Calculating the frequencies for a two-sided Fourier transform can initially be a bit of a challenge.
Try something like this —
Fs = 1234; % Sampling Frequency
Ls = 10;
t = linspace(0, Fs*Ls-1, Fs*Ls)/Fs; % Time Vector
s = sum(sin([1; 5; 50; 100; 200; 300]*2*pi*t)); % Signal Vector
figure
plot(t, s)
grid
xlabel('Time')
ylabel('Amplitude')
title('Signal')
Fn = Fs/2; % Nyquist Frequency
L = numel(t); % Signal Length
NFFT = 2^nextpow2(L); % For Efficiency
FTs = fft(s,NFFT)/L; % Fourier Transform
FTss = fftshift(FTs); % Shift For Central Symmetry
Fv = linspace(-Fn, Fn-Fs/NFFT, NFFT); % Frequency Vector For Two-Sided 'fft' With An Even Number Of Elements
figure
plot(Fv,abs(FTss))
grid
xlabel('Frequency')
ylabel('Magnitude')
title('Fourier Transform')
[pks,locs] = findpeaks(abs(FTss), 'MinPeakProminence',0.25); % Use 'findpeaks' To Return Peak Indices To Demonstrate Peak Frequencies
Frequencies = Fv(locs)
figure
plot(Fv,abs(FTss))
grid
axis([[-1 1]*10 0 0.7])
xlabel('Frequency')
ylabel('Magnitude')
title('Fourier Transform (Detail)')
txtstr = compose('\\leftarrow %+9.4f Hz',Frequencies);
text(Frequencies(5:8), pks(5:8), txtstr(5:8), 'Vert','top', 'Horiz','left', 'Rotation',45)
Using the ‘NFFT’ value as calculated here serves two purposes. First, it improves the efficiency of the fft calculation, and second it creates a fft vector with an even number of elements.
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