Topic: Digital Circuits
If A & B are 8 bit right shift registers connected as shown in fig. and the initial values of A&B in hexadecimal are B2 and OF respectively, the value of A in hexadecimal after 6 clock pulses is
- A. C8
- B. 7D
- C. FB
- D. 3E
Answer: D. 3E
Explanation: The registers form a 16-bit circular right shift register where B's LSB shifts into A's MSB, and A's LSB shifts into B's MSB.
Initial state of B is $0F_{16} = 00001111_2$.
Initial state of A is $B2_{16} = 10110010_2$.
Because it is a right shift register, after 6 clock pulses, the 6 LSBs of B are shifted sequentially into the MSB of A.
The 6 LSBs of B are $001111$. They will occupy the top 6 bits of A in the same order they shifted.
The remaining 2 bits of A will be its original top 2 bits, which are $10$.
So, the new 8-bit content of A becomes $00111110_2$. Converting to hexadecimal gives $3E_{16}$.