Help Simplifying This Logarithmic Function

[serdar]

Hi,

I'll use a function called A-weighting Filter for a spectrum analyzer I'm designing in Flash. Here is the equation of A-Weighting Filter:

I thought the function is a little heavy to compute (for hundreds of values) realtime in Actionscript.

I won't need a precise computation of the a-weighting filter.

Can you provide a simpler function example which produces more or less the same curve as below.

A javascript example to compute a-weigthing:

function AWeightLevel()
{
  var f = parseFloat(AWeightForm.Frequency.value);
  var K = 3.5041384 * Math.pow(10,16);
  var M1 = Math.pow(Math.pow(f,4),2);
  var N1 = Math.pow(Math.pow(20.598997,2) + Math.pow(f,2),2);
  var N2 = Math.pow(107.65265,2) + Math.pow(f,2);
  var N3 = Math.pow(737.86223,2) + Math.pow(f,2);
  var N4 = Math.pow(Math.pow(12194.217,2) + Math.pow(f,2),2);
  var Level = K * M1 / (N1 * N2 * N3 * N4);
  Level = 10 * Math.log(Level) / Math.log(10);
  Level = Math.round(Level * 1000) / 1000;
  AWeightForm.Level.value = "  " + Level;
}
 
 

Again, any equation that more or less matches with the curve is ok with me.

Thank you.

[JosAH]

You can reduce the calculations needed for that expression quite a bit by substituting x= f^2 and expanding the expressions; I'll do the first one:

ax^2/((x+b)*(x+c) ==

(-(ac^2)/(b-c)-ab-ac)/(x+b) + ac^2/(b-c)/(x+c) ==

A == ac^2/(b-c) ==>

(-A-ab-ac)/(x+b) + A/(x+c)

Note that the constant expressions only need to be calculated once. The second expression is similar but I'm far too lazy to expand that one as well ;-) I suspect the second expression to be dominant w.r.t. the first expression.

kind regards,

Jos

[serdar]

Thanks, I decided something simple like:

y=log(x)

would do the work with some adjustments. (There is no way to get exact frequency amplitudes in Flash so implementing a real a-weighting on the data would be hard anyway.)

I don't do maths since high school :) so can you tell me what to add to change the slope of such a curve ( y=log(x) )?

[Dormilich]

the second expression looks a bit strange considering the exponents. the numerator got exp 4 while the denominator got exp 8 giving an overall exp -4. as far as I know you need exp 0 with respect to the units to compute the logarithm.
@serdar
put a factor in front of x. you really want to go for a line (in the diagram)?

[serdar]

put a factor in front of x. you really want to go for a line (in the diagram)?

Something like y=2x ?

No, I did not mean a fixed slope. By slope of the curve I mean the curve being more oblique etc, an "accelerating slope" if this is the correct term.

[JosAH]

@serdar
If all you need is a crude approximation, have a look at your own graph: it's a sort of upside down parabola with a logarithmic x scale and a linear y scale. The roots of the parabola are at x == 1000 and x == 7000, so the parabola looks something like this: a*(x-log(1000))*(x-log(7000)). A third point, say at x == 1 (which is log(10)) and y= -70; solves for the constant a.

kind regards,

Jos

[serdar]

Thanks Jos,

I think I have what I need now. Sorry for the complicated equation I posted at first, but I was not aware that a simple log. function would be enough for me.

Btw, what I'm trying to match is this analyzer which produces the best results (most accurate weighting for my taste): Winamp Plug-in Details - Customize Winamp Media Player

I'm still not there but just a little closer.

[JosAH]

@serdar
You're welcome of course; I think that 'logarithmic parabola' does fine, especially because you don't need the range > 20,000Hz; the differences will be larger there.

kind regards,

Jos

[Dormilich]

@serdar
y = log(2x) is what I meant.

[franknagy]

Take in account that
log(a*f^4/((f^2+b)*(f^2+c))=log(a)+4*log(f)-log(f^2+b)-log(f^2+c).
So yo can count g=f^2 in advance before calculating each members like this then sum the [log(a)+4*log(f)-log(f^2+b)-log(f^2+c)] values.