How can i determine phase in FFT?

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Big dream 2016-10-24

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评论： Star Strider 2023-6-28

Hi everyone, right now im trying to calculate signal phases using angle(x) from FFT Function im Matlab. Noted that i've coded the program like below :

%%Plotting Grafik
%create a time vector 't', containing integers from 1 to n(summary of data)
count= length(data);
Ts=mean(diff(times1));
Fs=1/Ts;
NFFT=2^nextpow2(count);
y=fft(data,NFFT)/count;
f=Fs/2*linspace(0,1,NFFT/2+1);
figure(2)
plot(f,2*abs(y(1:NFFT/2+1)));
grid on
title('Single-Sided Amplitude Spectrum of y(t)');
xlabel('Frequency(Hz)');
ylabel('|Y(f)|');
phase=angle(y);

But i'm not really sure with the phase function. Does anyone ever has the same experience?

Thanks before for your answer.

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WH Wang 2023-6-28

it's atan2(imaginary,real) in matlab, or using phase() function of matlab.

Star Strider 2023-6-28

There is no phase function that I can find in the documentation, however there is the angle function that returns the phase angle (in radians) of a complex argument.

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Star Strider 2016-10-24

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编辑：Star Strider 2016-10-24

在 MATLAB Online 中打开

Your code appears to me to be correct, including the ‘phase’ assignment. Note that the units of ‘phase’ are in radians. Convert them to degrees (if desired) by multiplying them by 180/pi.

I’m not certain what you want, so also consider the unwrap function. Use it on the radian units, and convert to degrees later if necessary.

EDIT — If you want to plot it, the easiest way is to use the subplot function. Your plotting code changes to:

phase=angle(y);
figure(2)
subplot(2,1,1)
plot(f,2*abs(y(1:NFFT/2+1)));
grid on
title('Single-Sided Spectrum of y(t)');
ylabel('|Y(f)|');
subplot(2,1,2)
plot(f,phase)
grid on
xlabel('Frequency(Hz)');
ylabel('Phase (rad)')

I did not test that, but it should work.

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Star Strider 2016-10-25

编辑：Star Strider 2016-10-25

在 MATLAB Online 中打开

My pleasure.

This is correct:

y = fft(data,NFFT)/count;

You have to normalise the result by the length of the original data (the ‘energy’ in the original signal) to get the correct amplitudes for the double-sided Fourier transform. (To get the correct amplitudes for the single-sided Fourier transform, you coded it correctly in multiplying it by 2.)

To get the inverse transform, take the inverse of the transformed signal:

yinv = ifft(y);

I’ve never actually done that experiment before now, so I coded it to see the result. It was as I expected, the only challenge being to define correctly the new time vector for ‘yinv’ (although I named my variables differently so as not to confuse them with your code):

Ts = 0.01;
Fs = 1/Ts;
Fn = Fs/2;
t = linspace(0, 1, 1500)*Ts;
s = sin(2*pi*t*5000) .* cos(2*pi*t*150);                        % Create Original Data
NFFT = 2^nextpow2(length(s));
FTs = fft(s, NFFT);                                             % Fourier Transform Of Zero-Padded Original Data
IFTs = ifft(FTs);                                               % Inverse Fourier Transform Of Zero-Padded Original Data
t2 = linspace(0, 1, NFFT)*Ts*NFFT/length(t);                    % Inverse Fourier Transform Sampling Interval = Ts*NFFT/length(t)
figure(1)
subplot(2,1,1)
plot(t,s)
title('Original Signal')
grid
subplot(2,1,2)
plot(t2,IFTs)
grid
title('IFFT Of FFT Of Zero-Padded Original Signal')
xlabel('Time (sec)')

Note that the times for both signals are the same, with the zero-padding clearly apparent in the second signal.

I learned something from this, so I thank you!

Big dream 2016-12-6

thanks again for your advice.^^

Star Strider 2016-12-6

My pleasure.

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