how to find phase angle for the FFT data of a transient signal ?

currently i am using PSD to calculate my fft

[y(:,1),f] = pwelch(z1(:,1),hann(fsize),round(fsize/2),fsize,fs);% calculate PSD

yfft(:,1)=sqrt(1.5*y(:,1));% calculate fft

Answer 1

[Honglei Chen]

Why not just use fft? PSD contains only the power information so the phase information is lost.

[Mahesh]

i tried using the inbuild matlab function>>>

t = (0:99)/100; % Time vector x = sin(2*pi*15*t+pi/2) + sin(2*pi*40*t); % Signal y = fft(x); % Compute DFT of x p = unwrap(angle(y)); % Phase

Plot the phase: f = (0:length(y)-1)'/length(y)*100; % Frequency vector plot(f,p)

but the result you expect is different from what i expected.

so insted of this i used

L=length(data); fs=100; NFFT = 2^nextpow2(L); % Next power of 2 from length of y Y = fft(data,NFFT)/L; f = fs/2*linspace(0,1,NFFT/2+1); figure plot(f,2*abs(Y(1:NFFT/2+1))) title('Single-Sided Amplitude Spectrum of y(t)') xlabel('Frequency (Hz)') ylabel('|Y(f)|')

how do i get the phase angles in this case ?

[Honglei Chen]

What is your expected result and what is the result you get from FFT? Like I mentioned, you cannot expect to get phase from PSD since there is no phase information in it.

[John Petersen]

You may need to apply a window on the data to reduce the effects of nonzero starting and ending points of the data set.