5 ms·
I would like to see more of these articles cover the phase portion.
by frozenport 14y ago
I would like to see more of these articles cover the phase portion.
- alecst 14y agoWhat do you mean?
- frozenport 14y agoThe imaginary part of the transform.
- nphrk 14y agoThe article uses the magnitude of the coefficients, which is computed using both the real and the imaginary part.
- dbaupp 14y agoThe phase portion isn't actually just the imaginary part, it's just the piece of information lost when one goes from real+imag -> magnitude, i.e. it's the argument of the complex number.
- jjoonathan 14y agoPerhaps this was left as an exercise for the reader? It seems to follow quite closely from the same intuition: after you have averaged the points on the imaginary plane you get a point that you can can represent as (r,theta). Theta is the phase portion. It's just the direction of the constructively interfering peak.