How can bass/low end be added to a signal?

With a low-pass filtered sub-oscillator running at an octave or two below the signal, there still doesnt seem to be enough bass.

Is there a mathematical function to easily add some warmth or boom to the sub oscillator? Or possibly adding two in unison, slightly detuned?

  • $\begingroup$ Are you asking about reproducing spectrum that was part of the original source of the signal? Or about creating some sort of synthetic addition? $\endgroup$ – hotpaw2 Jul 28 '16 at 9:01
  • $\begingroup$ Good question. The first would be ideal, but synthetic addition could also work. $\endgroup$ – some_id Jul 28 '16 at 11:00
  • 2
    $\begingroup$ You might want to have a look at the way "exciters" work too. $\endgroup$ – A_A Jul 28 '16 at 11:48
  • 1
    $\begingroup$ "exciters" create high frequency components by inferring harmonics from lower frequency tones. they use a form of distortion of the high-frequency portion to create those higher-frequency components that did not exist in the original signal. but it doesn't work in the other direction, it terms of creating frequency components lower than the original content. $\endgroup$ – robert bristow-johnson Dec 26 '16 at 7:09

Let's say the audio is given by $ x $.
Let's define Low Pass Filter by $ h $.

Whet you need to do is:

$$ y = x + \alpha (h \ast x) $$

Where $ \alpha $ is the amplification level of the Low Pass data.
Once you set which frequencies you want to amplify it is easy to design.


You could use a low-pass filter and add this to the original signal to boost the bass. What kind of speakers are you using? Regardless of how much you boost the bass signal, you need to have speakers that can put out the bass you want.

  • $\begingroup$ Im testing on Sony headphones which can output the low end. $\endgroup$ – some_id Jul 29 '16 at 10:13

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.