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I was testing out the Fourier transform in numerical analysis software on some artificial synthesized waveforms from adding two or three sinusoids together. I was wondering what in the Fourier transform determines which frequency gets assigned a phase angle of zero.

For example, a waveform composed of a $\sin(2\pi60t) + \sin(2\pi 120t-\pi)$ spit out a phase angle of $-\pi$ for 60Hz and $-2\pi$ for 120Hz rather than 0 and $-\pi$ respectively. The interval fed into the FFT in this example was 1/60th of a second.

What in the transform determined which frequency received a phase angle of zero with which all other phase angles are measured with respect to?

I don't have access to the software right now but it just occurred to me that maybe it has something to do with only one fundamental cycle being run through the transform and maybe if I ran thousands of cycles through it or an "infinite" number of cycles then the phase angle might approach or end up being zero since there are reduced/no frequencies responsible for starting and stopping the waveform in infinite time.

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    $\begingroup$ For real signals, for any frequency component that you can express as $\cos(2\pi f_a t + 0)$ in the time domain, has a phase of $0$ in the frequency domain. I.e. the phase of the cosinusoid in the time domain, is the phase in the frequency domain. In your example, use trig identities to convert your sinusoids to cosinusoids, and see that the phases match in the time and frequency domains. $\endgroup$ – Andy Walls 2 days ago
  • $\begingroup$ @AndyWalls Thanks. I'll check it out when I get the chance $\endgroup$ – DKNguyen 2 days ago

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