Posted on April 10, 2003 at 03:57 Hi Nathan...
We want z=x*y where x,y are 16 bits long. If x=x1x2 (not a multiplication!!) and y=y1y2, where x1,x2,y1,y2 are 8 bits long, it comes with 8 bit multiplication based operations:
z = y2*x2 + (x2*y1+x1*y2)<
for a 16 bits result:
z = x1*y1 + trunc(x2*y1+x1*y2+trunc(x2*y2))
where: trunc(x*y) = MSB (x*y)
It's not clear, I know...an example:
E03A * A7D9 = ????
here: x1 = e0, x2 = 3a, y1 = a7, y2 = d9
well: x2*y2=3a*d9=312a=MSB.LSB with MSB=31 LSB=2a
So we keep the MSB 31 and :
x1*y2+x2*y1+MSB=e0*d9+3a*a7+31
=bde0+25d6+31
=e3e7 (result can be greater than 0xffff!! care in your code...)
Forget the first 8 bits => we keep 0xe3
Finally:
z=x1*y1+0xe3=e0*a7+e3=0x9303
This is easy to implement in assembly, coz those operations are based only on 8 bit multiplications (moreover the ST7 core returns the MSB in the index register) and 16 bit additions.
By the way, if you want to implement signed operation, you should add before this a testing procedure on the bit dedicated to the sign and reset those bits before going through the multiplication...
Depending of the accuracy, you can also truncate x and y and use a single 8 bits multiplication, but I guess that's not what you want!!
Hope it can help!
Florent
