Spectrum in 1/3 octave bands from FFT

Question

I’m looking to use an FFT to generate a frequency spectrum in 1/3 octave bands. After reading many posts on this site as well as others, I believe the appropriate approach is set out below. I’ve dry-run the calculations in Excel and the results look reasonable, but reasonable is not necessarily correct.

To ask a specific question, is the method I’m using valid? In particular, I question if the method applied in Step 2 is correct.

Background. I’m sampling the electric signal on the output of a pre-amp and my goal is a frequency spectrum in dB. My reference (denominator in the dB calc) is arbitrary and chosen to set an appropriate scale for the spectrum output display. I’m not attempting to take measurements from the output and don’t need to calibrate it against an objective reference. My project will be implemented in C++.

Step 1. Start with a 1024 sample FFT at 48,000 Hz (513 bins at 46.9 Hz each). Result is the DFT, F_n for n = 0 – 512. Units are in V.

Note, normally, I’d normalize the FFT output by multiplying each bin by 2/N (N is number of samples, i.e. 1024). However, because we eventually plot the spectra in dB relative to an arbitrary amplitude, any normalizing I do here will be washed away when I compute the results in dB so I don’t bother with it in my project.

Step 2. Convert the DFT into 1/3 octave bands.

I obtained the following formulas from here: https://www.ap.com/technical-library/deriving-fractional-octave-spectra-from-the-fft-with-apx/

The amplitude in V for each 1/3 octave band is:

Where L_b is the amplitude for each 1/3 octave band in V, for b = 1 to 32, and g_n,b is the gain multiplier for FFT bin n and 1/3 octave band b:

Where f_n is the frequency of bin n and f_b is the center frequency of band b. k is the octave bandwidth designator, 1 for full octave and 3 for 1/3 octave.

Result is L_b for b = 1 to 32 in V.

Step 3. Disregard lower bands.

Below about 250 Hz, for an FFT of length 1024, there are not enough FFT bins within each band to make the results meaningful. Therefore, we disregard the lower bands and keep only the 1/3 octave bands starting with b13 = 251 Hz.

Step 4. Convert the amplitude data in each band to dB.

For this computation, I’m computing the dB value relative to the approximate maximum amplitude likely to appear in any frequency band. This will be L_ref.

Final value in dB for each 1/3 octave frequency band above 250 Hz is:

Below is a sample of individual g_n,b terms (FFT bins down the left, 1/3 octave bands across the top):

Assistance, input, corrections and comments appreciated. Thanks!

Answer 1

It seems like you’ve by and large got it, but I did notice a couple things. You’ll probably want to include support for windowing and the corresponding correction factor as shown by APs article on FFT noise. Rectangular windows aren’t great for RMS type measurements (at least in my opinion) for all the side lobe type explanations that are common with windows. Second, I’m pretty sure you need to sum from 1 to N, not N/2. You may have realized that because the input signal is real, that there is redundant data that doesn’t need to be calculated twice. That is true, with the exception of the 0 and N/2 (if N is even) bins. Granted this probably wouldn’t make the output too wrong, but it would be off a bit.

Edit: you know what, AP may have designed those coefficients so that windowing either isn’t necessary, or would do more harm than good. You may still want to try it, since it’s easy enough to support, but it may not help. Also, I don’t see how this would be more computationally efficient than doing the calculations directly in the time domain.