On Mon, 17 Sep 2001, Kevin Moll wrote:

> 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.

You're right. It would appear that not only do I not read the Off Topic
messages, I don't read the On Topic ones either!

>                      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.

Or, you could choose to factor the larger of x and y then the difference
in the exponent would always be negative. The table only needs to be 16
entries (if linear interpolation is used).

Sorry about the foul up.

Scott

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