phase angle from discrte hilbert tranform
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bobby
am 2 Dez. 2011
Kommentiert: Fatma Keskin
am 20 Mai 2023
If I have time series data and I want to Hilbert transformation it to get the phase angle as a function of time what do I need to do? Sorry I'm completely stumped and brand new to matlab
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Wayne King
am 2 Dez. 2011
Hi Bobby, the derivative of the phase of the analytic signal is the instantaneous frequency that is true. But estimating the instantaneous frequency is tricky business. There are a number of papers on algorithms for that. It's very sensitive to noise for one thing. Yes, in the simple case I gave you, you can fit a least squares line to the unwrapped phase and get the frequency. But that is because the instantaneous frequency is the same everywhere.
x = cos(pi/4*(0:99));
y = hilbert(x);
sigphase = (unwrap(angle(y)))';
X = ones(length(sigphase),2);
X(:,2) = (1:length(sigphase))';
beta = X\sigphase;
beta(2)
Note beta(2) is very close to the frequency of pi/4. But this is a very simple example where the frequency is not changing.
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Wayne King
am 2 Dez. 2011
I can recommend:
B. Boashash, “Estimating and interpreting the instantaneous frequency of asignal. II. Algorithms and applications,” Proceedings of the IEEE, vol. 80, no. 4, pp. 540–568, 1992.
Wayne King
am 2 Dez. 2011
yea, it definitely won't be a simple process. If you're feeling ready to head into the thick of it, write back and I can suggest a few routes, but it's definitely tricky business. Don't get me wrong, it's a fruitful endeavor, but not straightforward.
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Wayne King
am 2 Dez. 2011
Hi, Do you really just want the phase function, or are you trying to get the instantaneous frequency. At any rate, you can get the phase with:
x = cos(pi/4*(0:100));
y = hilbert(x);
sigphase = atan2(imag(y),real(y));
% or
sigphase = angle(y);
3 Kommentare
Fatma Keskin
am 20 Mai 2023
If we have an amplitude modulated signal, what would the instantaneous phase give me? Phase of the envelope or the actual oscillation?
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