MATLAB: Initial phase angle calculation

initial phase angle

hello

I have this signal

clear all;clc;
t=0:0.0001:10;
tt=t(1:100000);%take 100 000 samples
a=30*pi/180; %phase (calculated form deg. to rad.)

b=90*pi/180; %phase (calculated form deg. to rad.)
ia=5*cos(2*pi*10*tt-a)+7*cos(2*pi*50*tt-b);
fs=1/0.0001; %sampling frequency
X=fft(ia); %FFT
df=fs/length(X); %frequency resolution
f=(0:1:length(X)/2)*df; %frequency axis
subplot(2,1,1);
M=abs(X)/length(ia)*2; %amplitude spectrum
plot(f,M(1:length(f)));
subplot(2,1,2);
P=angle(X)*180/pi; %phase spectrum (in deg.)
plot(f,P(1:length(f)));

this technique do not give a good results; I want any technique to get the right "a" an "d" the initial phase angle of 10 HZ and 50 HZ;

I know it is a hard question but please help me

Best Answer

I ran your code it and seems to be working as-is. -30 and -90 deg at 10 and 50Hz

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MATLAB: Problem with phase in FFT analysis

The problem is that there is not an exact whole number of cycles of the sine wave in the original time series. If you look at the values of x(1) and x(end), you'll find they are the same, because the final point is at time t=1, when a new cycle is just starting.

Unless you have an exactly whole number of cycles of a sine wave present in the initial series, there won't be a sample point in the frequency domain lying at exactly the frequency of the wave. In your case, the sample at X(101) (with its complex conjugate at X(902)) is close to, but not exactly at, the frequency f=100. This slight shift in frequency doesn't affect the amplitude too much, but it has a big effect on the phase.

A minimal change to get the answer you expect is to reduce the length of the input by 1, for example like this:

t=0 : 1/fs : 1-0.5/fs; %time

which will give a whole number of cycles provided f is an integer. (As Walter Roberson points out, linspace may be preferable - but rounding errors aren't in fact the problem in this case.) This change gives the phase as expected, as you can see by printing the value of P(101).

There's a general lesson: if you have real-world data, you are very unlikely to have an exactly whole number of cycles in the sequence, and so you can't get the phase easily from the FFT.

MATLAB: How to determine phase of a sine wave using FFT

The code hasn’t posted. Use the angle function on the output of the fft function:

t  = [...];                                                 % Time Vector
s  = [...];                                                 % Signal Vector
Ts = mean(diff(t));                                         % Sampling Time
Fs = 1/Ts;                                                  % Sampling Frequency
Fn = Fs/2;                                                  % Nyquist Frequency
L  = length(s);
fts = fft(s)/L;                                             % Normalised Fourier Transform
Fv = linspace(0, 1, fix(L/2)+1)*Fn;                         % Frequency Vector
Iv = 1:length(Fv);                                          % Index Vector
amp_fts = abs(fts(Iv))*2;                                   % Spectrum Amplitude
phs_fts = angle(fts(Iv));                                   % Spectrum Phase

Note — This is obviously UNTESTED CODE but it should work.

Best Answer

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MATLAB: Problem with phase in FFT analysis

MATLAB: How to determine phase of a sine wave using FFT

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