0
$\begingroup$

An equalizer we have consists of Loudness ,Bass ,Mid and Treble .The loudness is nothing but a Peaking filter with cutoff frequency 20Hz .I read this in a book that outlines the EQ design. I was wondering why the Loudness block which is basically a Peaking Filter should have a cut off frequency of 20Hz ?? Any intuition behind this

$\endgroup$
2
$\begingroup$

The purpose of the loudness filter is to compensate for the level dependency of the a human's "frequency response". Most music is mixed at pretty high levels, say at 80-90 dB SPL. If you play this back at much lower levels, say 40 dB SPL, it will sound very bass deficient. The loudness filter compensates for this by boosting the bass at lower levels.

In a "good" implementation the loudness adjusts automatically based playback volume setting and content energy: the lower the playback volume the more bass boost is needed to maintain the same perceived spectral balance between low and mid frequencies.

The shape of the loudness filter can be derived from the Equal Loudness contour of humans (see https://en.wikipedia.org/wiki/Equal-loudness_contour). In essence the loudness filter is the difference between the equal loudness curve at the playback level and the equal loudness curve at some reference level, all normalized to 0 dB at 1 kHz (or something like this).

The target filters have a continuously rising bass response. That's not easy to implement with "standard" EQ blocks. A very low frequency peak with the right choice of center frequency and Q can be a fairly good match in the region you care about. You use a 20 Hz peak filter because it has the right shape from 40Hz-100Hz and you don't care much what happens at 20 Hz since there almost no content there.

$\endgroup$
0
$\begingroup$
  1. Human cannot hear the sound under 20 Hz which is called infrasound.

  2. Loudspeakers have a poor amplitude frequency response at low frequency, meaning that they cannot reproduce sound under 20 Hz. A loud enough low frequency sound requires a large displacement, which may damage the loudspeaker.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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