<snip> Does this mean that if frequency and
phase of a sinusoid can br represented as A*sin(2*PI*f*t)+B*cos(2*PI*f*t), then the left output of fft~ is B and the right output is A?
Yes, if f is the sample position within the output block (i.e. which frequency it refers to) and t is the frequency of ffting, i.e. block size divided by sample rate. The equations don't have anything about phase, though, as this form represents the input as sin and cos waves (with phase of 0, starting at the beginning of each fft block).
If so, how do I get frequency and phase?
I think you mean magnitude and phase. The frequency of the sinusoid depends upon which sample of the output block you are looking at. E.g. if the first three samples output from the left outlet (the cos side) are 0.1, 0.2, 0.3, then the amplitude of the 0*f (direct, 0Hz) component is 0.1, of the 1*f component is 0.2, of 2*f component is 0.3, etc. The 2*f component, for example, will be that which makes 2 complete oscillations within each block. So, sample 2 (counting from 0, i.e. the 'third', counting from 1, sample) of the left output is the amplitude of the cos wave which makes 2 complete oscillations. There's only a direct component for the cos side, as sin of 0 is always 0.
You can convert them into magnitude and phase form, where instead of a pair of sin and cos waves of a certain frequency, you have one wave of the same frequency but a certain phase relative to the start of the fft block.
Magnitude = sqrt(sincomponentamp^2 + coscomponentamp^2)
Can't remember phase, but it should be in here somewhere, together with all of this explained better than I can: http://crca.ucsd.edu/~msp/techniques/latest/book-html/node152.html
Good luck, I'm still learning this stuff, but I think it's worthwhile once you get there!
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