Output of "x^2 + y^2" from 2's complement to binary (with sign)

I kind of knew that since it's a positive result, it would no longer be a two's complement, and your answer re-affirms this.

Thank you :).

x/y = 8 bits signed

x^2/y^2 = 16 bits signed

if we add these two numbers we get a 17 bit result. bit 17 is always zero because the number is always positive, so it can be dropped to give a 16 bit Unsigned number. If you need to add it to any -ve numbers you have to add the sign bit back on again (a '0') so it might be worth keeping it after all.

So, if x^2 and y^2 = 16-bits

Is it always the case that x^2 + y^2 = 33-bits?

And that bit 33 is THE sign bit (zero in this case)?

Sorry to be picky about this; I have been trying to understand this since last week.

No, that is not the case.

when adding two N bit wide numbers the largest possible result is N+1 bit wide.

let's assume both x² and y² are the largest possible 16 bit numbers (when using signed arithmetics):

0x7FFF and 0x7FFF

or binary: 0111.1111.1111.1111

Adding those two numbers is equivalent to multiplying one of the numbers with 2.

Multiplikation with 2 again is equivalent to shifting the number to the left by one, so we end up with:

0.1111.1111.1111.1110

which equals 0x0000FFFE

When using unsigned arithmetics from now on we will not need the zeros at the beginning: so we end up with unsigned 0xFFFE.

Hi guys,

There's actually a couple of possible answers to this problem.

If x and y are both 8-bit signed integers, then the range for x and y is -128 (1000_0000b) to 127 (0111_1111b), so the largest possible value of x^2 or y^2 is 128^2 = 16384 (4000h = 0100_0000_0000_0000b) which requires 16-bits to represent. If you then add x^2 and y^2, you need an extra bit to represent the sum, so 17-bits is required.

If however, you limit the input range to symmetric signed integers, i.e., inputs are limited -127 to 127 (by converting any -128 codes into -127), then the largest possible value for x^2 or y^2 is 127^2 = 16129 (3F01h = 0011_1111_0000_0001h), which is still a 16-bit number, but the MSB (the second binary digit, since the first is the sign) is not set. This means that you can perform the accumulate x^2 + y^2 and not require 17-bits for the sum.

Another reason for using symmetric integers is that you often want to flip the sign on data, and flipping the sign on -128 gives 128, which requires 1-bit more to represent, i.e., 9-bits.

These types of DSP tricks are discussed here:

http://www.ovro.caltech.edu/~dwh/correlator/pdf/esc-320paper_hawkins.pdf (http://www.ovro.caltech.edu/%7edwh/correlator/pdf/esc-320paper_hawkins.pdf)

Cheers,

Dave