How can i perform binary subtraction of two numbers that are already in 2's complement? I have to subtract 01010011 from 10100110,both numbers are in 2's complement. I know that 10100110 is -90,and 01010011 is 83,so the result should be -173. So if i use 8 bits that means that there's overflow. But i don't get how can i perform the subtraction when both the numbers are in 2's complement. If i just perform the subtraction i get the same number 01010011. How do i show the result and that there's overflow?
You cannot represent –173 using 8 bits, since the range is –128 to 127.
To subtract two numbers in two's complement, you use the identity $a - b = a + (-b)$. In your case, $a = 10100110$, $b = 01010011$, and $-b = 10101101$. Adding $a$ and $-b$, there is overflow, so you know that the number is outside the range.
You need to subtract 01010011 from 10100110.
10100110 - (01010011)
So find the 2's complement of 01010011 which is 10101101 and just add it to 10100110
01011000 <=carry bits 10100110 +10101101 ---------- 101010011 <= Result, but the extra 9th bit to the left of MSB is stored outside a register in a special carry out flip-flop
Now the carry-in and carry out of the MSB of the 8 bit register are different and so the result is overflow... (the carry in is 0 ,above the two 1's in the MSB and the carry out is 1 which is left to the MSB in the result)
If and only if the XOR of the carry-in and carry-out of the MSB is 1 then overflow.