Note that numbers are represented in 2's complement form. therefore an-1, bn-1 and sn-1 are sign bits
1) an-1 .............. a3 a2 a1 a0
2) bn-1 .............. b3 b2 b1 b0
cn cn-1 .............. c3 c2 c1 c0
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3) sn-1 .............. s3 s2 s1 s0
Yes, there are two formulas for checking is it overflow or not?
i) cn⊕cn-1 = 0 ===> No overflow otherwise there is overflow
ii) a'n-1 . b'n-1 sn-1 + an-1 bn-1 s'n-1 = 0 ===> No overflow otherwise there is overflow
Remembering these formulas easy But understanding the mechanism behind these points are somewhat tough but not impossible...
But as i commented
convert them into decimal and perform addition
if resultant decimal no, is in range of 6-bits, then no overflow, else overflow
those points you can easily understand.
1) A = 1 0 1 0 1 1 ====> -32 +8+2+1 = -21
B = 1 1 0 1 0 1 ====> -32+16+4+1 = -11
therefore A+B = -21-11=-32, what is the 2's complement range with 6 bits? -32 to +31 therefore no overflow
2) A = 0 1 1 1 1 1 ====> 16+8+4+2+1 = 31
B = 0 0 0 0 0 1 ====> 1 = 1
therefore A+B = 31+1=32, what is the 2's complement range with 6 bits? -32 to +31 therefore overflow