How does bandpower() round the input frequencies in variable 'freqrange'?

I am reviewing the documentation on the function 'bandpower' (Band power - MATLAB bandpower (mathworks.com))

My question is about the input variable 'freqrange':

If I want the range [4, 8), i.e. I want the power at 4 Hz and up to but not including 8 Hz, how would I communicate that to the bandpower function? I'm guessing that bandpower is based on FFT analysis, which would give signal power at discrete frequency bins and frequency resolution depends on signal length.

So if I told bandpower that freqrange=[4 7.99], would it round up to 8? or would it round down to the nearest frequency bin? (e.g. down to 7.75 if frequency resolution is .25Hz)

Answer 1

[Paul]

Hi Joey,

bandpower is an .m file, at least in 2022a, so you can inspect the code.

My reading is that it will compute the power from f1 to f2 where f1 is the largest periodogram frequency that satisfies f1 <= freqrange(1) and f2 is the smallest frequency that satisfies f2 >= freqrange(2).

rng(100)

t = 0:0.001:1-0.001;
x = cos(2*pi*100*t)+randn(size(t));

The frequency spacing is

1000/numel(t)  % Hz

ans = 1

Bandpower from the exact frequencies from 50-150

pband1 = bandpower(x,1000,[50 150])

pband1 = 0.6190

Bandpower still from 50-150 because those are the closest that bound freqrange from below and above at spacing of 1 Hz

pband2 = bandpower(x,1000,[50.9 149.1])

pband2 = 0.6190

isequal(pband1,pband2)

ans = logical
   1