How to find the frequency and amplitude of an oscillating signal?

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Rohan Kokate 2023-11-29

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https://ww2.mathworks.cn/matlabcentral/answers/2053867-how-to-find-the-frequency-and-amplitude-of-an-oscillating-signal

评论： Star Strider 2023-12-1

book1.xlsx

Hello,

I have a pressure signal which oscillates at a certain frequency and amplitude at steady-state. Is there a way to use the raw data (pressure and time) to find the frequency and amplitude of the oscillations. The raw data does have a little noise. I have attached an example image of the signal that I have along with an excel sheet which has the raw data. I believe FFT is the way to go but I have zero experience/background in using it.

Any help would be greatly appreciated.

Thank you in advance!

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Answer 1

Star Strider 2023-11-29

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https://ww2.mathworks.cn/matlabcentral/answers/2053867-how-to-find-the-frequency-and-amplitude-of-an-oscillating-signal#answer_1362522

编辑：Star Strider 2023-11-29

在 MATLAB Online 中打开

book1.xlsx

One approach —

T1 = readtable('book1.xlsx', 'VariableNamingRule','preserve')

T1 = 50×2 table

Time [s] P5 [psia] ________ _________ 3094.8 118.37 3097.1 116.52 3099.7 118.26 3102.2 120.05 3104.6 119.49 3107.1 116.54 3109.6 117.33 3112.1 119.37 3114.5 120.24 3117 117.57 3119.4 116.61 3121.9 118.29 3124.4 120.15 3126.8 119.49 3129.4 116.46 3131.8 117.45

VN = T1.Properties.VariableNames;

t = T1{:,1};

s = T1{:,2};

figure

plot(t,s)

grid

checkTime = [mean(diff(t)) std(diff(t))] % Check Sampling Time Variation

checkTime = 1×2

2.4724 0.0631

Fs = 1/mean(diff(t)); % Mean Sampling Frequency

[sr,tr] = resample(s,t,Fs); % Resample To Constant Sampling Intervals (Required For Signal Ptocessing)

Fn = Fs/2; % Nyquist Frequency

L = size(T1,1); % Signal LEngth

NFFT = 2^nextpow2(L); % Use This Value To Make The 'fft' More Efficient

FTs = fft((s-mean(s)).*hann(L), NFFT)/L; % Discrete Windowed Fourier Transform

Fv = linspace(0, 1, NFFT/2+1)*Fn; % Frequency Vector

Iv = 1:numel(Fv); % Index Vector

[pks,locs,width] = findpeaks(abs(FTs(Iv))*2, 'MinPEakProminence',0.25, 'WidthReference','halfheight');

Peak_Frequency = Fv(locs)

Peak_Frequency = 0.0885

Peak_Magnitude = pks

Peak_Magnitude = 0.8425

Peak_Width = width*Fv(2)

Peak_Width = 0.0168

idx = find(diff(sign(abs(FTs(Iv))*2-(pks/2))));

for k = 1:numel(idx)

idxrng = max(1,idx(k)-1) : min(idx(k)+1,Iv(end));

hpf(k) = interp1(abs(FTs(idxrng))*2, Fv(idxrng), pks/2);

end

figure

plot(Fv, abs(FTs(Iv))*2)

hold on

plot(hpf, [1 1]*pks/2, '-k')

hold off

grid

xlabel('Frequency (Hz)')

ylabel('Magnitude')

title('Fourier Transform')

text(Peak_Frequency, Peak_Magnitude, sprintf('\\leftarrow Frequency = %.3f Hz\n Magnitude = %.3f',Peak_Frequency, Peak_Magnitude), 'Horiz','left')

text(hpf(2), pks/2, sprintf('\\leftarrow FWHM = %.3f Hz', Peak_Width))

EDIT — (29 Nov 2023 at 18:28)

Added text calls to Forueir Transform plot.

.

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Star Strider 2023-11-30

在 MATLAB Online 中打开

book1.xlsx

The fft amplitude roughly corresponds to the peak-to-peak amplitude of the oscillations, with reservations for the fact that the peaks do not all have the same amplitudes, and the total energy of the waveform is distributed over about 0 to 0.2 Hz, and the peak itself has a full-width-half-maximum value of 0.017 Hz, meaning that the energy of the oscillation is distributed over that range. (That implies that even with the resampled signal, the oscillation frequency varies over a reasonable range.)

For a pure sine or cosine curve, the fft amplitude is almost exactly the peak-to-peak amplitude, and the width of the peak approaches zero. For real-world signals, it is usually different, and sometimes significantly so.

T1 = readtable('book1.xlsx', 'VariableNamingRule','preserve')

T1 = 50×2 table

Time [s] P5 [psia] ________ _________ 3094.8 118.37 3097.1 116.52 3099.7 118.26 3102.2 120.05 3104.6 119.49 3107.1 116.54 3109.6 117.33 3112.1 119.37 3114.5 120.24 3117 117.57 3119.4 116.61 3121.9 118.29 3124.4 120.15 3126.8 119.49 3129.4 116.46 3131.8 117.45

VN = T1.Properties.VariableNames;

t = T1{:,1};

s = T1{:,2};

Fs = 1/mean(diff(t)); % Mean Sampling Frequency

[sr,tr] = resample(s,t,Fs); % Resample To Constant Sampling Intervals (Required For Signal Ptocessing)

[vy,pk] = bounds(sr);

figure

plot(tr,sr, 'DisplayName','Signal')

yline(pk, '--g', 'DisplayName','Maximum Peak Amplitude')

yline(vy, '--r', 'DisplayName','Minimum Valley Amplitude')

grid

legend('Location','best')

xlabel(VN{1})

ylabel(VN{2})

pos = get(gca,'Position');

xapf = @(x,pos,xl) pos(3)*(x-min(xl))/diff(xl)+pos(1); % 'x' Annotation Position Function

yapf = @(y,pos,yl) pos(4)*(y-min(yl))/diff(yl)+pos(2); % 'y' Annotation Position Function

annotation('doublearrow', xapf([1 1]*3090, pos, xlim), yapf([pk vy], pos,ylim))

text(3091, mean([pk vy]), sprintf('Peak-to-Peak Maximum = %.3f',pk-vy), 'Horiz','left', 'Vert','middle')

.

