3)
The group of bits 101001 is serially shifted (LSB first) into a 6-bit parallel output shift register
with an initial state 011110. After fifth clock pulses, the register contains
?
a. 011101
b. 000011
c. 001010
d.010010
Answers
Answered by
0
Answer:
sorry for the inconveniences ok
Answered by
1
Answer:
Consider that,
101001 → 0 1 1 1 1 0
Therefore, after 2 clock pulses,
Content in the register is,
1st clock pulse: 10100 → 1 0 1 1 1 1
2st clock pulse: 1010 → 0 1 0 1 1 1
3rd clock pulse: 101 → 0 0 1 0 1 1
4th clock pulse: 10 → 1 0 0 1 0 1
5th clock pulse: 1 → 0 1 0 0 1 0
After fifth clock pulses, the register contains is 010010
Therefore, the correct option is (d) 010010
Similar questions