> > 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

Kevin, you are right.  I am just gathering information in my spare time for
a future project so haven't started to code yet.  I didn't notice that Scott
had mutated the expression.  Sorry Scott! ;-)

I am not sure if I completely follow your rearrangement above but if I can
do everything in one hit with a single table that would be excellent.  It
makes sense to pre-calculate as much as possible outside of the program.

The original expression is definitely Z = 10*log(2^(X/10)+2^(Y/10))

My first thoughts were along the lines of creating a 16-bit table containing
the result of 2^(X/10) for X=40 to 100 (using an index offset of 40). The
table values run from 16 to 1024.  An 8-bit table could be used instead with
the hi bytes calculated in line which might be more efficient.

The same table could also yield 2^(Y/10) since the range is the same.  Using
sixteen bit addition would then give me 2^(X/10)+2^(Y/10).  Z has to be
within the same range of 40 to 100 since the input X or Y values will have
to be limited by testing the result in Z.   I really wanted to avoid using a
log routine so I was toying with the idea of trying to use the same table in
reverse by using succesive approximation.  It all started to get very messy.

When I get a chance I will try your suggested method in Excel.  Any other
thoughts welcome!

Thanks

David

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