbox findpeaks function. It has a number of name-value pair arguments that make it extremely flexible.
  8 个评论
显示 6更早的评论
Thomas Leiw
Thomas Leiw 2015-9-13
The values that were mentioned in the previous post were measured magnitudes at those frequency bands using another software even if noise has contributed to it. I am curious why the fft function in MatLab returned different results. I drafted another code to demonstrate the magnitude of a superimposed signal as the following:
f1=14; % 1st frequency of 14 Hz
f2=40; % 2nd frequency of 40 Hz
fs=1000; % sampled frequency
L=1000; % length of the signal
T=1/fs; % sampled period
t=(0:L-1)*T; % total last time of the signal
signal = 0.7*sin(2*pi*f1*t)+sin(2*pi*f2*t); % superimposed sine waveforms
N = length(signal); % defines the total number of signal array
fax_bins = [0:N/2-1]; % fix the x-axis with the starting point of 0
half_N = ceil(N/2); % get the half number of samples for single side plot
fnyquist = fs/2; % Nyquist frequency of signal
signal_FFT = abs(fft(signal)/N);
% magnitude of single-sided signal after transform
signal_mag = 2*signal_FFT(1:N/2+1);
fax_Hz = fax_bins*fs/N; % convert bins to Hz using simple formula
[PKS,FRQS] = findpeaks(signal_mag(1:half_N),'MINPEAKHEIGHT',0.69)
plot(fax_Hz, signal_mag(1:half_N))
title('Single-side Magnitude Spectrum')
xlabel('Frequency (Hz)')
ylabel('Magnitude (Amps)');
Since fft has the mirrored image at both sides of nyquist frequency, the magnitude needs to be multiplied by 2 in order to get the correct magnitude of 0.7 and 1 at the defined frequencies.
I think this is also implemented on the fft documentation page: http://au.mathworks.com/help/matlab/ref/fft.html
where:
P2 = abs(Y/L);
P1 = P2(1:L/2+1);
P1(2:end-1) = 2*P1(2:end-1);
Do you think this is correct?
Thomas
Star Strider
Star Strider 2015-9-13
It is correct, and entirely consistent with the previous version of the documentation on the fft function. (The documentation for fft changed in R2015b. The previous versions seemed to me to be more straightforward.) The single-sided fft in the documentation is correct, since it seems to have been taken from the R2015b fft documentation.
For your data, the amplitudes in my code at 16.3 Hz even with slightly revised code (corresponding to the R2015a fft documentation), are about 0.1 of those stated.
The only significant change in my code is:
[Pks,Frq] = findpeaks(2*abs(Fd(Iv)), Fv, 'MinPeakDistance', 0.25, 'MinPeakHeight',0.01);

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Time-Frequency Analysis 的更多信息

标签

Community Treasure Hunt

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

Start Hunting!

Translated by