Overflow in signed magnitude for Negative numbers. Discussion with Example.

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Example: Add 2 Signed magnitude numbers.

Add two sign magnitude numbers -70 & -90  with previous carry = 0.
Sol. Load the values in two 8 bit registers R1 and R2. 
So, R1 = -70 (decimal) 
&   R2 = -90 (decimal)

The detailed steps to calculate results are listed in table below:-

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Table to list four steps to add registers R1 and R2. Result is shown in the last column, number 5.

Value (in decimal)
Value (in Hex)
Value (in decimal)
Value (in 2’s complement)
-70 -46 1 100 0110      1 011 1010
-90 -5A 1 101 1010      1 010 0110
1   0110 0000 

Following are the rules for overflow condition detection in signed magnitude. 
1. For signed numbers leftmost bit always represents sign.
2. Is there a carry into sign bit position?
3. Is there a carry out of sign bit position?
4. If step 2 and step 3 results are not equal then the overflow condition is detected.

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Lets apply the rules to our example:- Statement 2 is false and statement 3 is true. So we have an overflow. 
So the final result is:- 
Carry bit = sign 
Sum = 2’s complement of 8 bits

1 1010 0000 = 1A0 (in hex) = -160 (decimal)

We need an extra flop to store the overflow bit.  

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