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How to extract the correct phase of a sine wave from data

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bob98 on 24 Oct 2020
Commented: bob98 on 25 Oct 2020
Hi everyone
I'm trying to write a code which allows me to determine the phase of a sine wave from its vector.
given: the sinusoidal signal vector, its frequency
unknown: its phase
I have tried with this code but i don't understand why do i end up getting wrong/ unprecise results
a=0; b=800; %time intervalls
N=20000; %N number of samples
t=linspace(a,b,N); %time vector
ts=(b-a)/N; %sampling time
fs=1/ts; %sampling frequency
fn=fs/2; %Nyquist frequency
fx=1; phix= 180; %Frequency and phase of x(t)
x=sin(2*pi*fx.*t+phix* (pi/180)); %Original signal
binwidth = 1 / (b-a); % frequency binwidth in Hz
f_index = fx / binwidth +1; % frequency index
y = abs(fft(x)) / (N/2);
phase = angle(fft(x))* 180/pi+90;
for example here i get 187.2000 instead of 180
I'd be very grateful for your help and clarifications!


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Accepted Answer

Freewing on 24 Oct 2020
The reason you get imprecise results is because you are using discrete fourier transform. When you get the phase of the frequency bin from fft, you get, roughly speaking, the phase of a frequency bin, but not the phase of a single frequency. You can get much better estimation if you analyse a few neighboring bins, but that’s a complex process.
If you’d like to compute exact phase and you know the signal is exactly sinusoidal, you’d better just use asin(x(1)).

  1 Comment

bob98 on 25 Oct 2020
thank you for your clarification,but i can't use asin(x(1)) since i don't know the amplitude of the signal.

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More Answers (1)

Ameer Hamza
Ameer Hamza on 24 Oct 2020
That is because you are doing a discrete Fourier transform. There can be a bit of spectral leakage to other frequency components if the sampling rate is low. For example, run your code after setting.
and check the result.

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