Star Strider 2023-11-30

在 MATLAB Online 中打开

Not really. The Fourier transform calculates the energy in a signal by frequency, and if the frequency varies, that will produce a wide peak with the energy of the signal dispersed over those frequencies. Add to that varying amplitudes, and the Fourier transform will do its best to describe the signal taking all that into account.

I devised a signal to demonstrate that and did a similar analysis on it here. Note that ‘s’ has an amplitude of approximately ±1 throughout its range, however the frequency varies significantly (50 Hz, with a centre frequency of 150 Hz), and the Fourier transform has an amplitude that is less tha 1% of the original signal amplitude. Some of this is ‘spectral leakage’, however most of it is due to the fact that the signal energy is spread over ±25 Hz from the centre frequency. That dissipates the energy significantly.

Example —

Fs = 1000;

Ls = 120;

t = linspace(0, Fs*Ls, Fs*Ls+1).'/Fs;

s = sum(sin(2*pi*t.*(100+50*t/max(t))),2);

figure

plot(t, s)

grid

xlabel('t')

ylabel('Amplitude')

title('Whole Signal')

figure

plot(t, s)

grid

xlabel('t')

ylabel('Amplitude')

title('Zoomed Segment')

xlim([-0.025 0.025]+60)

Fn = Fs/2; % Nyquist Frequency

L = numel(t); % Signal LEngth

NFFT = 2^nextpow2(L); % Use This Value To Make The 'fft' More Efficient

FTs = fft((s-mean(s)).*hann(L), NFFT)/L; % Discrete Windowed Fourier Transform

Fv = linspace(0, 1, NFFT/2+1)*Fn; % Frequency Vector

Iv = 1:numel(Fv); % Index Vector

[pks,locs,width] = findpeaks(abs(FTs(Iv))*2, 'MinPeakProminence',0.005, 'WidthReference','halfheight');

Peak_Frequency = Fv(locs);

Peak_Magnitude = pks;

Peak_Width = width*Fv(2);

idx = find(diff(sign(abs(FTs(Iv))*2-(max(pks)/2))));

for k = 1:numel(idx)

idxrng = max(1,idx(k)-1) : min(idx(k)+1,Iv(end));

hpf(k) = interp1(abs(FTs(idxrng))*2, Fv(idxrng), max(pks)/2);

end

figure

plot(Fv, abs(FTs(Iv))*2)

hold on

plot([min(hpf) max(hpf)], [1 1]*max(pks)/2, '-k')

hold off

grid

xlabel('Frequency (Hz)')

ylabel('Magnitude')

title('Fourier Transform')

text(Peak_Frequency, Peak_Magnitude, sprintf('\\leftarrow Frequency = %.3f Hz\n Magnitude = %.3f',Peak_Frequency, Peak_Magnitude), 'Horiz','left')

text(hpf(2), pks/2, sprintf('\\leftarrow FWHM = %.3f Hz', Peak_Width))

.

Rohan Kokate 2023-12-1

Oh, I get it now. I will try to read a bit more to get more clarity. Thanks for all the help!

Star Strider 2023-12-1

As always, my pleasure!

Essentially, the more frequencies that exist in a signal, the more the total energy is distributed among them.

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Answer 2

Paul 2023-11-30

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https://ww2.mathworks.cn/matlabcentral/answers/2053867-how-to-find-the-frequency-and-amplitude-of-an-oscillating-signal#answer_1363419

编辑：Paul 2023-11-30

在 MATLAB Online 中打开

book1.xlsx

Hi Rohan,

For this signal, the output of fft without windowing gives a very good estimate of the amplitude and frequency of the signal.

Read and plot the data.

T = readtable('book1.xlsx', 'VariableNamingRule','preserve');

t = T{:,1};

p5 = T{:,2};

figure

plot(t,p5),grid

Get a rough estimate of its amplitude and frequency

[highpeaks,locs] = findpeaks(p5);
lowpeaks = -findpeaks(-p5);
amplitudeEstimate = (mean(highpeaks)-mean(lowpeaks))/2;
frequencyEstimate = 1./mean(diff(t(locs)));

The data is not equally spaced, but we'll assume it is for a first cut and use the mean dt to define a sampling frequency. Not necessarily a recommended approach, but may be good enough for a first cut because the deviation is not too large.

plot(1./diff(t))

Fs = 1/mean(diff(t));

Compute the FFT

P5 = fft(p5-mean(p5))/numel(p5);
f = (0:numel(P5)-1)/numel(P5)*Fs;

If you want to use a window, the scaling has to be modified

whann = hann(numel(p5));
P5hann = fft((p5-mean(p5)).*whann)/sum(whann);

Plot the results

plot(f,2*abs(P5),'-o',f,2*abs(P5hann),'-o'),grid

yline(amplitudeEstimate,'m'),xline(frequencyEstimate,'m')

xlim([0 Fs/2]),xlabel('Frequency (Hz)'),ylabel('Amplitude')

legend('Raw','Windowed')

You can probably sharpen this analysis by being more careful in dealing with the non-equally spaced samples, and maybe zero padding the fft's as well.

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Rohan Kokate 2023-12-1

Thanks Paul!

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How to find the frequency and amplitude of an oscillating signal?

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