An FFT bin width is fs/N, where fs is the sampling rate, and N is the number of points in the FFT.
Thus with 44.1kSPS and 1024 point FFT the bin width is about 43Hz. Thus 0..20 covers 0..860Hz
Note that you will need to choose a suitable FFT window, the Hann window (AudioWindowHanning1024)
is a good default to choose unless you need accurate measurements, which requires a flattop window.
You always seem some spreading of signal to adjacent bins, to an extent depending on properties
of the window, this is unavoidable(*). Without a window you get far worse behaviour if the signal is
not synchronized to the sample clock exactly - basically you must use a window for what you are doing.
(*) Thus that 43 Hz resolution is optimistic...
Thus with 44.1kSPS and 1024 point FFT the bin width is about 43Hz. Thus 0..20 covers 0..860Hz
Note that you will need to choose a suitable FFT window, the Hann window (AudioWindowHanning1024)
is a good default to choose unless you need accurate measurements, which requires a flattop window.
You always seem some spreading of signal to adjacent bins, to an extent depending on properties
of the window, this is unavoidable(*). Without a window you get far worse behaviour if the signal is
not synchronized to the sample clock exactly - basically you must use a window for what you are doing.
(*) Thus that 43 Hz resolution is optimistic...