I'm calculating the beat of song through the frequency domain. I have 60 seconds of audio sampled at 48000Hz, multiplied by a Hanning window in the time domain. This gives me a resolution in the frequency domain of 1BPM, and I'm interested in frequencies from ~10 BPM to ~300BPM, which is only the first 300 bins of the 48000*60 bin length FFT. How does the Hanning window affect the phase of the song's beat? Using no window (really a rectangular window) results in a different phase than the Hanning window does.
It't really more of a question of what the rectangular window is doing to your phase relationships, because it doesn't deemphasize the "edge effects" that occur at the beginning and end of your block of samples, which is the whole point of using a non-rectangular window.
Try this experiment: Do a number of FFTs with lengths ranging from 59*48000 samples to 60*48000 samples, using both window types. You'll find that the Hanning window gives much more consistent results than the rectangular window does. The two results will agree best when the length is close to an integer number of beats.
Recall that windowing in the time domain is the equivalent of convolution in the frequency domain. Multiplying a signal in the time domain with a Hanning window is the equivalent of convolving the frequency domain signal with the kernel [-1 2 -1]/4.
Thus, if $H(f)$ represents the (non-windowed) frequency response, a corresponding windowed frequency response would be: $$H_w(f) = -0.25H(f-\Delta f) + 0.5H(f) -0.25H(f+\Delta f)$$ where $\Delta f$ is the frequency resolution - else you could write this (with $n$ being a frequency index) as:
$$H_w[n] = -0.25H[n-1] + 0.5H[n] -0.25H[n+1]$$
( windowed frequency response in terms of 'non-windowed' frequency response )
So the phase difference you are looking for is:
delta_phase = angle(Hw(n)) - angle(H(n))
Hope this helps