1
$\begingroup$

The conventional MEL scale converts the frequency scale using the following formula:

$$M_f = 1127.01048 \cdot \log\left(\frac{f}{700} +1\right)$$

I want to use a reverse scale where resolution is better at higher frequencies only up to $10\textrm{ kHz}$.

In my opinion it will be a s-shaped curve. Any suggestions?

$\endgroup$
  • $\begingroup$ do you have any particular use/application for such kind of scale ? $\endgroup$ – Arpit Jain Jan 6 '17 at 6:06
  • $\begingroup$ Yes, I have some animal/aquatic recordings where information is mostly in higher bands. $\endgroup$ – Astro Jan 6 '17 at 7:21
  • $\begingroup$ exponential function based scale will definitely work out for this. you can define your own exponential based scale based on particular frequency requirements. $\endgroup$ – Arpit Jain Jan 6 '17 at 8:50
  • $\begingroup$ I have added the figure and the formulation of the exponential based filter bank which will give more resolution to frequencies around 10 KHz and less resolution for lower frequencies. Is this what you need or looking for something else ? $\endgroup$ – Arpit Jain Jan 6 '17 at 12:39
  • $\begingroup$ @arpitjain thanks for detailed description. In addition I want to know how to make sure my first filter bank starts from say 1kHz instead of zero. $\endgroup$ – Astro Jan 7 '17 at 10:47
0
$\begingroup$

The formulation for MEL frequency scale is logarithm scale based, it clubs(put together) the higher frequencies more closely(so less resolution), because logarithmic functions vary slowly for larger numbers(see the logarihmic functions plot).

So some function of exponential form will definitely work out if one wants to give more resolution to higher frequencies.

so based on particular requirements one can define exponential scale for effect that looks like reverse of MEL scale. enter image description here

Following is the filter bank I created using exponential function, similar to the formulation of MEL scale.

$fnew$ $=$ $exp(f / 7000)$.

I think this is what you are looking for, more resolution for higher frequencies as compared to lower frequencies.

| improve this answer | |
$\endgroup$
0
$\begingroup$

I found this paper on the topic.

Basically you're modifying the original Mel filter conversion formula to fit to a mirrored position in the high frequency region. You use the nyquist frequency to estimate the mirror position.

| improve this answer | |
$\endgroup$

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.