Question

A number N is stored in a 4-bit 2's complement representation as

$a_3$

$a_2$

$a_1$

$a_0$

It is copied into a 6-bit register and after a few operations. The final bit pattern is

$a_3$

$a_3$

$a_2$

$a_1$

$a_0$

1

The value of this bit pattern in 2's complement representation is given in terms of the original number N as

• A. $32a_3+2N+1$
• B. $32a_3-2N-1$
• C. $2N-1$
• D. $2N+1$

Answer 1

The correct option is D $2N+1$

Given number is

$a_3$

$a_2$

$a_1$

$a_0$

in 2's complement form.

We know that in 2's complement form if we copy MSB at left of MSB any times the number remains unchanged.

So, $a_3{a}_2{a}_1{a}_0=a_3{a}_3{a}_2{a}_1{a}_0=N$

When we left shift a number by 1 bit then it is multiplied by 2, So, $a_3{a}_3{a}_2{a}_1{a}_{0}0=2N$

Now,
$a_3{a}_3{a}_2{a}_1{a}_{0}0+1=a_3{a}_3{a}_2{a}_1{a}_{0}1=2N+1$