>         Here is my situation.  The input data is 8-bit, two's complement
> data.  I need to multiply two inputs together then multiply the result by
> 8 and still get the result to fit in 16 bits.
>
>         As I see it the max 8-bit value can be FFh (-1).  In order to
> make the multiply work for the 16-bit math I need to set all upper 8 bits
> (I think) so that my multiplier data is FFFFh.  This will cause an
> overflow if I have two such values to multiply so I was going to limit
> one of my inputs to 5 bits (ie, max value = 1Fh).  But if I do this how
> do I use this 5-bit representation of -1 in the 16-bit adder???
>
Hi Martin,

I'm not sure that I fully understand what you are trying to do.  If you
convert your data from two's complement, you can simply multiply the
two values and then left shift the 16-bit result three times for the 8*a*b
result.

If you leave the data in 2's comp. format, when you multiply you will
get the following:

    a+a2=h'100' or a2=h'100'-a  these are your 2's comp values
    b+b2=h'100' or b2=h'100'-b

Now multiplying a2*b2

    a2*b2 = h'10000' - h'100'*a - h'100'*b + a*b

For a 16-bit result, the h'10000' overflows.  To get your desired value
of a*b you must correct the a2*b2 result by the following

    a*b = a2*b2 + h'100'*a + h'100'*b - h'10000'

This simply means to add a and b to the high byte of the 2 byte a2*b2 result,
but your data is still in 2's comp.  So you'll need to substitute

    a*b = a2*b2 + h'100'*(h'100'-a2) + h'100'*(h'100'-b2) - h'10000'

or

    a*b = a2*b2 - h'100'*a2 - h'100'*b2 + h'10000'

So, you can get your a*b from you a2*b2 by subtracting a2 and b2 from the
high byte of a2*b2.  You may want to use the 16-bit subtraction with a2
and b2 as the high byte.

Now that you have a*b, just left shift three times for 8*a*b.

I hope this is correct.

Good luck,

Derrick Early
early@finite.nrl.navy.mil