How to find the exact value of the frequency by MATLAB?

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I want to find the exact value of the frequency at which the peak is highest (represented by purple circle in the figure)on the left hand side of the figure. Is their a command or a way by which i can find that value? Please help
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Fx = fft(x - mean(x));
plot(real(Fx))

Accepted Answer

Walter Roberson
Walter Roberson on 12 Jul 2021
No, there is no command in MATLAB that can tell you the exact frequency of a peak. The most you would be able to tell is the frequency bin involved and the relative width of the bin.
It is never possible to tell frequency without having some information that relates the data to time. That could be sample interval (such as 1 hour between samples), or sample frequency (such as 20 Hz), or total duration. fft just gets passed a list of sample values assumed to be uniformly sampled in time, and cannot tell whether the data was sampled nanoseconds apart or hours apart or millennia apart.
If you did have time information then it would be possible to calculate frequency, but only approximately. For example if the samples were taken at 100 Hz for 1/4 second then you would not be able to distinguish between 22.4 Hz and 22.8 Hz -- and you cannot be sure that both of them will land on the same bin: one might round into the next bin, especially if there is a notable imaginary (phase) component.

More Answers (2)

Chunru
Chunru on 12 Jul 2021
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11365.9 11366.6 11369.4 11386.2 11405.7 11349 11372.3 11370.5 11370.2 11402.6 11385 11401.8 11382.5 11362.5 11353.6 11368.9 11365.3 11377.3 11375.3 11385.2 11364.7 11357 11366.8 11337.1 11348.1 11316.3 11308.3 11322.2 11311 11341.8 11350.1 11340.1 11348.3 11338.4 11379.9 11395.1 11368.1 11363.4 11342.3 11329.5 11317.9 11318.7 11304.5 11256.4 11277.6 11247.3 11264.2 11247.9 11245.6 11243.5 11257.1 11289.5 11250.5 11252.9 11243.2 11278.9 11323.5 11299.4 11328.1 11306.9 11311.5 11317.8 11308.3 11302.3 11276.2 11295.2 11252.7 11267.4 11268.3 11265.8 11271.8 11266.6 11280.2 11273.3 11300.6 11280.5 11298.7 11341.9 11336.2 11354.5 11351 11344.7 11346.5 11321.8 11358.6 11340.5 11352.6 11333.3 11347.1 11369.5 11352.9 11359.9 11335.2 11308.9 11331.4 11319.5 11344.5 11337.1 11352.8 11371.9 ...
11359.4 11373.2 11339.4 11340.5 11343.4 11326.3 11332 11300.9 11335.2 11319.5 11327 11353.4 11358.1 11376.5 11356.3 11369.2 11381.7 11364.1 11392.4 11363.8 11376.8 11371.5 11366 11401 11366 11374.2 11347.6 11350.8 11354.1 11348.3 11377.9 11350.9 11384.9 11370.2 11378.9 11410.1 11399.8 11422.8 11390.8 11410.8 11377.9 11365.6 11390 11363 11373.6 11358.8 11350.5 11370.1 11377.4 11421.5 11392.7 11429.5 11429.2 11440.5 11469.2 11415.8 11402.9 11349.8 11338.3 11312.8 11319.9 11323.2 11316.9 11320.3 11304.7 11311 11360.6 11354.3 11396.2 11376.8 11390.6 11404.5 11413.7 11394.7 11377.3 11346.3 11358.1 11339.4 11379.6 11358.9 11372.3 11377.1 11367.6 11389.9 11360 11408.8 11376 11373.3 11344.2 11342.9 11355.1 11328.9 11348.1 11348.7 11350.8 11371.7 11362.2 11360.3 11338.7 11319.6 ...
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11356 11354.2 11335.5 11372.6 11370.7 11386.8 11363.4 11358.3 11397.9 11383.7 11396.8 11397.8 11385.2 11401.2 11395.9 11398.1 11394 11399.5 11373.6 11388.2 11377.5 11394.6 11414 11393.3 11406.4 11405.6 11410.5 11434.6 11413.3 11453.9 11439.2 11457.6 11458.4 11446.7 11439.1 11433.6 11441.8 11414.7 11436.4 11403.5 11450.9 11432.5 11439.6 11473.3 11449.6 11463.1 11452.3 11427.2 11401.4 11405.6 11360.7 11366.5 11336.1 11317.6 11289.2 11306.6 11306.7 11302.1 11350.6 11343.2 11368.7 11370 11365.4 11334.1 11317.6 11336 11305.5 11330.4 11306.5 11317.9 11321.9 11313.5 11344.5 11307.2 11335.5 11316.7 11334.7 11313.4 11326.3 11313.2 11307.2 11334.8 11321 11323.6 11317.9 11338.3 11330.7 11314.4 11334.7 11335.3 11353.3 11341.7 11366.5 11351.7 11353.6 11364.3 11361 11351.9 11337.9 11372.8];
Fx = fft(x - mean(x));
plot(real(Fx));
% specify a region to search for max
r = [10 600];
[Fmax, idx] = max(real(Fx([r(1):r(2)]))); % findpeaks are more sophiscated
idx = idx + (10-1);
hold on
plot(idx, real(Fx(idx)), 'ro')
idx, real(Fx(idx))
idx = 33
ans = 1.0703e+04
  6 Comments
Haya Ali
Haya Ali on 12 Jul 2021
Can you please elaborate what each of the step is doing?
r = [10 600];
[Fmax, idx] = max(real(Fx([r(1):r(2)]))); % findpeaks are more sophiscated
idx = idx + (10-1);
hold on
plot(idx, real(Fx(idx)), 'ro')

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Simon Chan
Simon Chan on 12 Jul 2021
Edited: Simon Chan on 12 Jul 2021
May use threshold, which find another peaks on the left hand side
plot(real(Fx));
hold on
threshold = 1e4;
pos = find(real(Fx)>threshold);
plot(pos, real(Fx(pos)), 'go');
pos =
3 33 1169 1199
real(Fx(pos))
ans =
1.0e+04 *
1.1762 1.0703 1.0703 1.1762
  9 Comments
Simon Chan
Simon Chan on 12 Jul 2021
Need to plot the spectrum first:
plot(abs(Fx));
hold on
threshold = max(abs(Fx));
pos = find(abs(Fx)==threshold);
plot(pos, abs(Fx(pos)), 'go');

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