Help Simplifying This Logarithmic Function | Member | | Join Date: Nov 2008
Posts: 80
| |
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()
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{
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var f = parseFloat(AWeightForm.Frequency.value);
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var K = 3.5041384 * Math.pow(10,16);
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var M1 = Math.pow(Math.pow(f,4),2);
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var N1 = Math.pow(Math.pow(20.598997,2) + Math.pow(f,2),2);
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var N2 = Math.pow(107.65265,2) + Math.pow(f,2);
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var N3 = Math.pow(737.86223,2) + Math.pow(f,2);
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var N4 = Math.pow(Math.pow(12194.217,2) + Math.pow(f,2),2);
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var Level = K * M1 / (N1 * N2 * N3 * N4);
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Level = 10 * Math.log(Level) / Math.log(10);
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Level = Math.round(Level * 1000) / 1000;
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AWeightForm.Level.value = " " + Level;
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}
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-
Again, any equation that more or less matches with the curve is ok with me.
Thank you.
|  | Expert | | Join Date: Mar 2007
Posts: 10,611
| | | re: Help Simplifying This Logarithmic Function
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
| | Member | | Join Date: Nov 2008
Posts: 80
| | | re: Help Simplifying This Logarithmic Function
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) )?
|  | Moderator | | Join Date: Aug 2008 Location: Leipzig, Germany
Posts: 3,640
| | | re: Help Simplifying This Logarithmic Function
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. Quote:
Originally Posted by serdar  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) )? put a factor in front of x. you really want to go for a line (in the diagram)?
| | Member | | Join Date: Nov 2008
Posts: 80
| | | re: Help Simplifying This Logarithmic Function
Quote:
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.
| | Expert | | Join Date: Mar 2007
Posts: 10,611
| | | re: Help Simplifying This Logarithmic Function
Quote:
Originally Posted by 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) )? 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
| | Member | | Join Date: Nov 2008
Posts: 80
| | | re: Help Simplifying This Logarithmic Function
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.
| | Expert | | Join Date: Mar 2007
Posts: 10,611
| | | re: Help Simplifying This Logarithmic Function
Quote:
Originally Posted by 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. 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
| | Moderator | | Join Date: Aug 2008 Location: Leipzig, Germany
Posts: 3,640
| | | re: Help Simplifying This Logarithmic Function
Quote:
Originally Posted by serdar Something like y=2x ? y = log(2x) is what I meant.
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