I want to find the magnitude and phase of electric power signal i-e x(t)=220.cos(2.pi.60+pi/5) Using DFT in MATLAB using sampling frequency 1kHz.
but I am confused about how to acquire one cycle of the sampled signal then apply DFT on that.
you don not need to take one period of signal
you should acquire an interval to be sure that one complete period of signal is selected, for example 1 second of signal(1000 sample) or 2 seconds, and apply FFT and easily find phase and magnitude you are looking for.
The very first problem is the function you wrote is not periodic, because it's not time dependent. It should be
x(t)=220.cos(2.pi.60.t+pi/5)
Where 60 is the signal frequency. I agree with Hossein that the sample is not needed to be one cycle, but if you want it, you need to take a time interval like
t = linspace(0,1/60, N);
where N is number of samples you choose.
Yes Gabor Sir t is missing, thanx for the correction...let me try ur method
You need to take a sampling frequency of 960Hz so that you have 16 samples per period. Then you apply DFT to disjoint segments of 16 samples. Voilá!
Serna Sir when i perform FFT on 960 samples then it give me correct magnitude and phase , but when i use only sixteen samples keeping the interval 1/960 than i get magnitude of 182 instead of 220 and .353 degree phase instead off 36.
could you tell me what im doing wrong actually i want to find phasor of the input signal
Your question is interesting. DFT is an improper name for the Fourier filter because DFT (or FFT) is defined not as the calculation of the Fourier coefficient, but as:
For length N input vector x, the DFT is a length N vector X,
with elements
N
X(k) = sum x(n)*exp(-j*2*pi*(k-1)*(n-1)/N), 1
Sir here is my codes... in normal_dft code i get is correct answer while when using just 16 samples in fft_for_synchrophasor i get the wrong answer
You are doing operations over the whole vector of the fft. You only need to work on the first component X(2). That is why you have wrong calculations.
I made some little changes in your code. It seems the inaccuracy calculating the phase is due to the small number of samples (and the small time window). If you change the sampling frequency (keeping the time interval at 1 cycle of the signal) you get:
Sampling freq. Calculated phase
960 Hz 58.5 deg
9600 Hz 38.25 deg
96000 Hz 36.225 deg
960000 Hz 36.0225 deg
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