Given an audio file (WAV), I need to find the $\rm L_{A_{eq}}$ (continuous A-weighted equivalent sound pressure level) of the audio file. For this purpose, I am following this workflow:

  1. Get a frame of audio (say $0.125$ seconds of samples)
  2. Scale the samples to find the $\rm SPL \ re \ 20 \ \mu Pa$ based on calibration factor obtained from a calibration tone.
  3. Apply A weighting to the frame of audio data.
  4. For each frame of audio data, find the SPL (dB) as:

$$ \begin{align} \textrm{pressure}_{\rm Ref} &= 20\times10^{-6};\\ \textrm{amplitude}_{\rm rms} &= \sqrt{\rm mean( audData_1frame.^2 )};\\ \textrm{dBspl}_{\rm perFrame} &= 20\times \log_{10}\left(\frac{\rm amplitude_{rms}}{\rm pressure_{Ref }}\right);\\ \end{align} $$

This gives the $\rm SPL \ re \ 20 \ \mu Pa$ of a frame of data. This would be the Short $\rm L_{eq}$.

How do I proceed further to get these 3 quantities:

  • $\rm L_{eq}$ = The total continuous equivalent sound pressure level. This wold be a scalar value for the entire audio file. If it has to be updated in a plot, how do we integrate the SPL of each frame with the previous frames?

  • $\rm L_{eq_{fast}}$ = Sound pressure level with $125\ ms$ time weighting

  • $\rm L_{eq_{slow}}$ = SPL with $1\ s$ time weighting

  • $\begingroup$ Note that A-weighting filter is loosely defined near 20 kHz, and bilinear transform warps the response, so it's probably best to oversample before filtering or use an equi-ripple type filter instead of BLT. I have an accurate A-weighting filter based on standards here github.com/endolith/waveform_analysis/blob/master/… $\endgroup$
    – endolith
    Jul 26, 2017 at 19:35
  • $\begingroup$ @endolith : Thanks for sharing the link. Currently, I am using the weighting filters from MathWork's Audio System Toolbox and they are giving pretty good initial results. I will test it out with your code too $\endgroup$ Jul 26, 2017 at 19:41
  • $\begingroup$ Oh, I didn't know they had that. Theirs should be fine, they're using the same tolerance limits as mine $\endgroup$
    – endolith
    Jul 26, 2017 at 19:47

1 Answer 1


how do we integrate the SPL of each frame with the previous frames?

You're measuring the RMS value of the (filtered) samples, which is sqrt(average(samples^2)), so if you're finding the RMS value in multiple chunks, it would just be sqrt(weighted_average(chunk1^2, chunk2^2, chunk3^2, ...)) where the average is weighted by the number of samples in each chunk (typically they are all the same length, but maybe a recording is not an exact multiple of chunk size, for instance).

To find the time weighting, the signal flow is:

  1. Apply A-weighting filter
  2. Square the samples
  3. Apply a low-pass filter with a real pole at 1/τ (where τ is 125 ms or 1 s in your case)
  4. Square root
  5. 20*log10() of the signal to convert to dB

It's the same as the RMS of the weighted signal, above, except that instead of an unweighted average over the whole signal, you're using a sliding-time average (same as https://en.wikipedia.org/wiki/Moving_average#Exponential_moving_average I think).

The time-weighted SPL will be a signal, not a scalar for each frame.

  • $\begingroup$ Thanks, your suggestion worked. I had to do some clever math because I'm calculating streaming LAeq values in real-time, but the sqrt(mean(frame1.^2, ..., frameN.^2)) gives me good results. Can you point me to some theoretical background of using low-pass filter with a real pole at 1/τ for this purpose. $\endgroup$ Jul 26, 2017 at 20:52
  • $\begingroup$ @ArnavMendiratta It's in the IEC 61672 standard (which happens to be identical to Indian Standard 15575), section 3.5 "time-weighted sound level" and 3.9 "time-average sound level" $\endgroup$
    – endolith
    Jul 26, 2017 at 21:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.