# Understanding FFT bin resolution

For a signal sampled at $f_s$, the frequency resolution (or "bin width") for an $N$ point FFT is $f_s/N$. Does this mean that the $k$th bin will contain energy from sinusoids within $0.5f_s/N$ on either side of the center frequency of that bin (ignoring spectral leakage for now)? Is this energy averaged? Is it the same as taking a $2N$ point FFT and then averaging every two bins?

• You can't "ignore spectral leakage for now", because that is exactly the point here. Commented Sep 9, 2013 at 2:21
• @chirlu Does a longer FFT length imply lesser leakage from $f_1$ into $f_2$?
– user5400
Commented Sep 9, 2013 at 4:02
• Not necessarily. See hotpaw2 answer below: the "leakage" is given by a Sinc function which in turn is determined by the FFT length. Since the Sinc function isn't monotonous, leakage at certain frequencies can actually increase with FFT length Commented Sep 9, 2013 at 14:47
• is N the number of samples in time series? Commented Oct 20, 2015 at 14:02