Question

3. Use 6 bits (2's complement representation) to solve the following problem. Indicate any overflow. a. Convert decimal +16 and +21 to binary. (5 pts.) b. Perform the binary equivalent of (+16) + (-21) (3 pt.) c. Perform the binary equivalent of (-16)+(-21) (3 pts.)

Answer 1

3) a) In all form of representation tve no's have got the same representation so +16 can be converted to binary by 2116 à 4o remainders 312 o i o so binary of +16 is 10000 in 6 biter it can be represented as 010000

+21 -> binary 2121 210 25 o 10 +21 binary is 10101 in 6 bit (010101 b) 2's Complement of +21 is step-i; Complement all the bits it will be 101010 Step-2; Add one to it. 101011 -21 Note: - in 2's complement form MSB has got a weight of -20-1

Now; +16+(-21) means +16+ (2's complement of 21) = 0 10000 + (+16) 101011 (-21) (-5) Iloil There is no overflow here. and the result is 111011 whose magnitude is -23 + 24 +2²+0+2²20 -82+ (16+8+4+1) -32+27= -5 which is the required result.

c) - 16 Step-1; complement all bits of +16. 101111 Step-2; Add I to it. 110000 => -16 Now; (-16)+(-21) is . FI10000 (-16) 010101! (-21) 6 011011 (-37) overflow is there. and our result is 011011 whose magnitude is 0+2++2+0+2'+2° = 16+8+2+1 = 27 (-37)

so here we need to take the 2's complement of the result to get the required Answer and put ve sign infront of it. step-1; complement the result 100100 step-2; Add I to it. 100101 it's magnitude is 4540+0+23+0+2°) (22+4+1) -37

so in terms of 6-bit the Answer is (100101) or if we put the overflow bit in the result of 011011 it will be 11011011 whose magnitude is -26+0+2+23+0+2+2° -64+27 = -37 (That is the presence & of overflow bit indicate that we Can't represent the result with the same number of bits as that of the added terms)

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