On Sun, 16 Sep 2001, Scott Dattalo wrote:

> On Sun, 16 Sep 2001, David J Binnington wrote:
>
> > Can anybody give me a kick start with this?
>
> David,
>
>
> You can re-write your equation like:
>
> Z = 10*lg(x^2 + y^2)...

This isn't the intial expression that was asked to be calculated.  It is not
x^2, but rather 2^x.  Actually, the correct equation is much easier to
calculate than the one you've been discussing.  Along the same lines as you
mentioned, the correct way to rewrite it is:

Z = 10*log(2^(x/10)+2^(y/10))
  = 10*log(2^(x/10) (1+2^[(y-x)/10]))
  = 10*log{2^(x/10)} + 10*log{1+2^[(y-x)/10]}
  = x+10*log{1 + 2^[w/10]}

where w = y-x.  Notice that this is a subtraction, not a division.  That
makes life much easier.  For this expression, let y be the larger of your 2
input values.  Now you only need one table lookup for f(w) =
10*log{1+2^[w/10]} and an addition.  I doubt that you can code a division,
power, and log routine in 100 instructions, which is what the table will
occupy since w can be at most 100 if x & y are limited to 100.  Plus,
execution cycles will consist of a subtraction to calculate w (and possible
negation if x>y), a table lookup and an addition.

Kevin

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