This is a digital circuits problem A 4bit register holds the

This is a digital circuits problem.

A 4-bit register holds the signed 2\'s complement description of (-6) base 10. A left-shift with Data input = 1 is performed on the register. The final state of the register is.

(a) 0011

(b) 0101

(c) 1010

(d) 1101

(e) 1100

Also could you solve the same question for right-shift for the same data input and the answer options are also the same as mentioned above. WITH AS DETAILED METHOD AS POSSIBLE. I HAVE AN EXAM SHORTLY AND THIS QUESTION IS BUGGING ME!

Thanks.

Solution

Please follow the data and description :

Given decimal number is -6.

The binary representation for the number is given by 0110(2). But moving on to the signed representation the sign bit or the MSB is given by 1. So the resultant binary number is given by 10110.

The two\'s complement for the representation is given by,

-6 = 10110
negation = 11001
two\'s complement = 11010

But as the register is of 4-bit the data is saved regardless of the sign bit which is 1010.

After the execution of the left shift on the register with the input data as 1 we get the 0101.

So the answer is OPTION B (0101).

For the right shift oif the same number the data 1010 is given by 1101.

So the answer is OPTION D (1101).

Hope this is helpful.